To evaluate this integral $$\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2} \text{d}x\,,$$

Let us consider the following function. $$f(z)=\frac{\ln^2 z}{(1+z^2)^2}=\frac{\ln^2 z}{(z+i)^2(z-i)^2}\,.$$ The integral over $C_R$ and $C_\epsilon$ vanish as $R$ approaches $\infty$ and $\epsilon$ approaches $0$. So we only need to evaluate the residues $$\oint\limits_C f(z)dz = 2\pi i \sum \text{Res}\big(f(z)\big)\,.$$

The poles are $\pm i$ and of second order. Here is our chosen contour.

Let us recall the residue of an $m$-th order of pole is $$\text{Res}[f(z);z_0]=\frac{1}{(m-1)!}\lim_{z\to z_0} \frac{\text{d}^{m-1}}{\text{d}z^{m-1}}\,(z-z_0)^m\,f(z)\,.$$

Consequently the residues are $$\text{Res}[f(z);i]=\lim_{z \to i}\Bigg(\frac{2\ln z}{z(z+i)^2}-\frac{2\ln^2 z}{(z+i)^3}\Bigg)=\frac{2\cdot \left(\frac{i\pi}{2}\right)}{-4i}-\frac{2\cdot (-\frac{\pi^2}{4})}{-8i}=-\frac{\pi}{4}+i\frac{\pi^2}{16}$$ As for the other pole $$\text{Res}[f(z);-i]=\lim_{z \to -i}\Bigg(\frac{2\ln z}{z(z-i)^2}-\frac{2\ln^2 z}{(z-i)^3}\Bigg)=\frac{2\cdot \left(\frac{i3\pi}{2}\right)}{4i}-\frac{2\cdot \left(-\frac{9\pi^2}{4}\right)}{8i}=\frac{3\pi}{4}-i\frac{9\pi^2}{16}\,.$$

Adding these two together and remembering the residue theorem, we have $$\oint\limits_C f(z)\,\text{d}z = 2\pi i \cdot \left(\frac{\pi}{2}-i\frac{\pi^2}{2}\right)=i\pi^2+\pi^3$$ On the other hand, for the integrals over $C_+$ and $C_-$, $z=x$ and $z=xe^{2\pi i}$ respectively. Hence, we can write $$\int\limits_\epsilon^R\frac{\ln^2 x}{(1+x^2)^2} \,\text{d}x + \int\limits_R^\epsilon \frac{(\ln x +2\pi i)^2}{(1+x^2)^2} \,\text{d}x=i\pi^2+\pi^3\,.$$ If we let $R \to \infty$ and $\epsilon \to 0$ and change the direction of the second integral, we get $$-4\pi i\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2}\,\text{d}x + 4\pi^2\int\limits_0^\infty\frac{\text{d}x}{(1+x^2)^2}=i\pi^2+\pi^3\,.$$

Comparing the real and the imaginary parts $$\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2}\,\text{d}x = -\frac{\pi}{4}$$ $$\int\limits_0^\infty\frac{\text{d}x}{(1+x^2)^2} = \frac{\pi}{4}$$

  • $\begingroup$ Aren't you ignoring the singularity of $\ln z$ at zero? $\endgroup$ – uniquesolution Jul 26 '18 at 7:18
  • $\begingroup$ @Skiver You have a mistake, but it's kind of hard to explain. See my edit in your question. $\endgroup$ – Batominovski Jul 26 '18 at 8:09
  • $\begingroup$ No I don't think ignored the singularity of $\ln z$ at zero. And @Batominovski your suggestion works. But as to why $e^{-i\pi/2}$ and $e^{i3\pi/2}$ are not the same in this case, I don't understand. I'm self teaching complex variables. Thank you for your answer. :) $\endgroup$ – Skiver Jul 26 '18 at 9:19
  • $\begingroup$ There is not a unique complex logarithm. From your contour, you want your $\ln(z)$ to be holomorphic on $\mathbb{C}\setminus \mathbb{R}_{\geq 0}$. Hence, the branch cut is the positive real line. In this case, $\ln(z)=\ln(r)+\text{i}\theta$ if $z=r\exp(\text{i}\theta)$ with $\theta\in[0,2\pi)$. If you use the usual $\ln(z)$ where the argument is between $-\pi$ and $\pi$, then you cannot use the Residue Theorem, at least not with your contour, since the contour crosses the branch cut. $\endgroup$ – Batominovski Jul 26 '18 at 9:37
  • $\begingroup$ Oh, I see now. Silly me. I understand my mistake completely now. Thanks a lot ;) $\endgroup$ – Skiver Jul 26 '18 at 9:40

