# How to evaluate $\int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx$

Evaluate $$\int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx$$ $$(a>0)$$

How can I use contour appropriately?

What is the meaning of this integral?

I tried to solve this problem.

First, I take a branch $$\Omega=\mathbb C - \{z|\text{Re}(z)=0\; \text{and} \; \text{Im}(z)\le0\}$$

Then ${\log_\Omega z}=\log r +i\theta (-\frac{\pi}{2}\lt\theta\lt\frac{3\pi}{2})$

Now, $\frac{\log z}{(z^2+a^2)^2}$ is holomorphic in $\Omega - \{ai\}$ with double poles at $ai$.

Now I'll take the contour which forms an indented semicircle.

For any $0\lt\epsilon\lt{a}$, where $\max (1,a)\lt R$, $\Gamma_{R,\epsilon}\subseteq\Omega - \{ai\}$ and in $\Omega$, $i=e^{i\pi/2}$.

Now using the residue formula, $$2\pi{i}\operatorname*{Res}_{z=ai}\frac{\log_\Omega{z}}{(z^2+a^2)^2}=2\pi{i}\operatorname*{lim}_{z\to ai}\frac{d}{dz}(z-ai)^2\frac{\log_\Omega{z}}{(z^2+a^2)^2}=\frac{\pi}{2a^3}(\log_\Omega{ai}-1)$$

Now, the last part, take $i=e^{i\pi/2}$, then is equal to $\frac{\pi}{2a^3}(\log{a}-1+i\pi/2)$

So, I can split integrals by four parts,

$$\int_{\epsilon}^R dz + \int_{\Gamma_R} dz + \int_{-R}^{-\epsilon} dz + \int_{\Gamma_\epsilon} dz$$

First, evaluate the second part,

$$\left|\int_{\Gamma_R} dz\right|\le\int_0^{\pi}\left|\frac{\log_\Omega{Re^{i\theta}}}{(R^2e^{2i\theta}+a^2)^2}iRe^{i\theta}\right|d\theta$$

Note that

$$\left|\log_\Omega{Re^{i\theta}}\right|=\left|\log R+i\theta\right|\le\left|\log R\right|+|\theta|$$ $$\left|R^2e^{2i\theta}+a^2\right|\ge R^2-a^2\quad (R\gt a)$$

Then, 2nd part $\le\frac{R(\pi R+\frac{\pi^2}{2})}{(R^2+a^2)^2}\to 0\; \text{as} \; R \to \infty\quad \left|\log R\right|\lt R\;\text{where}\;(R\gt 1)$

So, 4th part similarly, goes to $\;0$.

Then 3rd part, substitute for $\;t=-z$,

$$\int_\epsilon^{R}\frac{\log t}{(t^2+a^2)^2}dt + i\pi\int_\epsilon^{R}\frac{dt}{(t^2+a^2)^2}$$

And $\;i\pi\lim\limits_{{\epsilon \to 0},\;{R\to\infty}}\int_\epsilon^{R}\frac{dt}{(t^2+a^2)^2}=\frac{\pi}{4a^3}$

With tedious calculations, I got $\frac{\pi}{4a^3}(\log a -1)$.

One thing you can do when confronted with integrals of the form

$$\int_0^{\infty} dx \, f(x) \log{x}$$

is to consider a contour integral of the form

$$\oint_C dz \, f(z) \, \log^2{z}$$

where $C$ is a keyhole contour about the positive real axis, as pictured below.

To evaluate the contour integral, we parametrize about each piece of the contour. There are four such pieces: a large arc of radius $R$, a small arc of radius $\epsilon$, and lines above and below the positive real axis.

This contour allows us to derive the integral of interest by exploiting the multivalued behavior of the log at a branch point. In this case, we define the argument of the complex numbers above the positive real axis to be zero and below to be $2 \pi$. Thus, above the real axis $z=x$ while below $z=x e^{i 2 \pi}$. This difference is crucial when taking logs.

I will let the reader perform the analysis as the outer radius $R \to \infty$ and inner radius $\epsilon \to 0$; the contour integral is then equal to

$$\int_0^{\infty} dx \, f(x) \log^2{x} - \int_0^{\infty} dx \, f(x) (\log{x}+i 2 \pi)^2 = -i 4 \pi \int_0^{\infty} dx \, f(x) \log{x} + 4 \pi^2 \int_0^{\infty} dx \, f(x)$$

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z_k$ of $f$ in the complex plane outside of the origin. Thus,

$$\int_0^{\infty} dx \, f(x) \log{x} = -i \pi \int_0^{\infty} dx \, f(x) - \frac12 \sum_k \operatorname*{Res}_{z=z_k} [f(z) \log^2{z}]$$

In the OP's case,

$$f(z) = \frac1{(z^2+a^2)^2}$$

so the poles are of order two and the residues must be computed accordingly. The OP should be able to derive

$$\operatorname*{Res}_{z=\pm i a} \frac{\log^2{z}}{(z^2+a^2)^2} = \left[\frac{d}{dz} \frac{\log^2{z}}{(z\pm i a)^2} \right ]_{z=\pm i a}$$

Note also that the poles must have their arguments between $[0,2 \pi]$ for the residue calculation to come out correctly. In this case, we may say that the poles are at $z_{\pm}=\pm i a$, but it is important to note that $z_+ = a e^{i \pi/2}$ and $z_-=a e^{i 3 \pi/2}$.

