I am trying to verify that $$ \int_0^\infty \frac{\log x}{(x^2+1)^2}dx = -\frac{\pi}{4}$$
using the above contour. I would appreciate detailed steps as I want to advance to problems that have roots on the positive real axis.
edit: Metamorphy recommended a wiki page that discusses this exact problem with loose steps, and I want to expand the problem for others.
To calculate this integral, one uses the function $$f(z) = \left( \frac{\log z}{z^2+1} \right)^2$$ and the branch of the logarithm corresponding to −π < arg z ≤ π (here I think $0$ < arg z < $2\pi$ is the same logic)
We will calculate the integral of f(z) along the keyhole contour shown above. As it turns out this integral is a multiple of the initial integral that we wish to calculate and by the Cauchy residue theorem we have $$\left(\int_{\gamma_1}+\int_{\gamma_2}+\int_{\Gamma_1}+\int_{\Gamma_2} \right)f(z)dz = 2\pi i\left(\sum Res \right) = 2\pi i\left( Res\left[\left( \frac{\log z}{z^2+1} \right)^2,i\right]+Res\left[\left( \frac{\log z}{z^2+1} \right)^2,-i\right] \right)$$ Let $\Gamma_2$ be the segment of radius R of the large circle, and $\Gamma_1$ be the segment of radius r of the small one. We will denote the upper line by $\gamma_1$, and the lower line by $\gamma_2$. As before we take the limit when R → ∞ and r → 0. The contributions from the two circles vanish. We can easily show $$Res\left[\left( \frac{\log z}{z^2+1} \right)^2,i\right] = \lim_{z\to i}\left[ \frac{d}{dz}\left( \frac{(z-i)^2}{(z-i)^2(z+i)^2}\log^2z \right) \right]$$ $$= \lim_{z\to i}\left[ \frac{d}{dz}\left( \frac{\log^2z}{(z+i)^2} \right) \right]$$ $$= \lim_{z\to i}\left[ \frac{2\log z}{z(z+i)^2}-\frac{2\log^2 z}{(z+i)^3} \right]$$ $$= \frac{2\log i}{i(2i)^2}-\frac{2\log^2 i}{(2i)^3} $$ $$= \frac{2\frac{\pi}{2}}{(-4)}+\frac{2\frac{i\pi^2}{4}}{8} $$ $$= -\frac{\pi}{4}+\frac{i\pi^2}{16}$$ Similarly, we can show $$Res\left[\left( \frac{\log z}{z^2+1} \right)^2,-i\right] =-\frac{\pi}{4}-\frac{i\pi^2}{16}$$ where here you might be tempted to write $\log z = \log (-i) = \log (exp(i\frac{3\pi}{2})) = i\frac{3\pi}{2}$ but it is better to write $\log z = \log (-i) = \log (exp(-i\frac{\pi}{2})) = -i\frac{\pi}{2}$, which will give you the above equation. Using $z = exp(i\frac{3\pi}{2})$ we get $$Res\left[\left( \frac{\log z}{z^2+1} \right)^2,-i\right] =\frac{3\pi}{4}-\frac{9i\pi^2}{16}$$ and it's obvious which residue for $z=-i$ we should choose, though you would think both should work. We will proceed with the odd case; the other case can be followed in the wiki page.
We get $$\left(\int_{\gamma_1}+\int_{\gamma_2}+\int_{\Gamma_1}+\int_{\Gamma_2} \right)f(z)dz =2\pi i ( -\frac{\pi}{4}+\frac{i\pi^2}{16}+\frac{3\pi}{4}-\frac{9i\pi^2}{16})$$ $$=2\pi i(\frac{\pi}{2}-\frac{i\pi^2}{2})$$ $$=i\pi^2+\pi^3$$
In order to compute the contributions of $\gamma_1$ and $\gamma_2$ we set z = x + iε on $\gamma_1$ and z = x − iε on $\gamma_2$ (notice the difference in z values since our contour is different than in the example), with 0 < x < ∞: $$i\pi^2+\pi^3 = \left(\int_{\gamma_1}+\int_{\gamma_2}+\int_{\Gamma_1}+\int_{\Gamma_2} \right)f(z)dz$$ $$= \left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)dz$$ $$ = \int_0^\infty \left( \frac{\log (x+i\epsilon)}{1+(x+i\epsilon)^2} \right)^2dx +\int_{\infty}^0 \left( \frac{\log (x-i\epsilon)}{1+(x-i\epsilon)^2} \right)^2dx $$ $$ = \int_0^\infty \left( \frac{\log (x+i\epsilon)}{1+(x+i\epsilon)^2} \right)^2 dx-\int_{0}^\infty \left( \frac{\log (x-i\epsilon)}{1+(x-i\epsilon)^2} \right)^2dx $$ $$ = \int_0^\infty \left( \frac{\log (x)}{1+x^2} \right)^2 dx -\int_{0}^\infty \left( \frac{\log (x)+2\pi i}{1+x^2} \right)^2dx$$ $$=\int_0^\infty \frac{4\pi^2}{(1+x^2)^2}dx - 4\pi i\int_0^\infty \frac{\log x}{(1+x^2)^2}dx$$ $$=4\pi^2(\frac{\pi}{4}) - 4\pi i\int_0^\infty \frac{\log x}{(1+x^2)^2}dx$$ $$=\pi^3 - 4\pi i I$$ so $$i\pi^2 + \pi^3 = \pi^3 -4\pi i I$$ $$i\pi^2 = -4\pi i I$$ $$I = -\frac{\pi}{4}$$
Note: With the chosen contour, it appears you would not be able to use $\log -i = -i\frac{\pi}{2}$. This may be investigated in the future.