3
$\begingroup$

$$\int_0^\infty\frac{\log x}{(1+x^2)^2}\,dx$$

I have tried to evaluate this by trying I.B.P. on various different products, and have also tried the substitution $x\mapsto1/x$. None of these work. I have put this into an online integration tool and it gives a solution using some function called "Li", but I have never been taught this, so do not want to use it. Is there some way of evaluating this definite integral, perhaps using complex analysis and contour integrals?


I have been able to evaluate similar integrals, such as $\int_0^\infty\frac{\log x}{1+x^2}\,dx=0$ and $\int_0^\infty\frac{1}{(1+x^2)^2}\,dx=\pi/4$, but it is this one which is proving to be a struggle.

$\endgroup$
  • $\begingroup$ i think it must be $$-\frac{\pi}{4}$$ $\endgroup$ – Dr. Sonnhard Graubner May 21 '17 at 11:13
  • $\begingroup$ @Dr.SonnhardGraubner Yes, I think wolfram alpha gave this answer, but how did you obtain that? $\endgroup$ – John Doe May 21 '17 at 11:17
  • $\begingroup$ I also have tried parts with $u=\log x$, $v'=\frac1{(1+x^2)^2}$, but I believe this leads to a term $$\int_0^\infty\frac{\arctan x}{x}\,dx$$which again looks troublesome. $\endgroup$ – John Doe May 21 '17 at 11:20
  • $\begingroup$ integrate $\log^2(x)/(1+x^2)^2$ over (for example) a keyhole in the complex plane. search the site properly this was done hundreds of times before here on this site (Especially user @Mark Viola posted something similar these days) $\endgroup$ – tired May 21 '17 at 11:28
  • $\begingroup$ btw $\int dx \arctan(x)/x$ is a very good candiate for differentiation under the integral sign.... $\endgroup$ – tired May 21 '17 at 11:30
3
$\begingroup$

Take the keyhole contour of $f(z)=\left[\frac{\ln(z)}{1+z^2}\right]^2$ with $\gamma$ the radius of the inner circle and $\Gamma$ the radius of the outer circle to get

$$\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_1}f(z)\ dz=\int_0^{+\infty}\left[\frac{\ln(x)}{1+x^2}\right]^2\ dx$$

$$\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_3}f(z)\ dz=-\int_0^{+\infty}\left[\frac{\ln(x)+2\pi i}{1+x^2}\right]^2\ dx$$

$$\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_2}f(z)\ dz=\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_4}f(z)\ dz=0$$

And thus, we can see that

$$\begin{align}\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\oint_Cf(z)\ dz&=\int_0^{+\infty}\left[\frac{\ln(x)}{1+x^2}\right]^2\ dx-\int_0^{+\infty}\left[\frac{\ln(x)+2\pi i}{1+x^2}\right]^2\ dx\\&=\int_0^{+\infty}\frac{[\ln(x)]^2-[\ln(x)+2\pi i]^2}{(1+x^2)^2}\ dx\\&=\int_0^{+\infty}\frac{[\ln(x)]^2-[\ln(x)]^2-4\pi i\ln(x)+4\pi^2}{(1+x^2)^2}\ dx\\&=\int_0^{+\infty}\frac{-4\pi i\ln(x)+4\pi^2}{(1+x^2)^2}\ dx\\&=2\pi i\left(\operatorname{Res}_{z=i}(f(z))+\operatorname{Res}_{z=-i}(f(z))\right)\\&=\pi^2i\end{align}$$

By taking imaginary parts, we deduce that

$$\int_0^{+\infty}\frac{\ln(x)}{(1+x^2)^2}\ dx=-\frac\pi4$$

$\endgroup$
  • $\begingroup$ In the third line of evaluating the integral, should it not be this: $$\int_0^{+\infty}\frac{[\ln(x)]^2-[\ln(x)]^2-4\pi i\ln(x)+4\pi^2}{(1+x^2)^2}\ dx$$? And if so, does this change the answer, or was this a typo (or am I missing something)? $\endgroup$ – John Doe May 21 '17 at 11:58
  • $\begingroup$ @JohnDoe Ah yes, missed a negative, though it does not change the end result. $\endgroup$ – Simply Beautiful Art May 21 '17 at 12:02
5
$\begingroup$

Try the following integration by parts:

$$\begin{align} \int_{0}^{\infty}\frac{\ln{\left(x\right)}}{\left(1+x^{2}\right)^{2}}\,\mathrm{d}x &=\int_{0}^{\infty}\frac{2x}{\left(1+x^{2}\right)^{2}}\cdot\frac{\ln{\left(x\right)}}{2x}\,\mathrm{d}x\\ &=\left[\left(1-\frac{1}{1+x^{2}}\right)\frac{\ln{\left(x\right)}}{2x}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(1-\frac{1}{1+x^{2}}\right)\frac{1-\ln{\left(x\right)}}{2x^{2}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\ &=0-0+\int_{0}^{\infty}\frac{x^{2}}{1+x^{2}}\cdot\frac{\ln{\left(x\right)}-1}{2x^{2}}\,\mathrm{d}x\\ &=\frac12\int_{0}^{\infty}\frac{\ln{\left(x\right)}-1}{1+x^{2}}\,\mathrm{d}x.\\ \end{align}$$

Incidentally, the same method can be used to build a reduction formula for integrals of the form $\int_{0}^{\infty}\frac{\ln{\left(x\right)}}{\left(1+x^{2}\right)^{n}}\,\mathrm{d}x$.

