7
$\begingroup$

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?


(additionally posted)

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 $\frac{\pi}{4a^3}(\log a -1)$.

$\endgroup$
7
$\begingroup$

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.

enter image description here

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.

$\endgroup$
  • $\begingroup$ Cool. All the tools for the answer but the answer is not given...+1 $\endgroup$ – imranfat Oct 15 '15 at 2:50
  • $\begingroup$ 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. $\endgroup$ – Crescendo Mar 29 '18 at 1:12
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$\dfrac{\pi(\log(a)-1)}{4a^3}$$

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

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.