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Thread Subject:
Is it possible to evaluate this integral?

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 2 Aug, 2006 15:53:55

Message: 1 of 29

I couldn't get Matlab to compute it(Matlab uses Maple engine...)

int((exp(1i * t) + x * (1-exp(1i * t)))^r*exp(-1i*n*t), t, -pi, pi)

How to do it?

Thanks a lot!

Subject: Is it possible to evaluate this integral?

From: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson)

Date: 3 Aug, 2006 00:42:21

Message: 2 of 29

In article <earadh$deu$1@news.Stanford.EDU>,
Michael <michael.monkey.in.the.jungle@gmail.com> wrote:
>I couldn't get Matlab to compute it(Matlab uses Maple engine...)

>int((exp(1i * t) + x * (1-exp(1i * t)))^r*exp(-1i*n*t), t, -pi, pi)

>How to do it?


I'm rusty on such matters, but try a change of variables, Q=exp(I*t) .
Then the integral becomes

int((Q+x*(1-Q))^r/Q^n,Q=exp(-I*Pi)..exp(I*Pi));

But exp(-I*Pi) = exp(I*Pi) = -1 so no matter what the integral
comes out as, when one takes the definite form of it, the result will
be 0.

--
  Is there any thing whereof it may be said, See, this is new? It hath
  been already of old time, which was before us. -- Ecclesiastes

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 3 Aug, 2006 02:06:49

Message: 3 of 29

In article <eargpd$7jm$1@canopus.cc.umanitoba.ca>,
roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote:

> In article <earadh$deu$1@news.Stanford.EDU>,
> Michael <michael.monkey.in.the.jungle@gmail.com> wrote:
> >I couldn't get Matlab to compute it(Matlab uses Maple engine...)
> >int((exp(1i * t) + x * (1-exp(1i * t)))^r*exp(-1i*n*t), t, -pi, pi)
> >How to do it?
>
> I'm rusty on such matters, but try a change of variables, Q=exp(I*t) .
> Then the integral becomes
>
> int((Q+x*(1-Q))^r/Q^n,Q=exp(-I*Pi)..exp(I*Pi));
>
> But exp(-I*Pi) = exp(I*Pi) = -1 so no matter what the integral
> comes out as, when one takes the definite form of it, the result will
> be 0.
------------------------
  No, I'm afraid it isn't that easy, Walter. After your transformation,
the integration path becomes a closed unit circle in the complex plane
with Q = 0 at its center. However, if n is a non-negative integer, there
will be a singularity at Q = 0 and this will give the contour integral a
presumably non-zero value. The residue at this singularity could be used
in computing the integral, however there is the prospect of another
singularity at Q = (x-1)/x if r is negative and x is positive. I think we
need a bit more information from the original poster as to n, x, and r
parameter values.

  By the way, there should be an additional division by (Q*i) in your
integrand to take into account the change of variables (dQ=i*Q*dt). It
should read:

 (Q+x*(1-Q))^r/Q^(n+1)*(-i) .

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 2 Aug, 2006 22:30:05

Message: 4 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0208061906490001@dialup-4.232.228.45.dial1.losangeles1.level3.net...
> In article <eargpd$7jm$1@canopus.cc.umanitoba.ca>,
> roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote:
>
>> In article <earadh$deu$1@news.Stanford.EDU>,
>> Michael <michael.monkey.in.the.jungle@gmail.com> wrote:
>> >I couldn't get Matlab to compute it(Matlab uses Maple engine...)
>> >int((exp(1i * t) + x * (1-exp(1i * t)))^r*exp(-1i*n*t), t, -pi, pi)
>> >How to do it?
>>
>> I'm rusty on such matters, but try a change of variables, Q=exp(I*t) .
>> Then the integral becomes
>>
>> int((Q+x*(1-Q))^r/Q^n,Q=exp(-I*Pi)..exp(I*Pi));
>>
>> But exp(-I*Pi) = exp(I*Pi) = -1 so no matter what the integral
>> comes out as, when one takes the definite form of it, the result will
>> be 0.
> ------------------------
> No, I'm afraid it isn't that easy, Walter. After your transformation,
> the integration path becomes a closed unit circle in the complex plane
> with Q = 0 at its center. However, if n is a non-negative integer, there
> will be a singularity at Q = 0 and this will give the contour integral a
> presumably non-zero value. The residue at this singularity could be used
> in computing the integral, however there is the prospect of another
> singularity at Q = (x-1)/x if r is negative and x is positive. I think we
> need a bit more information from the original poster as to n, x, and r
> parameter values.
>
> By the way, there should be an additional division by (Q*i) in your
> integrand to take into account the change of variables (dQ=i*Q*dt). It
> should read:
>
> (Q+x*(1-Q))^r/Q^(n+1)*(-i) .
>
> Roger Stafford

You are right Roger. The x is in [1, +inf), the n is integer, r is negative
non-integer and may even by non-rational... any more thoughts?

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 3 Aug, 2006 07:13:15

Message: 5 of 29

In article <eas1ka$832$1@news.Stanford.EDU>, "Michael"
<michael.monkey.in.the.jungle@gmail.com> wrote:

> You are right Roger. The x is in [1, +inf), the n is integer, r is negative
> non-integer and may even by non-rational... any more thoughts?
-------------------------
Hello Michael,

  If you know that x is real and >= 1, then the problem is easier than I
thought. It doesn't matter that r might be irrational, but it is
important that n be an integer. I did make a mistake in my algebra
earlier. Let's use z = exp(t*i) instead of Walter's Q. The second
singularity would occur at z = x/(x-1) which would fortunately lie outside
the circle, and therefore the integrand's only singularity is at z = 0.
That means your integral can be determined entirely by the residue at z =
0. It's getting rather late for me now, so I'll just say that the
integrand needs to be expanded in a series of powers of z so as to
determine the coefficient of the one that is 1/z. We need to do a
binomial expansion of (z+x*(1-z))^r to accomplish this. That will give us
the residue and allow evaluation of the integral. I'll have to do the
rest of this for you tomorrow (unless of course you beat me to it and
solve it all yourself.) You can probably see at this point why Maple was
unable to produce a solution.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 3 Aug, 2006 16:26:54

