# Proving that $\int_{0}^{\infty} \frac{\log(x) }{(1+x^2)^2}dx =- \frac{\pi}{4}$ using residues.

I need to prove that $$\displaystyle\int_{0}^{\infty} \frac{\log(x) }{(1+x^2)^2}\,dx = -\frac{\pi}{4}$$ using the Residue theorem. I'm trying to solve using the function $$f(z)=\dfrac{\log(z)}{(1+z^2)^2}$$ (branch of $$\log(z)$$ with argument between $$-\pi/2$$ and $$3\pi/2$$) and integrate over the curve $$\gamma=[-R,-r] + \gamma_r +[r,R] + \gamma_R$$, where $$\gamma_r(t)=re^{-it}, t\in (0,\pi)$$ and $$\gamma_R(t)=Re^{it}, t\in (0,\pi)$$. And I've already computed that $$\operatorname{Res}(f,i)= -\pi/2 -\pi^2/4$$. Also I think that limit when $$R$$ goes to $$\infty$$, the integral over $$\gamma_R$$ goes to $$0$$.

My big problem is the integral over $$\gamma_r$$.

• For whoever is interested, here's an elementary way to solve it. – Zacky Dec 14 '19 at 15:04

We work with

$$f(z) = \frac{\mathrm{log}(z)}{(z+i)^2(z-i)^2}$$

where $$\log(z)$$ is the principal branch with argument in $$(-\pi,\pi].$$ We use a semicircular contour indented at the origin wih radius $$R$$ in the upper half plane. Let $$\Gamma_0$$ be the segment on the positive real axis up to $$R$$, $$\Gamma_1$$ the big semicircle of radius $$R$$, $$\Gamma_2$$ the segment on the real axis coming in from $$-R$$ and finally $$\Gamma_3$$ the small semicircle of radius $$\epsilon$$ enclosing the origin. We then have

$$\left(\int_{\Gamma_0} + \int_{\Gamma_1}+ \int_{\Gamma_2}+ \int_{\Gamma_3}\right) f(z) \; dz = 2\pi i \times \mathrm{Res}_{z=i} f(z).$$

We have

$$\mathrm{Res}_{z=i} f(z) = \left. \left(\frac{\mathrm{log}(z)}{(z+i)^2}\right)'\right|_{z=i} = \left. \left(\frac{1}{z} \frac{1}{(z+i)^2} - 2 \frac{\mathrm{log}(z)}{(z+i)^3}\right)\right|_{z=i} \\ = \frac{1}{4i^3} - \frac{\pi i}{8i^3} = \frac{\pi}{8} + \frac{i}{4}$$

Observe that in the limit

$$\int_{\Gamma_0} f(z) \; dz = \int_0^\infty \frac{\log(x)}{(x^2+1)^2} \; dx = J.$$

For $$\Gamma_1$$ we have by ML estimate $$\lim_{R\to\infty} \pi R \times \sqrt{\log^2 R+ \pi^2} / (R-1)^4 = 0,$$ so this vanishes.

Furthermore for $$\Gamma_2$$

$$\int_{\Gamma_2} f(z) \; dz = \int_{-\infty}^0 \frac{\log(|x|)+ \pi i}{(x^2+1)^2} \; dx = \int_{-\infty}^0 \frac{\log(|x|)}{(x^2+1)^2} \; dx + \pi i \int_{-\infty}^0 \frac{1}{(x^2+1)^2} \; dx \\ = \int_0^{\infty} \frac{\log(|x|)}{(x^2+1)^2} \; dx + \pi i \int_0^{\infty} \frac{1}{(x^2+1)^2} \; dx = J + \pi i K.$$

For $$\Gamma_3$$ we again use ML and get $$\lim_{\epsilon\to 0} \pi \epsilon \times \sqrt{\log^2 \epsilon +\pi^2} = 0$$, so this too vanishes. We have shown that in the limit

$$2 J + \pi i K = 2\pi i\times \left(\frac{\pi}{8} + \frac{i}{4}\right) = - \frac{\pi}{2} + \frac{\pi^2}{4} i.$$

We know that $$J$$ and $$K$$ are real numbers, hence,

$$\bbox[5px,border:2px solid #00A000]{ \int_0^\infty \frac{\log(x)}{(x^2+1)^2} \; dx = - \frac{\pi}{4} \quad\text{and}\quad \int_0^\infty \frac{1}{(x^2+1)^2} \; dx = \frac{\pi}{4}.}$$

Your parametrization for $$\gamma_r(t)$$ doesn't do what you think it does. It goes below the singularity (thereby crossing the branch cut it shouldn't) instead of going backwards. I'll use the correct version that doesn't include the singularity at $$0$$ by letting

$$\gamma_r(t) = re^{it} \hspace{20 pt} t\in (0,\pi)$$

but just have the integral go backwards. Using the branch that you have, the integral over $$\gamma_r$$ can be shown to be bounded by the following:

$$\Biggr| \int_\pi^0 \frac{\log r + it}{(1+r^2e^{i2t})^2}\left(ire^{it}dt\right)\Biggr| \leq (r\log r + \pi r)\int_0^\pi \frac{dt}{|1+r^2e^{i2t}|^2} \leq \frac{\pi r\log r + \pi^2 r}{(1-r^2)^2}$$

by triangle inequality. The limit goes to $$0$$ because $$r$$ dominates $$\log r$$. Thus the integral over $$\gamma_r$$ vanishes.

• Ok, thank you! But how I can finish the excersice now? Because now I can't compute the integral over $[-R,-r]$. – Peter Languilla Dec 12 '19 at 23:04
• @PeterLanguilla Do the whole problem instead with the function $\frac{1}{2}\frac{\log(x^2)}{(1+x^2)^2}$. It makes the function even and is equivalent to the original. – Ninad Munshi Dec 12 '19 at 23:25