Solving Complex Integral: Evaluating \int_{0}^{\infty}\frac{(x^a-1)}{x^3-1}\,dx

[Belgium 12]

Hi,

problrm with complex integral.Consider the integral

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

use the branch 0<phi<3pi/2 and an idented contour at z=0 and z=1.(circular contour in the

upper half plane)

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

$$\pi\frac{(e^{2ia\pi/3}-1)}{(e^{2i\pi/3}-1)sin(2\pi/3)}+\int_{0}^{\infty}\frac{(x^ae^{ia\pi}-1)}{x^3+1}$$


b)evaluate the second integral in part a)and find the value of the orginal integral

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

For the last integral I used a contour z=x(0<x<R)the sector circle in the upper half plane

0<phi<2pi/3)and the line z=xe^2pi/3(0<x<R)

but I can't find the integral question b)

Can somebody give me a help.Thanks

 

[mighty2000]

Hi there,

Thank you for sharing your problem with us. I can see that you have already made some progress in solving this complex integral by using the indent contour method. However, it seems like you are having trouble evaluating the second integral in part a) and finding the value of the original integral in question b). Let me offer some guidance to help you move forward.

Firstly, it is important to note that the integral in question b) can be simplified by using the identities sin(2\pi/3) = sin(\pi/3) = \sqrt{3}/2. This will help you to simplify the expression and make it easier to evaluate.

Next, for the second integral in part a), you can use the substitution x = e^{i\pi/3}t to transform the integral into a form that can be evaluated using the residue theorem. This will involve finding the poles of the integrand and calculating the residues at those poles.

Once you have evaluated the second integral, you can substitute the result into the expression in question b) and use the identities mentioned earlier to simplify it further.

I hope this helps you to make progress in solving your problem. If you encounter any further difficulties, don't hesitate to reach out for more assistance. Good luck!