$\endgroup$
  • $\begingroup$ That's a nice result! I did encounter the expression $\int\frac{2x}{(1+x^2)^2}$ when I was solving this, but I just integrated it to $C-\frac{1}{(1+x^2)^2}$, but then I automatically set the constant to be $0$, since I figured that would be easier. This is very nice $\endgroup$ – John Doe May 21 '17 at 12:38
  • $\begingroup$ @JohnDoe It usually is easier to set the constant to $0$, but for many improper integrals there will only be one value for $C$ where the integral converges. $\endgroup$ – David H May 21 '17 at 14:21
3
$\begingroup$

Change variable to $y = \frac1x$, we have

$$I = \int_0^\infty \frac{\log x}{(1+x^2)^2} dx = \int_\infty^0 \frac{-\log y}{(1+y^{-2})^2}\left(-\frac{dy}{y^2}\right) = - \int_0^\infty \frac{y^2\log y}{(1+y^2)^2}dy$$ This leads to $$\begin{align} I &= -\frac12\int_0^\infty \log(x) \frac{x^2-1}{(x^2+1)^2}dx = -\frac12\int_0^\infty\log(x) \frac{x-x^{-1}}{(x+x^{-1})^2}\frac{dx}{x}\\ &= -\frac12\int_0^\infty\log(x)\frac{x-x^{-1}}{(x+x^{-1})^2}\frac{d(x+x^{-1})}{x-x^{-1}} = -\frac12\int_0^\infty \log(x) \frac{d(x+x^{-1})}{(x+x^{-1})^2}\\ &= \frac12\int_0^\infty \log(x) \left(\frac{1}{x+x^{-1}}\right)' dx = \frac12\int_0^\infty \left[ \left(\frac{\log x}{x+x^{-1}}\right)' - \frac{(\log x)'}{x+x^{-1}} \right] dx\\ &= \left[ \frac{\log x}{x+x^{-1}}\right]_0^\infty -\frac12\int_0^\infty \frac{dx}{1+x^2}\\ &= -\frac{\pi}{4} \end{align} $$

$\endgroup$
  • $\begingroup$ You missed a small thing at the end with your $?\to\infty$ $\endgroup$ – Simply Beautiful Art May 21 '17 at 11:42
  • $\begingroup$ Sorry, could you elaborate on the integration by parts at the end? You let $u=\frac1{x+x^{-1}}$, then what? The $x$ doesn't rearrange nicely to put the $\log x$ into the form $\log f(u)$ $\endgroup$ – John Doe May 21 '17 at 11:52
  • $\begingroup$ @John Doe, there is no change of variable at that part. I rewrite the last part of answer and I hope it is clearer. $\endgroup$ – achille hui May 21 '17 at 14:19
  • $\begingroup$ Ohhh, ok I see, I think it was just unfamiliar with the notation, in that you wrote $d(x+x^{-1})$, but it makes sense now. Thanks :) $\endgroup$ – John Doe May 21 '17 at 19:46
  • $\begingroup$ @JohnDoe When you see an expression like $\int fdg$, it doesn't always mean change variable to $g$. It can mean $\int f(x(t))g'(x(t))x'(t)dt$ for any parametrization of the integral. It can also mean it is a Riemann–Stieltjes integral. $\endgroup$ – achille hui May 21 '17 at 19:57
2
$\begingroup$

The substitution $x= \frac{1}{y}$ on the interval $(1,\infty)$ followed by the substitution $y=\tan(t)$ and integration by parts yield \begin{gather*} \int_{0}^{\infty}\dfrac{\ln(x)}{(1+x^2)^2}\, dx = \int_{0}^{1}\dfrac{\ln(x)}{(1+x^2)^2}\, dx - \int_{0}^{1}\dfrac{y^2\ln(y)}{(1+y^2)^2}\, dy = \int_{0}^{1}\dfrac{(1-y^2)\ln(y)}{(1+y^2)^2}\, dy = \\[2ex] \int_{0}^{\pi/4}\cos(2t)\ln(\tan(t))\, dt = \left[\dfrac{1}{2}\sin(2t)\ln(\tan(t))\right]_{0}^{\pi/4}- \int_{0}^{\pi/4}\underbrace{\dfrac{\sin(2t)}{2\tan(t)\cos^2(t)}}_{=1}\, dt = -\dfrac{\pi}{4}. \end{gather*}

$\endgroup$
2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x & = \left.-\, \partiald{}{a}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + a}\,\dd x \,\right\vert_{\ a=\ 1} \\[5mm] & = -\,\partiald{}{a} \bracks{a^{-1/2}\int_{0}^{\infty}{\ln\pars{a^{1/2}x} \over x^{2} + 1}\,\dd x} _{\ a=\ 1} \\[5mm] & = -\,\partiald{}{a} \bracks{{1 \over 2}\,a^{-1/2}\ln\pars{a}\int_{0}^{\infty}{\dd x \over x^{2} + 1} + a^{-1/2}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x}\label{1}\tag{1} _{\ a=\ 1} \\[5mm] & = -\,{\pi \over 4}\,\partiald{}{a} \bracks{a^{-1/2}\ln\pars{a}}_{\ a\ =\ 1} \\[5mm] & = -\,{\pi \over 4} \bracks{-\,{1 \over 2}\,a^{-3/2}\ln\pars{a} + a^{-3/2}}_{\ a\ =\ 1} = \bbx{-\,{\pi \over 4}} \end{align}

The last integral in \eqref{1} vanishes outIt just changes its sign under $\ds{x \to 1/x}$.

$\endgroup$

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.