Further, it should not escape notice that the final result is in terms of an integral over the function $f$ without the log term. You should be able to see that the integral may be evaluated in exactly the same way as the original integral by introducing a log and integrating over the keyhole contour $C$. The result is

$$\int_0^{\infty} dx \, f(x) = -\sum_k \operatorname*{Res}_{z=z_k} [f(z) \log{z}]$$

At this point the OP has everything needed to carry out the computation.

• Cool. All the tools for the answer but the answer is not given...+1 – imranfat Oct 15 '15 at 2:50
• Sorry if this is a dumb question, but why do we have the $+2\pi i$ in the natural log? Shouldn't it be $+\pi i$ because the of a $\pi$ radian flip? Again, sorry if this is a fool's error; I'm just starting to learn contour integration. – Crescendo Mar 29 '18 at 1:12

I thought that it might be instructive to add to the answer posted by @RonGordon. We note that the integral of interest $I_1(a^2)$ can be written

$$I_1(a^2)=\int_0^\infty \frac{\log^2 x}{(x^2+a^2)^2}\,dx=-\frac{dI_2(a^2)}{d(a^2)}$$

where

$$I_2(a^2)=\int_0^\infty\frac{\log^2x}{x^2+a^2}\,dx$$

Now, we can evaluate the integral $J(a^2)$

$$J(a^2)=\oint_C \frac{\log^2z}{z^2+a^2}\,dz$$

where $C$ is the key-hole contour defined in the aforementioned post. There, we have

\begin{align} J(a^2)&=-4\pi i\,I_2(a^2)+4\pi^2\int_0^\infty \frac{1}{x^2+a^2}\,dx \\\\ &=-4\pi i\, I_2(a^2)+\frac{2\pi^3}{a}\\\\ &=2\pi i \left(\text{Res}\left(\frac{\log^2 z}{z^2+a^2},ia\right)+\left(\text{Res}\left(\frac{\log^2 z}{z^2+a^2},-ia\right)\right)\right) \end{align}

Finally, after calculating the residues, and simplifying, we obtain the integral $I_2(a^2)$ whereupon differentiating with respect to $a^2$ recovers the integral of interest $I_1(a^2)$. And we are done.

Let $x=at$. We then have \begin{align} I & = \int_0^{\infty} \dfrac{\log(x)}{(x^2+a^2)^2}dx = \dfrac1{a^3}\cdot\int_0^{\infty} \dfrac{\log(at)}{(t^2+1)^2}dt = \dfrac1{a^3}\left(\int_0^{\infty} \dfrac{\log(a)}{(t^2+1)^2}dt + \int_0^{\infty} \dfrac{\log(t)}{(t^2+1)^2}dt\right)\\ & = \dfrac{J+K}{a^3} \end{align} where $J=\displaystyle\int_0^{\infty} \dfrac{\log(a)}{(t^2+1)^2}dt$ and $K = \displaystyle\int_0^{\infty} \dfrac{\log(t)}{(t^2+1)^2}dt$.

$$J = \int_0^{\pi/2}\dfrac{\log(a)}{(\tan^2(y)+1)^2}\sec^2(y)dy = \log(a)\int_0^{\pi/2}\cos^2(y)dy = \dfrac{\pi\log(a)}4$$

\begin{align} K & = \displaystyle\int_0^1 \dfrac{\log(t)}{(t^2+1)^2}dt + \displaystyle\int_1^{\infty} \dfrac{\log(t)}{(t^2+1)^2}dt\\ & = \displaystyle\int_0^1 \dfrac{\log(t)}{(t^2+1)^2}dt + \displaystyle\int_1^0 \dfrac{\log(1/t)}{(1/t^2+1)^2}\dfrac{-dt}{t^2}\\ & = \int_0^1 \dfrac{(1-t^2)}{(1+t^2)^2}\cdot\log(t)dt\\ & = \sum_{k=0}^{\infty}(-1)^k (2k+1) \int_0^1 t^{2k}\log(t)dt\\ & = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac1{2k+1}\\ & = -1 + \dfrac13 - \dfrac15 + \dfrac17 \mp \cdots = -\dfrac{\pi}4 \end{align} Hence, the integral is $$\dfrac{\pi(\log(a)-1)}{4a^3}$$

• Thanks for answering, but my result for OP was $\frac{\pi}{4a^3}(\log{a}-1)$. Is it wrong??? – Sh7 Dec 16 '15 at 10:51
• Oh, you've edited. Does this approach come from Fourier series? – Sh7 Dec 16 '15 at 10:55
• @Sh7 This approach is just based on series expansion. – Adhvaitha Dec 16 '15 at 10:56