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Thread Subject:
integral equation

Subject: integral equation

From: Sulaiman Rabbaa

Date: 14 Oct, 2011 09:38:29

Message: 1 of 5

Dear all
I need help to solve integral equation with complex constants,
example:
a+6i=integral from 1 to 5 of (x+a)dx
This is simple to be solved directly, but actually I need a solution method to apply it for complicated integrals.
Thanks

Subject: integral equation

From: Torsten

Date: 14 Oct, 2011 10:30:45

Message: 2 of 5

On 14 Okt., 11:38, "Sulaiman Rabbaa" <sulr...@yahoo.com> wrote:
> Dear all
> I need help to solve integral equation with complex constants,
> example:
> a+6i=integral from 1 to 5 of (x+a)dx
> This is simple to be solved directly, but actually I need a solution method to apply it for complicated integrals.
> Thanks

This can be written as
a+6i-integral from 1 to 5 of (x+a)dx = 0

Define
f(a) =a+6i-integral from 1 to 5 of (x+a)dx

Then you want to find a zero of f. For this purpose, you can use
FZEROor fSOLVE.
To evaluate the integral for a value of the unknown parameter _a_
suggested by fzero or fsolve, use QUAD.

Best wishes
Torsten.

Subject: integral equation

From: Greg Heath

Date: 14 Oct, 2011 18:11:21

Message: 3 of 5

On Oct 14, 5:38 am, "Sulaiman Rabbaa" <sulr...@yahoo.com> wrote:
> Dear all
> I need help to solve integral equation with complex constants,
> example:
> a+6i=integral from 1 to 5 of (x+a)dx

I don't get it. Is this supposed to be an integral equation for a?

> This is simple to be solved directly,

a+6i = 25/2-1/2+a*(5-1)
-3*a = 12-6i
a = -4+2i

but actually I need a solution method to apply it for complicated
integrals.
> Thanks

Still confused. Can you provide more details and/or examples?

Greg

Subject: integral equation

From: Sulaiman Rabbaa

Date: 14 Oct, 2011 20:14:27

Message: 4 of 5

Greg Heath <heath@alumni.brown.edu> wrote in message <299161fb-8997-43d4-a457-2268740cc616@r1g2000yqm.googlegroups.com>...
> On Oct 14, 5:38 am, "Sulaiman Rabbaa" <sulr...@yahoo.com> wrote:
> > Dear all
> > I need help to solve integral equation with complex constants,
> > example:
> > a+6i=integral from 1 to 5 of (x+a)dx
>
> I don't get it. Is this supposed to be an integral equation for a?
>
> > This is simple to be solved directly,
>
> a+6i = 25/2-1/2+a*(5-1)
> -3*a = 12-6i
> a = -4+2i
>
> but actually I need a solution method to apply it for complicated
> integrals.
> > Thanks
>
> Still confused. Can you provide more details and/or examples?
>
> Greg
I have actually the following equation:
1+5/(3-a^2-2 a i )=integral of x^2/[(1+exp(2x^2))*(a+2+x)*(a-2+x)] dx from x=0 to x= infinity , where a is complex number
I need help to evaluate this integral using matlab
Thanks in advanced

Subject: integral equation

From: Jonas Lundgren

Date: 31 Oct, 2011 09:39:10

Message: 5 of 5

"Sulaiman Rabbaa" wrote in message <j7a573$7ld$1@newscl01ah.mathworks.com>...
> Greg Heath <heath@alumni.brown.edu> wrote in message <299161fb-8997-43d4-a457-2268740cc616@r1g2000yqm.googlegroups.com>...
> > On Oct 14, 5:38 am, "Sulaiman Rabbaa" <sulr...@yahoo.com> wrote:
> > > Dear all
> > > I need help to solve integral equation with complex constants,
> > > example:
> > > a+6i=integral from 1 to 5 of (x+a)dx
> >
> > I don't get it. Is this supposed to be an integral equation for a?
> >
> > > This is simple to be solved directly,
> >
> > a+6i = 25/2-1/2+a*(5-1)
> > -3*a = 12-6i
> > a = -4+2i
> >
> > but actually I need a solution method to apply it for complicated
> > integrals.
> > > Thanks
> >
> > Still confused. Can you provide more details and/or examples?
> >
> > Greg
> I have actually the following equation:
> 1+5/(3-a^2-2 a i )=integral of x^2/[(1+exp(2x^2))*(a+2+x)*(a-2+x)] dx from x=0 to x= infinity , where a is complex number
> I need help to evaluate this integral using matlab
> Thanks in advanced

A Newton-Raphson solution. Another root is close to (-2.6 - 1i).

f = @(x,a) x.^2./(x+a+2)./(x+a-2)./(1+exp(2*x.^2));
df = @(x,a) -2*x.^2.*(x+a)./(x+a+2).^2./(x+a-2).^2./(1+exp(2*x.^2));

g = @(a) quadgk(@(x)f(x,a),0,12);
dg = @(a) quadgk(@(x)df(x,a),0,12);

h = @(a) g(a) - 1 - 5/(3-a*a-2i*a);
dh = @(a) dg(a) - 5*(2*a+2i)/(3-a*a-2i*a)^2;

a = 2.6 - 1i;
da = 1;
while abs(da) > 1e-14
    da = -h(a)/dh(a);
    a = a + da;
end

/Jonas

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