Message: 6 of 29

In article
<ellieandrogerxyzzy-0308060013150001@dialup-4.232.228.196.dial1.losangeles1.level3.net>,
ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford) wrote:

> In article <eas1ka$832$1@news.Stanford.EDU>, "Michael"
> <michael.monkey.in.the.jungle@gmail.com> wrote:
>
> > You are right Roger. The x is in [1, +inf), the n is integer, r is negative
> > non-integer and may even by non-rational... any more thoughts?
> -------------------------
> Hello Michael,
>
> If you know that x is real and >= 1, then the problem is easier than I
> thought. It doesn't matter that r might be irrational, but it is
> important that n be an integer. I did make a mistake in my algebra
> earlier. Let's use z = exp(t*i) instead of Walter's Q. The second
> singularity would occur at z = x/(x-1) which would fortunately lie outside
> the circle, and therefore the integrand's only singularity is at z = 0.
> That means your integral can be determined entirely by the residue at z =
> 0. It's getting rather late for me now, so I'll just say that the
> integrand needs to be expanded in a series of powers of z so as to
> determine the coefficient of the one that is 1/z. We need to do a
> binomial expansion of (z+x*(1-z))^r to accomplish this. That will give us
> the residue and allow evaluation of the integral. I'll have to do the
> rest of this for you tomorrow (unless of course you beat me to it and
> solve it all yourself.) You can probably see at this point why Maple was
> unable to produce a solution.
>
> Roger Stafford
-----------------------------
Hello again Michael,

  Now that I've had a good night's sleep, I'll come back to your
integration problem.

  Performing Walter's change of variable, z = exp(t*i), leads to the
closed contour integral running counterclockwise around the unit circle in
the complex plane with center at z = 0 with the integrand

 f(z) = (z+x*(1-z))^r/z^n/(z*i) = x^r/i*(1-p*z)^r/z^(n+1)

where p = (x-1)/x. You stated that x >= 1, so this means that 0 <= p <
1. Assuming n is a non-negative integer, f(z) is then a single-valued
function analytic inside the unit circle except for a single pole of order
n+1 at z = 0. The factor, (1-p*z)^r, can be expressed via binomial
expansion as:

 (1-p*z)^r = 1 - r*p*z + r*(r-1)/2*p^2*z^2
   - r*(r-1)*(r-2)/6*p^3*z^3 + r*(r-1)*(r-2)*(r-3)/24*p^4*z^4 + ...
   + (-1)^k*r*(r-1)*..*(r-k+1)/(k!)*p^k*z^k + ...

If r is a non-negative integer, this series terminates and is a
polynomial. Otherwise it is an infinite series convergent inside the unit
circle. Hence the Laurent expansion of f(z) is

 f(z) = x^r/i*(1/z^(n+1) - r*p/z^n + r*(r-1)/2*p^2/z^(n-1) ....
  + (-1)^k*r*(r-1)*...*(r-k+1)/(k!)*p^k/z^(n-k+1) + ... )

The residue at the pole is the coefficient of the 1/z term which occurs at
k = n:

 b1 = x^r/i*(-1)^n*r*(r-1)*...*(r-n+1)/(n!)*p^n

and the contour integral around the unit circle by Cauchy's theorem is then:

 I = 2*pi*i*b1 = 2*pi*x^r*(-1)^n*r*(r-1)*(r-2)*...*(r-n+1)/(n!)*p^n
               = 2*pi*(1-x)^n/x^(n-r)*r*(r-1)*(r-2)*...*(r-n+1)/(n!)

This last is the definite integral you seek. In matlab terminology you
could express it using the 'prod' function as:

 I = 2*pi*(1-x)^n/x^(n-r)*prod((r-[0:n-1])./[1:n]); % <-- The answer

  If n had been a negative integer, f(z) would have been analytic and the
integral would be zero. Note that if n is not an integer, f(z) would not
be single-valued and this entire analysis becomes invalid.

  You would be wise to test this answer out with a few sets of specific
parameters, n, r, and x, using numerical integration, to guard against any
algebraic errors I might have made here.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Axel Vogt

