"Sulaiman Rabbaa" wrote in message <j7a573$7ld$1@newscl01ah.mathworks.com>...
> Greg Heath <heath@alumni.brown.edu> wrote in message <299161fb899743d4a4572268740cc616@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 = 25/21/2+a*(51)
> > 3*a = 126i
> > 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/(3a^22 a i )=integral of x^2/[(1+exp(2x^2))*(a+2+x)*(a2+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 NewtonRaphson solution. Another root is close to (2.6  1i).
f = @(x,a) x.^2./(x+a+2)./(x+a2)./(1+exp(2*x.^2));
df = @(x,a) 2*x.^2.*(x+a)./(x+a+2).^2./(x+a2).^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/(3a*a2i*a);
dh = @(a) dg(a)  5*(2*a+2i)/(3a*a2i*a)^2;
a = 2.6  1i;
da = 1;
while abs(da) > 1e14
da = h(a)/dh(a);
a = a + da;
end
/Jonas
