Complex integral

**Gilbert** · June 3rd 2008, 01:05 AM

Hi,

Problem with complex integral.Consider the integral:

$\int_{0}^{\infty}\frac{(x^a-1)}{x^3-1}\,dx$

with the branch -pi/0<phi<3pi/2 and the idented contour at z=0 and z=1

(circular contour in the upper half plane)

a)show that this integral can be written in terms of the integral:

$\pi\frac{(e^{2ia\pi/3}-1)}{(e^{2i\pi/3}-1)sin2\pi/3}$

$+\int_{0}^{\infty}\frac{(x^ae^{ia\pi}-1)}{x^3+1}$

no problem with this.I have found it.

b)Evaluate the second integral in part a)and fiond the value of the original

integral.

$\frac{\pi sin(a\pi/3)}{3sin\pi/3)sin¨[\pi/3(a+1)]}$

with this last integral I hav a problem.I used the contour z=x(0<x<R) the sectorcircle z=Re^i.phi(0<phi<2pi/3)and the line z=xe^2pî/3(0<x<R).

I can't find the result.
The choosen contour????????????
Can somebody give me a suggestion?????????Thanks

**ThePerfectHacker** · June 3rd 2008, 04:49 PM

Originally Posted by Gilbert

Hi,

Problem with complex integral.Consider the integral:

$\int_{0}^{\infty}\frac{(x^a-1)}{x^3-1}\,dx$

I do not see how to do this, not even sure if it converges.
It seems to me you are actually asking,
$\int_0^{\infty} \frac{x^a - 1}{x^3+1}dx$ .
So that the denominator is defined.

Also it might help if you say what $a$ is? Is it $0<a<1$ ?

If it is like how I describe then write,
$\int_0^{\infty} \frac{x^a}{x^3+1} dx - \int_0^{\infty} \frac{dx}{x^3+1}$ .
Now use the fact,
$\int_0^{\infty} \frac{dx}{x^3+1} = \frac{2}{3}\pi$ .
And we just have to compute,
$\int_0^{\infty} \frac{x^a}{x^3+1} dx$ .

I think this is done by using a keyhole contour.

**Gilbert** · June 4th 2008, 01:51 PM

Hi,

yes it is 0<a<1.I don't think it can be done by the keyhole contout???????????

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