Date: 3 Aug, 2006 21:32:24

Message: 7 of 29

Roger Stafford wrote:
>
> In article
> <ellieandrogerxyzzy-0308060013150001@dialup-4.232.228.196.dial1.losangeles1.level3.net>,
> ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford) wrote:
>
> > In article <eas1ka$832$1@news.Stanford.EDU>, "Michael"
> > <michael.monkey.in.the.jungle@gmail.com> wrote:
> >
> > > You are right Roger. The x is in [1, +inf), the n is integer, r is negative
> > > non-integer and may even by non-rational... any more thoughts?
> > -------------------------
> > Hello Michael,
> >
> > If you know that x is real and >= 1, then the problem is easier than I
> > thought. It doesn't matter that r might be irrational, but it is
> > important that n be an integer. I did make a mistake in my algebra
> > earlier. Let's use z = exp(t*i) instead of Walter's Q. The second
> > singularity would occur at z = x/(x-1) which would fortunately lie outside
> > the circle, and therefore the integrand's only singularity is at z = 0.
> > That means your integral can be determined entirely by the residue at z =
> > 0. It's getting rather late for me now, so I'll just say that the
> > integrand needs to be expanded in a series of powers of z so as to
> > determine the coefficient of the one that is 1/z. We need to do a
> > binomial expansion of (z+x*(1-z))^r to accomplish this. That will give us
> > the residue and allow evaluation of the integral. I'll have to do the
> > rest of this for you tomorrow (unless of course you beat me to it and
> > solve it all yourself.) You can probably see at this point why Maple was
> > unable to produce a solution.
> >
> > Roger Stafford
> -----------------------------
> Hello again Michael,
>
> Now that I've had a good night's sleep, I'll come back to your
> integration problem.
>
> Performing Walter's change of variable, z = exp(t*i), leads to the
> closed contour integral running counterclockwise around the unit circle in
> the complex plane with center at z = 0 with the integrand
>
> f(z) = (z+x*(1-z))^r/z^n/(z*i) = x^r/i*(1-p*z)^r/z^(n+1)
>
> where p = (x-1)/x. You stated that x >= 1, so this means that 0 <= p <
> 1. Assuming n is a non-negative integer, f(z) is then a single-valued
> function analytic inside the unit circle except for a single pole of order
> n+1 at z = 0. The factor, (1-p*z)^r, can be expressed via binomial
> expansion as:
>
> (1-p*z)^r = 1 - r*p*z + r*(r-1)/2*p^2*z^2
> - r*(r-1)*(r-2)/6*p^3*z^3 + r*(r-1)*(r-2)*(r-3)/24*p^4*z^4 + ...
> + (-1)^k*r*(r-1)*..*(r-k+1)/(k!)*p^k*z^k + ...
>
> If r is a non-negative integer, this series terminates and is a
> polynomial. Otherwise it is an infinite series convergent inside the unit
> circle. Hence the Laurent expansion of f(z) is
>
> f(z) = x^r/i*(1/z^(n+1) - r*p/z^n + r*(r-1)/2*p^2/z^(n-1) ....
> + (-1)^k*r*(r-1)*...*(r-k+1)/(k!)*p^k/z^(n-k+1) + ... )
>
> The residue at the pole is the coefficient of the 1/z term which occurs at
> k = n:
>
> b1 = x^r/i*(-1)^n*r*(r-1)*...*(r-n+1)/(n!)*p^n
>
> and the contour integral around the unit circle by Cauchy's theorem is then:
>
> I = 2*pi*i*b1 = 2*pi*x^r*(-1)^n*r*(r-1)*(r-2)*...*(r-n+1)/(n!)*p^n
> = 2*pi*(1-x)^n/x^(n-r)*r*(r-1)*(r-2)*...*(r-n+1)/(n!)
>
> This last is the definite integral you seek. In matlab terminology you
> could express it using the 'prod' function as:
>
> I = 2*pi*(1-x)^n/x^(n-r)*prod((r-[0:n-1])./[1:n]); % <-- The answer
>
> If n had been a negative integer, f(z) would have been analytic and the
> integral would be zero. Note that if n is not an integer, f(z) would not
> be single-valued and this entire analysis becomes invalid.
>
> You would be wise to test this answer out with a few sets of specific
> parameters, n, r, and x, using numerical integration, to guard against any
> algebraic errors I might have made here.
>
> Roger Stafford

Can one not consider it as the loop integral over -I*n*(z^(-1/n)+x*(1-z^(-1/n)))^r
using the path exp(-I*n*t), t in [-Pi ... Pi] (using Int( f(z) over path ) =
Int( f(path(t))*path'(t) ) and carry your reasoning over to cases where n is not
an integer?

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 3 Aug, 2006 13:21:47

Message: 8 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0308060926530001@dialup-4.232.6.14.dial1.losangeles1.level3.net...
> In article
> <ellieandrogerxyzzy-0308060013150001@dialup-4.232.228.196.dial1.losangeles1.level3.net>,
> ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford) wrote:
>
>> In article <eas1ka$832$1@news.Stanford.EDU>, "Michael"
>> <michael.monkey.in.the.jungle@gmail.com> wrote:
>>
>> > You are right Roger. The x is in [1, +inf), the n is integer, r is
>> > negative
>> > non-integer and may even by non-rational... any more thoughts?
>> -------------------------
>> Hello Michael,
>>
>> If you know that x is real and >= 1, then the problem is easier than I
>> thought. It doesn't matter that r might be irrational, but it is
>> important that n be an integer. I did make a mistake in my algebra
>> earlier. Let's use z = exp(t*i) instead of Walter's Q. The second
>> singularity would occur at z = x/(x-1) which would fortunately lie
>> outside
>> the circle, and therefore the integrand's only singularity is at z = 0.
>> That means your integral can be determined entirely by the residue at z =
>> 0. It's getting rather late for me now, so I'll just say that the
>> integrand needs to be expanded in a series of powers of z so as to
>> determine the coefficient of the one that is 1/z. We need to do a
>> binomial expansion of (z+x*(1-z))^r to accomplish this. That will give
>> us
>> the residue and allow evaluation of the integral. I'll have to do the
>> rest of this for you tomorrow (unless of course you beat me to it and
>> solve it all yourself.) You can probably see at this point why Maple was
>> unable to produce a solution.
>>
>> Roger Stafford
> -----------------------------
> Hello again Michael,
>
> Now that I've had a good night's sleep, I'll come back to your
> integration problem.
>
> Performing Walter's change of variable, z = exp(t*i), leads to the
> closed contour integral running counterclockwise around the unit circle in
> the complex plane with center at z = 0 with the integrand
>
> f(z) = (z+x*(1-z))^r/z^n/(z*i) = x^r/i*(1-p*z)^r/z^(n+1)
>
> where p = (x-1)/x. You stated that x >= 1, so this means that 0 <= p <
> 1. Assuming n is a non-negative integer, f(z) is then a single-valued
> function analytic inside the unit circle except for a single pole of order
> n+1 at z = 0. The factor, (1-p*z)^r, can be expressed via binomial
> expansion as:
>
> (1-p*z)^r = 1 - r*p*z + r*(r-1)/2*p^2*z^2
> - r*(r-1)*(r-2)/6*p^3*z^3 + r*(r-1)*(r-2)*(r-3)/24*p^4*z^4 + ...
> + (-1)^k*r*(r-1)*..*(r-k+1)/(k!)*p^k*z^k + ...
>
> If r is a non-negative integer, this series terminates and is a
> polynomial. Otherwise it is an infinite series convergent inside the unit
> circle. Hence the Laurent expansion of f(z) is
>
> f(z) = x^r/i*(1/z^(n+1) - r*p/z^n + r*(r-1)/2*p^2/z^(n-1) ....
> + (-1)^k*r*(r-1)*...*(r-k+1)/(k!)*p^k/z^(n-k+1) + ... )
>
> The residue at the pole is the coefficient of the 1/z term which occurs at
> k = n:
>
> b1 = x^r/i*(-1)^n*r*(r-1)*...*(r-n+1)/(n!)*p^n
>
> and the contour integral around the unit circle by Cauchy's theorem is
> then:
>
> I = 2*pi*i*b1 = 2*pi*x^r*(-1)^n*r*(r-1)*(r-2)*...*(r-n+1)/(n!)*p^n
> = 2*pi*(1-x)^n/x^(n-r)*r*(r-1)*(r-2)*...*(r-n+1)/(n!)
>
> This last is the definite integral you seek. In matlab terminology you
> could express it using the 'prod' function as:
>
> I = 2*pi*(1-x)^n/x^(n-r)*prod((r-[0:n-1])./[1:n]); % <-- The answer
>
> If n had been a negative integer, f(z) would have been analytic and the
> integral would be zero. Note that if n is not an integer, f(z) would not
> be single-valued and this entire analysis becomes invalid.
>
> You would be wise to test this answer out with a few sets of specific
> parameters, n, r, and x, using numerical integration, to guard against any
> algebraic errors I might have made here.
>
> Roger Stafford

