1. ## Complex Analysis

I need to show that $\int_{0}^{\infty}\frac{\log(x)}{(1+x^2)^2}=-\frac{\pi}{4}$

If we complexify the integrand and make a sort of pacman-contour $\mu$

$\mu_1 = t+ ic_0$ with $0\leq t\leq R$

$\mu_2 = Re^{i\pi t}$ with $t_0\leq t \leq 2\pi-t_0$

$\mu_3 = R-t-ic_0$ with $0\leq t\leq R$

$\mu_4 = re^{i\pi t}$ with $2\pi- t_0\leq t \leq t_0$

As $R\to \infty, r\to 0$ we let $t_0,c_0\to 0$
This contour contains the 2 second order singularities $p= \pm i$

And we can evaluate $\int_\mu f(z)dz$ with the Cauchy integral formula. I was somewhat surprised when I found $\frac{-\pi}{4}$. It's easy to check that

$|\int_{\mu_2}f(z)dz|\to 0$and $|\int_{\mu 4}f(z)dz|\to 0$

As a result we get $\int_{\mu_1}f(z)dz+\int_{\mu_3}f(z)dz = -\frac{\pi}{4}$

If all steps were ok, somehow I want to conclude that $\int_{\mu_1\cup\mu_3}f(z)dz$ tends to $\int_{0}^{\infty}f(x)dx$

But I'm not sure how. Also I found working with log(z) somewhat trouble-some. I believe I need to use some nice property of log(z)

Can anyone offer some suggestions?

2. Originally Posted by Dinkydoe
I need to show that $\int_{0}^{\infty}\frac{\log(x)}{(1+x^2)^2}=-\frac{\pi}{4}$

If we complexify the integrand and make a sort of pacman-contour $\mu$

$\mu_1 = t+ ic_0$ with $0\leq t\leq R$

$\mu_2 = Re^{i\pi t}$ with $t_0\leq t \leq 2\pi-t_0$

$\mu_3 = R-t-ic_0$ with $0\leq t\leq R$

$\mu_4 = re^{i\pi t}$ with $2\pi- t_0\leq t \leq t_0$

As $R\to \infty, r\to 0$ we let $t_0,c_0\to 0$
This contour contains the 2 second order singularities $p= \pm i$
I don't understand what your contour is; there may be several mistakes in your parameterization... Furthermore, what are $t_0, c_0$? Could you describe your contour in words, or with a sketch?

3. I don't understand what your contour is; there may be several mistakes in your parameterization... Furthermore, what are ? Could you describe your contour in words, or with a sketch?
I was afraid so. I believe it's called the key-hole contour.

When you see this picture, you understand why I called it a 'pacman' contour ;p

With $c_0,t_0$ I meant small numbers, actually dependant on R, but it's somewhat hard to define this contour it terms of R

I made a pretty sad mistake though. I forgot to multiply $\frac{-\pi}{4}$ with $2\pi i$.

4. Thanks for the precision. Note that $\log z$ is not defined for all $z\neq 0$; you must exclude a half-line, usually $\mathbb{R}_-$. Then the contour includes values where the function is not defined.

However, it seems to works if you symmetrize the picture, so that Pacman "looks to the left". This way the contour avoids the negative real axis $\mathbb{R}_-$.

We have $-x=e^{i\pi+\log x}=e^{-i\pi+\log x}$. When $z$ converges toward $-x$ from the upper-half plane, then $\log z$ converges toward $i\pi+\log x$ and when it converges from the lower-half plane, $\log z$ converges toward $-i\pi+\log x$. Thus, the integral along $\mu_1\cup \mu_3$ is almost equal to $\int_0^R \frac{2i\pi}{(1+x^2)^2}dx$ (real parts cancel). My computation shows that the residue at $i$ is $\frac{\pi}{8}+\frac{i}{4}$, and $\frac{\pi}{8}-\frac{i}{4}$ at $-i$. The conclusion is $\int_0^\infty \frac{dx}{(1+x^2)^2}=\frac{\pi}{4}$, which is cool, but not what we need...

Another contour will be necessary. For instance only the upper-half of the Pacman. Like a parliamentary hemicycle. Integral on the horizontal part on the right of 0 converges to your integral, and the integral on the horizontal part on the left converges to $\int_0^\infty \frac{\log x+i\pi}{(1+x^2)^2}dx$. By the way you'll recover the value of the above integral by looking at the imaginary part.

5. Hi. I don't think Cauchy's Integral Theorem can be used on this problem since the contour contains two poles. Also, just skip that "almost on the real axis thing" with the $\pm c$ and just go for "exactly on the real axis" in the limit. Then instead of using $\log(z)$, use $\log(z)^2$ and the same contour so that when you sum the branch-cut contours, the $\log(x)^2$ parts cancel and you're left with:

$-4\pi i\int_0^{\infty}\frac{\log(x)}{(1+x^2)^2}dx+4\pi^2 \int_0^{\infty}\frac{1}{(1+x^2)^2}dx=2\pi i\mathop\text{Res}\limits_{z=z_n}\left\{\frac{\log (z)^2}{(1+z^2)^2}\right\},\quad z_n=\pm i$

but keep in mind of the branch you're using: over the contour above the real axis you have $\log(x)$ and over the contour below the real axis you have $\log(x)+2\pi i$ and $\log(-i)=3\pi i/2$

I did it quick. May need a little fixin' up but I think it's close. That's my way of sayin', you work through it and in the process, learn how to do it .

6. That's my way of sayin', you work through it and in the process, learn how to do it .
Haha so true,