Let $f(z):=\dfrac{\big(\ln(z)\big)^2}{\left(1+z^2\right)^2}$ for $z\in\mathbb{C}\setminus\mathbb{R}_{\leq 0}$ (i.e., the chosen branch cut of $\ln(z)$ is the negative real line). For a real number $\epsilon$ such that $0<\epsilon<1$, define $\gamma_\epsilon$ to be the contour $$\left[\epsilon,\frac{1}{\epsilon}\right]\cup A_\epsilon \cup \left[\frac{1}{\epsilon}\,\exp\big(\text{i}(\pi-\epsilon)\big),\epsilon\,\exp\big(\text{i}(\pi-\epsilon)\big)\right]\cup A'_\epsilon\,,$$ where $A_\epsilon$ is the arc in the upper half-plane (namely, the set of complex numbers with nonnegative imaginary parts) of the circle centered at $0$ with radius $\frac{1}{\epsilon}$ starting from $\frac{1}{\epsilon}$ to $\frac{1}{\epsilon}\,\exp\big(\text{i}(\pi-\epsilon)\big)$ (i.e., in the counterclockwise direction), and $A'_\epsilon$ is the arc in the upper half-plane of the circle centered at $0$ with radius $\epsilon$ starting from $\epsilon\,\exp\big(\text{i}(\pi-\epsilon)\big)$ to $\epsilon$ (i.e., in the clockwise direction).

We note that $$\lim_{\epsilon\to0^+}\,\oint_{\gamma_\epsilon}\,f(z)\,\text{d}z=2\pi\text{i}\,\text{Res}_{z=\text{i}}\big(f(z)\big)=2\pi\text{i}\,\left(-\frac{\pi}{4}+\frac{\pi^2\text{i}}{16}\right)=-\frac{\pi^3}{8}-\frac{\pi^2\text{i}}{2}\,.$$ Furthermore, $$\lim_{\epsilon\to0^+}\,\oint_{\gamma_\epsilon}\,f(z)\,\text{d}z=2\pi\text{i}\,\int_0^\infty\,\frac{\ln(x)}{\left(1+x^2\right)^2}\,\text{d}x+\,\int_0^\infty\,\frac{2\big(\ln(x)\big)^2-\pi^2}{\left(1+x^2\right)^2}\,\text{d}x\,.$$ Consequently, $$\int_0^\infty\,\frac{\ln(x)}{\left(1+x^2\right)^2}\,\text{d}x=\frac{1}{2\pi\text{i}}\left(-\frac{\pi^2\text{i}}{2}\right)=-\frac{\pi}{4}\,.$$

In fact, with the same contour, we can see that $$\int_0^\infty\,\frac{1}{\left(1+x^2\right)^2}\,\text{d}x=\pi\text{i}\,\text{Res}_{z=\text{i}}\left(\frac{1}{\left(1+z^2\right)^2}\right)=\pi\text{i}\,\left(-\frac{\text{i}}{4}\right)=\frac{\pi}{4}\,.$$ Therefore, $$\int_0^\infty\,\frac{\big(\ln(x)\big)^2}{\left(1+x^2\right)^2}\,\text{d}x=\frac{\pi^2}{2}\,\int_0^\infty\,\frac{1}{\left(1+x^2\right)^2}\,\text{d}x-\frac{\pi^3}{16}=\frac{\pi^3}{8}-\frac{\pi^3}{16}=\frac{\pi^3}{16}\,.$$


Although the OP is specifically asking to evaluate the integral using the residue theorem, I thought it might be instructive and of interest to present a way forward using real analysis only. It is to that end that we proceed.

Enforcing the substitution $x\mapsto \sqrt x$ reveals

$$\begin{align} \int_0^\infty \frac{\log(x)}{(1+x^2)^2}\,dx&=\frac14\int_0^\infty \frac{\log(x)}{\sqrt{x}(1+x)^2}\,dx\\\\ &=\frac14\left.\left(\frac{d}{dt}\int_0^\infty \frac{x^t}{(1+x)^2}\,dx\right)\right|_{t=-1/2}\\\\ &=\frac14\left.\left(\frac{d}{dt}B\left(t+1,1-t\right)\right)\right|_{t=-1/2}\\\\ &=\frac14\left.\left(\frac{d}{dt}\frac{\Gamma(1+t)\Gamma(1-t)}{\Gamma(2)}\right)\right|_{t=-1/2}\\\\ &=\frac14\left.\frac{d}{dt}\left(t\Gamma(t)\Gamma(1-t)\right)\right|_{t=-1/2}\\\\ &=\frac14\left.\frac{d}{dt}\left(\frac{t\pi}{\sin(\pi t)}\right)\right|_{t=-1/2}\\\\ &=\frac\pi4 \end{align}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.