Hi Roger,

What a wonderful work and what a wonderful help you've given me.

I am digesting your solution and picking up my complex analysis from old
days.

My question is that why Maple, or Matlab or Mathematica(I've tried it also,
but the newsgroup for Mathematica has censorship so I could not post my
question-- btw, that's why I've never been a "loyal" user to the Mathematica
software-- I deem efficiency in obtaining help a great factor in reducing
learning curve and allowing me focus on the main objective), why cann't
these software evaluate it?

So numerical evaluation is the only way to check errors?

At first I thought I could use these softwares to help me derive something
useful.

Thanks again Roger. I will feedback to you if I find something...

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 3 Aug, 2006 21:14:48

Message: 9 of 29

In article <44D24F48.92D90C11@axelvogt.de>, Axel Vogt <test3@axelvogt.de> wrote:

> Can one not consider it as the loop integral over
-I*n*(z^(-1/n)+x*(1-z^(-1/n)))^r
> using the path exp(-I*n*t), t in [-Pi ... Pi] (using Int( f(z) over path ) =
> Int( f(path(t))*path'(t) ) and carry your reasoning over to cases where
n is not
> an integer?
--------------------
  You have me a bit confused, Axel. Suppose, as you say, n is not an
integer. It looks as though you have made the change of variable, z =
exp(-n*t*i). However, in this new z-plane, as t varies from -pi to +pi,
that would take this z in a circular, possibly overlapping, path from z =
exp(n*pi*i) to z = exp(-n*pi*i), in which the endpoint must differ from
the startpoint. The looping path is not closed. Also the quantity
z^(-1/n) in the integrand would not be single-valued. How would we make
use of Cauchy's theorem of residues in this kind of situation?

  The problem seems to be that, with n irrational, the exp(-n*t*i) factor
is not periodic over the given interval, and this poses a very different
kind of problem in integration. It may not be accessible to the ordinary
methods of contour integration.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 3 Aug, 2006 21:53:16

Message: 10 of 29

In article <eatlsr$gp$1@news.Stanford.EDU>, "Michael"
<michael.monkey.in.the.jungle@gmail.com> wrote:

> So numerical evaluation is the only way to check errors?
---------------------
  It's the only way I know for this problem at present. It would be
wonderful if Maple were astute enough to get through such an analysis as
this. From my own experience Maple does marvels in solving many
fearsome-looking indefinite integrals. However, I have noted that it does
have a difficult time making solutions appropriate to the varied ranges of
certain parameters. As you undoubtedly know, some complicated indefinite
integrals often have two or more quite different-appearing solutions
depending on the value of a parameter. Maple usually contents itself with
giving only one, even if that results in somewhat confusing complex-valued
expressions. It's a matter of difficulty in transmitting user information
about parameter ranges to the integration logic. The same situation is
surely true for definite integrals that require contour integration, and
that is what is involved with your problem. Your n must be an integer and
your x must be greater than or equal to 1. How can one relay this very
necessary information to the cold, unfeeling logic of Maple, which insists
on issuing a single possible answer covering all cases, and which in this
case may be impossible?

  Incidentally, I took time out to run some tests myself on that formula I
gave you. I used five different sets of n, x, r values, and the formula
values agree with the 'quad8' numerical integrations out to about the 15th
decimal place in each case (I had the tolerance set to very tight
values.) That is a fairly good indication that the errors were due to
roundoff and integration errors, not to errors in algebra or method.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 3 Aug, 2006 15:39:34

Message: 11 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0308061453160001@dialup-4.232.228.88.dial1.losangeles1.level3.net...
> In article <eatlsr$gp$1@news.Stanford.EDU>, "Michael"
> <michael.monkey.in.the.jungle@gmail.com> wrote:
>
>> So numerical evaluation is the only way to check errors?
> ---------------------
> It's the only way I know for this problem at present. It would be
> wonderful if Maple were astute enough to get through such an analysis as
> this. From my own experience Maple does marvels in solving many
> fearsome-looking indefinite integrals. However, I have noted that it does
> have a difficult time making solutions appropriate to the varied ranges of
> certain parameters. As you undoubtedly know, some complicated indefinite
> integrals often have two or more quite different-appearing solutions
> depending on the value of a parameter. Maple usually contents itself with
> giving only one, even if that results in somewhat confusing complex-valued
> expressions. It's a matter of difficulty in transmitting user information
> about parameter ranges to the integration logic. The same situation is
> surely true for definite integrals that require contour integration, and
> that is what is involved with your problem. Your n must be an integer and
> your x must be greater than or equal to 1. How can one relay this very
> necessary information to the cold, unfeeling logic of Maple, which insists
> on issuing a single possible answer covering all cases, and which in this
> case may be impossible?
>
> Incidentally, I took time out to run some tests myself on that formula I
> gave you. I used five different sets of n, x, r values, and the formula
> values agree with the 'quad8' numerical integrations out to about the 15th
> decimal place in each case (I had the tolerance set to very tight
> values.) That is a fairly good indication that the errors were due to
> roundoff and integration errors, not to errors in algebra or method.
>
> Roger Stafford

Hi Roger,

Yes I've checked everything is right and great!q

Could you please recommend me a good reference book on complex analysis? I
have learned this before and then occasionally I found I need a reference on
this subject. I guess it is time for me to add some complex analysis books
on my bookshelf.

