Integrating Functions with Double Poles: A Contour Integral Approach

[Ancient_Nomad]

Hi,

Can someone tell me how to integrate functions which have a branch point and a pole (of order > 1) on the x-axis.
Specifically, I ran into the following problem while playing around with contour integrals, which has a double pole at x = -2 . I tried to do this with a keyhole contour, but that doesn't work out, as the integrals over the horizontal lines cancel out.

\int^{\inf}_0 \frac{ln(x)}{(x+2)^2}\, dx

Thanks.

 

[jostpuur]

I don't know how to make the contour approach work either. It could be that integration by parts might work, but the origo seems to produce some problems. If the integration is first restricted to [\epsilon,\infty[, one could substitute

<br /> \frac{\log(x)}{(x+2)^2} = -D_x \frac{\log(x)}{x+2} + \frac{1}{x(x+2)},<br />

work out the integrals (I didn't do the details yet, but it looks like a simple exercise at quick glance), and then try to deal with the limit \epsilon\to 0.

 

[Ancient_Nomad]

Thanks for the reply. I worked it out using the way you suggested. It comes out correctly and easily. But I would still like to find a way using contour integration.

For those who are interested,
the answer is \frac{ln(2)}{2} .

Edit:

The given integral can be found by solving the following integral over a keyhole contour, and dividing the complete integral into real and imaginary parts.

\oint_C \frac{(ln(z))^2}{(z+2)^2} \, dz

 

[maverick280857]

Edit:

The given integral can be found by solving the following integral over a keyhole contour, and dividing the complete integral into real and imaginary parts.

\oint_C \frac{(ln(z))^2}{(z+2)^2} \, dz

Ok, so it worked. Nice. Please post some details here.