Thread Subject:

Matlab's Quadl results wrong

I need help to figure out why using matlab's quad function to solve a simple double inetgral results in a wrong answer.

the integral is:

int_0^1 log int_0^1 (x+y) dx dy,

using the following function

function z=intlog(y) n=length(y); z=zeros(size(y)); for j=1:n a=0;b=0; fy=@(x)(x+y(j)) z(j)=quadl(fy,a,b) end return

the correct answer is -1.3333... but running the above function in the command window as:q=quad(@intlog,0,1) gives the solution -0.0452.

"Innocent " <2400163@uwc.ac.za> wrote in message <j14dlb$kqq$1@newscl02ah.mathworks.com>...
> I need help to figure out why using matlab's quad function to solve a simple double inetgral results in a wrong answer.
>
> the integral is:
>
> int_0^1 log int_0^1 (x+y) dx dy,
>
> using the following function
>
> function z=intlog(y) n=length(y); z=zeros(size(y)); for j=1:n a=0;b=0; fy=@(x)(x+y(j)) z(j)=quadl(fy,a,b) end return
>
> the correct answer is -1.3333... but running the above function in the command window as:q=quad(@intlog,0,1) gives the solution -0.0452.
- - - - - - - - - - - -
  The correct solution is actually 1/2*log(27/4)-1 = -0.045228748 .

  The integrand of the outer integral is:

 log(int_0^1 x+y dx) = log( (1^2/2+1*y)-(0^2/2+0*y) ) = log(1/2+y) .

Then doing integration by parts:

 int_0^1 log(1/2+y) dy =
  (1/2+1)*log(1/2+1)-(1/2+0)*log(1/2+0) - int_0^1 (1/2+y)/(1/2+y) dy
  = 3/2*log(3/2) - 1/2*log(1/2) - 1
  = 1/2*log((3/2)^3/(1/2)) -1 = 1/2*log(27/4 - 1 = -0.045228748

Roger Stafford

Dear Innocent,

> I need help to figure out why using matlab's quad function to solve a simple double inetgral results in a wrong answer.

There are two questions about this at MATLAB Answers and one here. Three people and Matlab replied the correct correct result.
This is not an efficient method to solve an integral.

Kind regards, Jan

Thanks Roger, The answer from matlab is correct, my working was wrong as i missed inetgrating by parts, my calculus is rust i guess.

Thanks indeed.
Innocent



"Roger Stafford" wrote in message <j14hoi$m1$1@newscl02ah.mathworks.com>...
> "Innocent " <2400163@uwc.ac.za> wrote in message <j14dlb$kqq$1@newscl02ah.mathworks.com>...
> > I need help to figure out why using matlab's quad function to solve a simple double inetgral results in a wrong answer.
> >
> > the integral is:
> >
> > int_0^1 log int_0^1 (x+y) dx dy,
> >
> > using the following function
> >
> > function z=intlog(y) n=length(y); z=zeros(size(y)); for j=1:n a=0;b=0; fy=@(x)(x+y(j)) z(j)=quadl(fy,a,b) end return
> >
> > the correct answer is -1.3333... but running the above function in the command window as:q=quad(@intlog,0,1) gives the solution -0.0452.
> - - - - - - - - - - - -
> The correct solution is actually 1/2*log(27/4)-1 = -0.045228748 .
>
> The integrand of the outer integral is:
>
> log(int_0^1 x+y dx) = log( (1^2/2+1*y)-(0^2/2+0*y) ) = log(1/2+y) .
>
> Then doing integration by parts:
>
> int_0^1 log(1/2+y) dy =
> (1/2+1)*log(1/2+1)-(1/2+0)*log(1/2+0) - int_0^1 (1/2+y)/(1/2+y) dy
> = 3/2*log(3/2) - 1/2*log(1/2) - 1
> = 1/2*log((3/2)^3/(1/2)) -1 = 1/2*log(27/4 - 1 = -0.045228748
>
> Roger Stafford

Thanks for pointing me in the right direction.

regards
Innocent


"Jan Simon" wrote in message <j14t8k$qj4$1@newscl02ah.mathworks.com>...
> Dear Innocent,
>
> > I need help to figure out why using matlab's quad function to solve a simple double inetgral results in a wrong answer.
>
> There are two questions about this at MATLAB Answers and one here. Three people and Matlab replied the correct correct result.
> This is not an efficient method to solve an integral.
>
> Kind regards, Jan