Thanks a lot!

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 4 Aug, 2006 00:09:33

Message: 12 of 29

In article <eattv5$89i$1@news.Stanford.EDU>, "Michael"
<michael.monkey.in.the.jungle@gmail.com> wrote:

> Could you please recommend me a good reference book on complex analysis? I
> have learned this before and then occasionally I found I need a reference on
> this subject. I guess it is time for me to add some complex analysis books
> on my bookshelf.
--------------------------
  Just about all the textbooks I am familiar with in complex analysis are
likely to be out of print by now (which should tell you something about my
age.) I'll just mention one of them which I value most highly. It is the
very famous book, "A Course of Modern Analysis" by E. T. Whittaker and G.
N. Watson. It is not light reading but contains some marvelous stuff. I
have used it on many an occasion answering questions on
comp.soft-sys.matlab.

  I checked on the internet about its availability and, amazingly, there
appear to still be some for sale currently, though I think it's been out
of print for many, many years. I saw some offered, for example, at:
http://www.alibris.com/search/books/qwork/1370872/used/A%20Course%20of%20Modern%20Analysis
.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Axel Vogt

Date: 4 Aug, 2006 20:07:54

Message: 13 of 29

Roger Stafford wrote:
>
> In article <44D24F48.92D90C11@axelvogt.de>, Axel Vogt <test3@axelvogt.de> wrote:
>
> > Can one not consider it as the loop integral over
> -I*n*(z^(-1/n)+x*(1-z^(-1/n)))^r
> > using the path exp(-I*n*t), t in [-Pi ... Pi] (using Int( f(z) over path ) =
> > Int( f(path(t))*path'(t) ) and carry your reasoning over to cases where
> n is not
> > an integer?
> --------------------
> You have me a bit confused, Axel. Suppose, as you say, n is not an
> integer. It looks as though you have made the change of variable, z =
> exp(-n*t*i). However, in this new z-plane, as t varies from -pi to +pi,
> that would take this z in a circular, possibly overlapping, path from z =
> exp(n*pi*i) to z = exp(-n*pi*i), in which the endpoint must differ from
> the startpoint. The looping path is not closed. Also the quantity
> z^(-1/n) in the integrand would not be single-valued. How would we make
> use of Cauchy's theorem of residues in this kind of situation?
>
> The problem seems to be that, with n irrational, the exp(-n*t*i) factor
> is not periodic over the given interval, and this poses a very different
> kind of problem in integration. It may not be accessible to the ordinary
> methods of contour integration.
>
> Roger Stafford

Thanks for your kind reply, Roger, you are right ...

--
"please use mail at ... instead of test3 at ... to send me a mail"

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 4 Aug, 2006 13:26:18

Message: 14 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0308061709330001@dialup-4.232.3.31.dial1.losangeles1.level3.net...
> In article <eattv5$89i$1@news.Stanford.EDU>, "Michael"
> <michael.monkey.in.the.jungle@gmail.com> wrote:
>
>> Could you please recommend me a good reference book on complex analysis?
>> I
>> have learned this before and then occasionally I found I need a reference
>> on
>> this subject. I guess it is time for me to add some complex analysis
>> books
>> on my bookshelf.
> --------------------------
> Just about all the textbooks I am familiar with in complex analysis are
> likely to be out of print by now (which should tell you something about my
> age.) I'll just mention one of them which I value most highly. It is the
> very famous book, "A Course of Modern Analysis" by E. T. Whittaker and G.
> N. Watson. It is not light reading but contains some marvelous stuff. I
> have used it on many an occasion answering questions on
> comp.soft-sys.matlab.
>
> I checked on the internet about its availability and, amazingly, there
> appear to still be some for sale currently, though I think it's been out
> of print for many, many years. I saw some offered, for example, at:
> http://www.alibris.com/search/books/qwork/1370872/used/A%20Course%20of%20Modern%20Analysis
> .
>
> Roger Stafford

Roger,

Amazon.com shows that this book is a 5-star book. One comment touts "it
solves applied math problems"... It rocks! I am happy to put it on my
bookshelf. Thank you very much for your wonderful recommendation!

M.

Subject: Is it possible to evaluate this integral?

From: Axel Vogt

Date: 7 Aug, 2006 21:35:11

Message: 15 of 29

Axel Vogt wrote:
>
> Roger Stafford wrote:
> >
> > In article <44D24F48.92D90C11@axelvogt.de>, Axel Vogt <test3@axelvogt.de> wrote:
> >
> > > Can one not consider it as the loop integral over
> > -I*n*(z^(-1/n)+x*(1-z^(-1/n)))^r
> > > using the path exp(-I*n*t), t in [-Pi ... Pi] (using Int( f(z) over path ) =
> > > Int( f(path(t))*path'(t) ) and carry your reasoning over to cases where
> > n is not
> > > an integer?
> > --------------------
> > You have me a bit confused, Axel. Suppose, as you say, n is not an
> > integer. It looks as though you have made the change of variable, z =
> > exp(-n*t*i). However, in this new z-plane, as t varies from -pi to +pi,
> > that would take this z in a circular, possibly overlapping, path from z =
> > exp(n*pi*i) to z = exp(-n*pi*i), in which the endpoint must differ from
> > the startpoint. The looping path is not closed. Also the quantity
> > z^(-1/n) in the integrand would not be single-valued. How would we make
> > use of Cauchy's theorem of residues in this kind of situation?
> >
> > The problem seems to be that, with n irrational, the exp(-n*t*i) factor
> > is not periodic over the given interval, and this poses a very different
> > kind of problem in integration. It may not be accessible to the ordinary
> > methods of contour integration.
> >
> > Roger Stafford
>
> Thanks for your kind reply, Roger, you are right ...
>
> --
> "please use mail at ... instead of test3 at ... to send me a mail"

