Understanding Branch Cuts of Ln(x) for Integrating (x^2+1)^2 from 0 to Infinity

[barrybaker]

1.I am trying to evaluate the $$\int$$$$\frac{ln(x)}{(x^{2}+1)^{2}}$$ from 0 to inf2. I think I am having problems defining what my branch cut should be.

3. Here is what I have tried to no avail. First set the branch cut from -infty to 0. Then changing it to a contour integral gives us ($$\int$$$$\frac{ln(x)}{(x^{2}+1)^{2}}$$) = 2 $$\pi$$ i (a_-1(i)+a_-1(-i))=$$\frac{i \pi^{2}}{4}$$

I get somewhat lost after this. Please help. I'm studying for finals but i don't really understand this

 

[squidsoft]

I'd use an upper half-washer with an indentation around the origin. The branch-cut can then be anywhere in the lower half-plane including on the real axis. Then consider the following contour integral:

$$\mathop\oint\limits_{W_u}\frac{\text{Log}(z)}{(1+z^2)^2} dz$$

Assuming you can show the integral goes to zero over both large and small semi-circles of the contour, we're left with:

$$\mathop\oint\limits_{W_u}\frac{\text{Log}(z)}{(1+z^2)^2} dz=2\int_0^{\infty}\frac{\ln(r)}{(1+r^2)^2}dr+\pi i\int_0^{\infty}\frac{1}{(1+r^2)^2}dr=2\pi i \frac{d}{dz} \frac{\log(z)}{(z+i)^2}\Bigg|_{z=i}$$

where I let $$z=r e^{0i}$$ on the leg from $$0$$ to $$\infty$$ and $$z=re^{\pi i}$$ on the leg from $$-\infty$$ to $$0$$ to arrive at that expression.

 

[barrybaker]

Thanks so much. I worked it out as you have laid it out and received the correct answer. but how did you know to put the contour only in the upper half plane?

 

[squidsoft]

Thanks so much. I worked it out as you have laid it out and received the correct answer. but how did you know to put the contour only in the upper half plane?

You could use the lower half-plane and adjust the analysis accordingly. You just have to work a bunch of them to gain experience to analyze it and try things and be willing to fail at it from time to time. Remember, good cooks try again after failing a recipe, bad ones give up. :)