I think the general case goes through a hypergeometric 2F1, for which I
use Maple to get 2*x^r/n*sin(Pi*n)*hypergeom([-r, -n],[1-n],-(x-1)/x)
=: theSol(n,r). Here is a sketch (using some of Roger's steps):

Write the integral theI(n,r) as x^r*Int(S*exp(-I*t*n),t = -Pi .. Pi),
S = (1-rho*exp(t*I))^r, rho = (x-1)/x and take the Taylor series for S,
S = Sum(pochhammer(-r,k)/k!*(rho*exp(t*I))^k,k = 0 .. infinity).

Now use Maple to integrate each summand and compute the resulting series.
Maple spits it out as the stated solution - for which n must not be an
integer >= 0 (for negative integers sin brings it to 0).

For n=0 use L'Hospital's Rule to get theI(0,r) = 2*x^r*Pi and for positive
integers theI(n,r) = ((2*(-1+2*x)^r*sin(n*Pi) +r*(1-x)*theI(n-1,r-1))/n)
is true, as one shows by partial integration (differentiating the factor
with power r).

sin = 0 for n in IN, so one gets 2*pochhammer(-r,n)*((x-1)/x)^n/n!*x^r*Pi,
which is Roger's result, where pochhammer(z,n) = z*(z+1)*...*(z+n-1).

For a test one can even take complex values, using 14 Digits of exactness
with tstData:=[ x=1.123+0.456*I, r=0.345-1.23*I, n=-1.4-1.3*I], i get
-59.428502841874+42.371655732337*I resp -59.428502841876+42.371655732336*I
for theI(n,r) resp theSol(n,r).

Note that for x close to 1 the argument for the r-th power approaches 0 and
one may needs high accuracy to compare results from integration and the other
ways for general r.

Subject: Is it possible to evaluate this integral?

From: Axel Vogt

Date: 7 Aug, 2006 22:21:58

Message: 16 of 29

Axel Vogt wrote:
...
> Note that for x close to 1 the argument for the r-th power approaches 0 and
> one may needs high accuracy to compare results from integration and the other
> ways for general r.

grrr ... one just needs high accuracy ...

Subject: Is it possible to evaluate this integral?

From: Martin Eisenberg

Date: 8 Aug, 2006 01:18:10

Message: 17 of 29

Axel Vogt wrote:

> expr =: theSol(n,r)

Is that recent-ish Maple notation, or are you speaking
purely mathematically there?


Martin

--
Quidquid latine scriptum sit, altum viditur.

Subject: Is it possible to evaluate this integral?

From: israel@math.ubc.ca (Robert Israel)

Date: 8 Aug, 2006 05:41:29

Message: 18 of 29

In article <1154999935.405283@ostenberg.wh.uni-dortmund.de>,
Martin Eisenberg <martin.eisenberg@udo.edu> wrote:
>Axel Vogt wrote:
>
>> expr =: theSol(n,r)
>
>Is that recent-ish Maple notation, or are you speaking
>purely mathematically there?


Maple never uses "=:", only ":=".
Didn't ":=" for assignment originate with Algol?

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Subject: Is it possible to evaluate this integral?

From: Paul Abbott

Date: 8 Aug, 2006 23:21:21

Message: 19 of 29

In article <eatlsr$gp$1@news.Stanford.EDU>,
 "Michael" <michael.monkey.in.the.jungle@gmail.com> wrote:

> My question is that why Maple, or Matlab or Mathematica(I've tried it also,
> but the newsgroup for Mathematica has censorship

It is not censored -- it is moderated ...

> so I could not post my
> question-- btw, that's why I've never been a "loyal" user to the Mathematica
> software-- I deem efficiency in obtaining help a great factor in reducing
> learning curve and allowing me focus on the main objective), why cann't
> these software evaluate it?

With the assistance of Mathematica, using series expansion, one can show
that

  Integrate[((1 - E^(I t)) x + E^(I t))^r E^(-I n t), {t, -Pi, Pi}]

evaluates to

   2 (-1)^n Pi ((x - 1)/x)^n x^r Gamma[r + 1] / (Gamma[r + 1 - n] n!)

for x > 1.

> So numerical evaluation is the only way to check errors?

No.

> At first I thought I could use these softwares to help me derive something
> useful.

You can.

Cheers,
Paul

_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 8 Aug, 2006 10:44:25

Message: 20 of 29


"Paul Abbott" <paul@physics.uwa.edu.au> wrote in message
news:paul-D78CB1.23212108082006@news.uwa.edu.au...
> In article <eatlsr$gp$1@news.Stanford.EDU>,
> "Michael" <michael.monkey.in.the.jungle@gmail.com> wrote:
>
>> My question is that why Maple, or Matlab or Mathematica(I've tried it
>> also,
>> but the newsgroup for Mathematica has censorship
>
> It is not censored -- it is moderated ...
>
>> so I could not post my
>> question-- btw, that's why I've never been a "loyal" user to the
>> Mathematica
>> software-- I deem efficiency in obtaining help a great factor in reducing
>> learning curve and allowing me focus on the main objective), why cann't
>> these software evaluate it?
>
> With the assistance of Mathematica, using series expansion, one can show
> that
>
> Integrate[((1 - E^(I t)) x + E^(I t))^r E^(-I n t), {t, -Pi, Pi}]
>
> evaluates to
>
> 2 (-1)^n Pi ((x - 1)/x)^n x^r Gamma[r + 1] / (Gamma[r + 1 - n] n!)
>
> for x > 1.
>


Hi Paul,

Could you post your Mathematica code for computing and evaluating it into
that nice expression with Gamma function? I've tried many times and it just
did not work.

Thanks,

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 8 Aug, 2006 18:12:58

Message: 21 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0308060013150001@dialup-4.232.228.196.dial1.losangeles1.level3.net...
> In article <eas1ka$832$1@news.Stanford.EDU>, "Michael"
> <michael.monkey.in.the.jungle@gmail.com> wrote:
>
>> You are right Roger. The x is in [1, +inf), the n is integer, r is
>> negative
>> non-integer and may even by non-rational... any more thoughts?
> -------------------------
> Hello Michael,
>
> If you know that x is real and >= 1, then the problem is easier than I
> thought. It doesn't matter that r might be irrational, but it is
> important that n be an integer. I did make a mistake in my algebra
> earlier. Let's use z = exp(t*i) instead of Walter's Q. The second
> singularity would occur at z = x/(x-1) which would fortunately lie outside
> the circle, and therefore the integrand's only singularity is at z = 0.
> That means your integral can be determined entirely by the residue at z =
> 0. It's getting rather late for me now, so I'll just say that the
> integrand needs to be expanded in a series of powers of z so as to
> determine the coefficient of the one that is 1/z. We need to do a
> binomial expansion of (z+x*(1-z))^r to accomplish this. That will give us
> the residue and allow evaluation of the integral. I'll have to do the
> rest of this for you tomorrow (unless of course you beat me to it and
> solve it all yourself.) You can probably see at this point why Maple was
> unable to produce a solution.
>
> Roger Stafford

Hi Roger,

I have one more question: when taking contour integral, do we consider the
branch cuts and branching points, since this "r" is a negative possibly
irrational number?

Thanks again!

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 9 Aug, 2006 02:04:18

Message: 22 of 29

In article <ebbcqa$1lh$1@news.Stanford.EDU>, "Michael"
<michael.monkey.in.the.jungle@gmail.com> wrote:

> I have one more question: when taking contour integral, do we consider the
> branch cuts and branching points, since this "r" is a negative possibly
> irrational number?
>
> Thanks again!
------------------------
  As long as you have x > 1 (and n an integer), then with the substitution
z = exp(t*i), we have

 (exp(t*i)+x*(1-exp(t*i))) = x*(1-(x-1)/x*z)

and this can never be zero when extended inside the unit circle since
(x-1)/x < 1. Consequently there is no branch point for your integrand
inside the circle regardless of what value r has, positive or negative,
rational or irrational. Just the pole at z = 0 mentioned earlier. Of
course, as I also mentioned, if n is not integral-valued, the integrand
would not be single-valued and contour integration would not work.

  As you may have noticed, Axel has worked on the idea of separately
integrating each term of the binomial expansion obtained with
(1-(x-1)/x*z)^r and expressing the sum of these in terms of hypergeometric
series, rather than performing contour integration. I don't seem to get
the same results as he does but the general idea sounds valid. I have
emailed him asking for more details.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 8 Aug, 2006 23:45:40

Message: 23 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0808061904180001@dialup-4.232.228.222.dial1.losangeles1.level3.net...
> In article <ebbcqa$1lh$1@news.Stanford.EDU>, "Michael"
> <michael.monkey.in.the.jungle@gmail.com> wrote:
>
>> I have one more question: when taking contour integral, do we consider
>> the
>> branch cuts and branching points, since this "r" is a negative possibly
>> irrational number?
>>
>> Thanks again!
> ------------------------
> As long as you have x > 1 (and n an integer), then with the substitution
> z = exp(t*i), we have
>
> (exp(t*i)+x*(1-exp(t*i))) = x*(1-(x-1)/x*z)
>
> and this can never be zero when extended inside the unit circle since
> (x-1)/x < 1. Consequently there is no branch point for your integrand
> inside the circle regardless of what value r has, positive or negative,
> rational or irrational. Just the pole at z = 0 mentioned earlier. Of
> course, as I also mentioned, if n is not integral-valued, the integrand
> would not be single-valued and contour integration would not work.
>
> As you may have noticed, Axel has worked on the idea of separately
> integrating each term of the binomial expansion obtained with
> (1-(x-1)/x*z)^r and expressing the sum of these in terms of hypergeometric
> series, rather than performing contour integration. I don't seem to get
> the same results as he does but the general idea sounds valid. I have
> emailed him asking for more details.
>
> Roger Stafford

Hi Roger,

Thanks for your explanation. I've got a bunch of complex analysis books on
my hand. The concept of branching points and branching cuts are confusing.
Esp. when it couples with contour integrations.

As a learning process and learning from you, let's say the problem is such
that for (f(z))^r there is a point that f(z)=0. And that is the branching
point. What would the contour integration change to in that case?

Thanks again!

-M

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 9 Aug, 2006 08:51:31

Message: 24 of 29

In article <ebc0a6$fe4$1@news.Stanford.EDU>, "Michael"
<michael.monkey.in.the.jungle@gmail.com> wrote:

> Hi Roger,
>
> Thanks for your explanation. I've got a bunch of complex analysis books on
> my hand. The concept of branching points and branching cuts are confusing.
> Esp. when it couples with contour integrations.
>
> As a learning process and learning from you, let's say the problem is such
> that for (f(z))^r there is a point that f(z)=0. And that is the branching
> point. What would the contour integration change to in that case?
>
> Thanks again!
>
> -M
-----------------------
  In my experience, contour integration is ordinarily performed using only
single-valued analytic functions defined in the interior of closed
curves. The trouble with functions possessing multiple branches is that
to traverse what one could reasonably call a "closed" path, one would have
to encircle a branch point a number of times traversing each of the
branches in turn in order to apply Cauchy's theorem. That is not
convenient when applying contour integration to solving definite
integrals. For example, traveling around the origin twice with f(z) =
z^(-1/2) over the same path would traverse branches of opposite sign and
would clearly cancel each other out giving a zero result and no useful
information about a definite integral.

  In the case of an analytic f(z) with a zero which is taken to a power r,
multiple branches occur if r is not integer-valued but only rational. If
r is irrational, there will be infinitely many branches.

  That is why I breathed a sigh of relief when you placed a condition on x
that avoided any zero values in z+x*(1-z) within the unit circle. It is
my impression that otherwise, different methods than contour integration
would have to be used to solve your integral.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Michael

Date: 9 Aug, 2006 10:14:36

Message: 25 of 29


"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message
news:ellieandrogerxyzzy-0908060151320001@dialup-4.232.228.178.dial1.losangeles1.level3.net...
> In article <ebc0a6$fe4$1@news.Stanford.EDU>, "Michael"
> <michael.monkey.in.the.jungle@gmail.com> wrote:
>
>> Hi Roger,
>>
>> Thanks for your explanation. I've got a bunch of complex analysis books
>> on
>> my hand. The concept of branching points and branching cuts are
>> confusing.
>> Esp. when it couples with contour integrations.
>>
>> As a learning process and learning from you, let's say the problem is
>> such
>> that for (f(z))^r there is a point that f(z)=0. And that is the branching
>> point. What would the contour integration change to in that case?
>>
>> Thanks again!
>>
>> -M
> -----------------------
> In my experience, contour integration is ordinarily performed using only
> single-valued analytic functions defined in the interior of closed
> curves. The trouble with functions possessing multiple branches is that
> to traverse what one could reasonably call a "closed" path, one would have
> to encircle a branch point a number of times traversing each of the
> branches in turn in order to apply Cauchy's theorem. That is not
> convenient when applying contour integration to solving definite
> integrals. For example, traveling around the origin twice with f(z) =
> z^(-1/2) over the same path would traverse branches of opposite sign and
> would clearly cancel each other out giving a zero result and no useful
> information about a definite integral.
>
> In the case of an analytic f(z) with a zero which is taken to a power r,
> multiple branches occur if r is not integer-valued but only rational. If
> r is irrational, there will be infinitely many branches.
>
> That is why I breathed a sigh of relief when you placed a condition on x
> that avoided any zero values in z+x*(1-z) within the unit circle. It is
> my impression that otherwise, different methods than contour integration
> would have to be used to solve your integral.
>
> Roger Stafford

Roger,

Thanks a lot for your explanation. Could Axel's alternative method be used
for the case of branching points? I really want to be able to integrate it
for r being irrational number and the x is such that there will be branching
points in the Contour integral? Please give me some pointers! Thanks a lot!

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 9 Aug, 2006 19:32:14

Message: 26 of 29

In article
<ellieandrogerxyzzy-0808061904180001@dialup-4.232.228.222.dial1.losangeles1.level3.net>,
ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford) wrote:

> ....
> As you may have noticed, Axel has worked on the idea of separately
> integrating each term of the binomial expansion obtained with
> (1-(x-1)/x*z)^r and expressing the sum of these in terms of hypergeometric
> series, rather than performing contour integration. I don't seem to get
> the same results as he does but the general idea sounds valid. I have
> emailed him asking for more details.
> Roger Stafford
--------------------------
  After getting past an error of mine, I now agree with Axel's solution,
"theSol", involving a hypergeometric series. This allows the quantity n
to be non-integer.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford)

Date: 9 Aug, 2006 20:17:29

Message: 27 of 29

In article <ebd55b$jo4$1@news.Stanford.EDU>, "Michael"
<michael.monkey.in.the.jungle@gmail.com> wrote:

> Thanks a lot for your explanation. Could Axel's alternative method be used
> for the case of branching points? I really want to be able to integrate it
> for r being irrational number and the x is such that there will be branching
> points in the Contour integral? Please give me some pointers! Thanks a lot!
---------------------
Hello Michael,

  If your x has the property that abs((x-1)/x) > 1, which is what will
give you a branch point inside the unit circle, then Axel's hypergeometric
series will be divergent and you won't be able to evaluate it. Look in
your newly acquired Whittaker and Watson in the chapter on hypergeometric
series and you will see the requirement abs(z) < 1 on its first page.

Roger Stafford

Subject: Is it possible to evaluate this integral?

From: Axel Vogt

Date: 9 Aug, 2006 22:42:57

Message: 28 of 29

Roger Stafford wrote:
>
> In article
> <ellieandrogerxyzzy-0808061904180001@dialup-4.232.228.222.dial1.losangeles1.level3.net>,
> ellieandrogerxyzzy@mindspring.com.invalid (Roger Stafford) wrote:
>
> > ....
> > As you may have noticed, Axel has worked on the idea of separately
> > integrating each term of the binomial expansion obtained with
> > (1-(x-1)/x*z)^r and expressing the sum of these in terms of hypergeometric
> > series, rather than performing contour integration. I don't seem to get
> > the same results as he does but the general idea sounds valid. I have
> > emailed him asking for more details.
> > Roger Stafford
> --------------------------
> After getting past an error of mine, I now agree with Axel's solution,
> "theSol", involving a hypergeometric series. This allows the quantity n
> to be non-integer.
>
> Roger Stafford

Thank you for cross-checking the statements!

Subject: Is it possible to evaluate this integral?

From: Axel Vogt

Date: 9 Aug, 2006 22:45:53

Message: 29 of 29

Michael wrote:
>
> I couldn't get Matlab to compute it(Matlab uses Maple engine...)
>
> int((exp(1i * t) + x * (1-exp(1i * t)))^r*exp(-1i*n*t), t, -pi, pi)
>
> How to do it?
>
> Thanks a lot!

As the thread is on-going I would like to ask:
where does that question come from, why you
want to have it?

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