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Question: Show that$$\int\limits_0^{\infty}\mathrm dx\,\frac {\arctan x\log(1+x^2)}{x(1+x^2)}=\frac {\pi}2\log^22.$$

I can't tell if I'm being an idiot, or if this is a lot more difficult than it looks. First, I tried integration by parts using the fact that $$\frac 1{x(1+x^2)}=\frac 1x-\frac x{1+x^2}.$$

But quickly I gave up as I wasn't sure what to do with the result. I then decided to make the substitution $t=\arctan x$ to get rid of the $1+x^2$ term in the denominator. Therefore$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\mathrm dt\,t\cot t\log\sec^2t=-2\int\limits_0^{\pi/2}\mathrm dt\, t\cot t \log\cos t.\end{align*}$$

However, I'm not exactly sure what to do after this. Should I use integration by parts? Differentiation under the integral sign? I'm having trouble getting started with this integral. Any ideas?

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    $\begingroup$ If I may ask, but why the close vote? I doubt my question is off topic, and furthermore, I included my two main attempts. Other attempts would be expanding $\log$ in its taylor expansion series and a similar thing with the $\arctan$ function. $\endgroup$ – Frank Jul 14 '18 at 0:49
  • $\begingroup$ After an hour I’m not optimistic for a closed form. I think you need something fancy here. $\endgroup$ – Randall Jul 14 '18 at 1:28
  • $\begingroup$ Side note: The integral occurs in this question as well. $\endgroup$ – ComplexYetTrivial Jul 14 '18 at 21:07
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We can use differentiation under the integral sign and a trick to evaluate this. First define $$ I(a,b) = \int_0^{\infty} \frac{\arctan{ax}}{x} \frac{\log{(1+b^2 x^2)}}{1+x^2} \, dx , $$ so $I(a,0)=I(0,b)=0$ and $I(1,1)$ is what we want. Differentiating one with respect to $a$ and once wrt $b$ gives $$ \partial_a\partial_b I = \int_0^{\infty} \frac{2bx^2 \, dx}{(1+x^2)(1+a^2x^2)(1+b^2x^2)}, $$ which can be done by using partial fractions and the arctangent integral a few times. When the dust settles, $$ \partial_a\partial_b I = \frac{b\pi}{(1+a)(1+b)(a+b)}, $$ and thus $$ I(1,1) = \int_0^1 \int_0^1 \frac{b\pi}{(1+a)(1+b)(a+b)} \, da \, db $$ But we can swap $a$ and $b$ and will get the same result for this integral by the symmetry of the region of integration, so we also have $$ I(1,1) = \int_0^1 \int_0^1 \frac{a\pi}{(1+a)(1+b)(a+b)} \, da \, db. $$ Adding gives $$ I(1,1) = \frac{\pi}{2}\int_0^1 \int_0^1 \frac{1}{(1+a)(1+b)} \, da \, db, $$ but this splits into a product of two copies of $\int_0^1 dy/(1+y) = \log{2}$, so $$ I(1,1) = \frac{\pi}{2}(\log{2})^2 $$ as desired.

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    $\begingroup$ Nice answer! +1 $\endgroup$ – Tom Himler Jul 14 '18 at 2:02
  • $\begingroup$ Oh lord, I’ve completely forgotten about Feynman’s Trick! Why didn’t I think of that! Although I’m not very familiar with differentiation with two variables. How did you know to use two variables instead of say, just setting the $a$ there? To me, having $\arctan ax$ and $\log(1+b^2x^2)$ is an overcomplication. $\endgroup$ – Frank Jul 14 '18 at 3:21
  • $\begingroup$ I have forgotten: when is differentiation under the integral allowed? $\endgroup$ – Randall Jul 14 '18 at 13:22
  • $\begingroup$ @Frank I did try only using one derivative, but it resulted in a rather nasty log that would have required a polylogarithm to integrate. Indeed, this trick only works because $a=b$ in the limit we care about, so we can apply a symmetry to make the integral much nicer; the general case will involve polylogarithms. $\endgroup$ – Chappers Jul 14 '18 at 20:10
  • $\begingroup$ @Randall $\partial_a f(x,a)$ integrable and $\frac{1}{h}\int | f(x,a+h)-f(x,a) - h \partial_a f(x) | dx \to 0$ should be sufficient in general. Dominated Convergence will cover most explicit cases, including this one since there's always a $1/(1+x^2)$ involved. But everything's positive here as well, so one can also view it as an application of Tonelli to permute the order of integration. $\endgroup$ – Chappers Jul 14 '18 at 20:14
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Define $$ f(a,b) = \int \limits_0^\infty \frac{\arctan(a x) \ln (1+ b^2 x^2)}{x (1+x^2)} \, \mathrm{d} x $$ for $0 \leq a , b \leq 1$ . Then $f(1,1) = \mathfrak{I}$ and $f(0,b) = f(a,0) = 0 $ . For $0< a,b<1$ we can differentiate under the integral sign to find $$ \partial_a \partial_b f(a,b) = 2 b \int \limits_0^\infty \frac{x^2}{(1+a^2 x^2)(1+b^2 x^2)(1+x^2)} \, \mathrm{d} t = \frac{\pi b}{(1+a)(1+b)(a+b)} \, . $$ The integral can be evaluated using the residue theorem, for example. Now integrate again and exploit the symmetry of the derivative to obtain \begin{align} \mathfrak{I} &= f(1,1) = \pi \int \limits_0^1 \int \limits_0^1 \frac{b}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b = \pi \int \limits_0^1 \int \limits_0^1 \frac{a}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b \\ &= \frac{\pi}{2} \int \limits_0^1 \int \limits_0^1 \frac{a+b}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b = \frac{\pi}{2} \int \limits_0^1 \int \limits_0^1 \frac{1}{(1+a)(1+b)} \, \mathrm{d}a \, \mathrm{d}b \\ &= \frac{\pi}{2} \ln^2 (2) \, . \end{align}

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Notice:

$$ \Im(\log^2(1+ix))=\Im\left(\left(\frac{1}{2}\log(1+x^2)+i\arctan(x)\right)^2\right)=\frac{1}{2}\log(1+x^2)\arctan(x) $$

The integral in question is therefore (use parity) $$ I=\Im\int_{\mathbb{R}}\underbrace{\frac{\log^2(1+ix)}{x(1+x^2)}}_{f(x)} $$

Integrate around a big semicircle in the lower halfplane (to avoid the branchcut) yields

$$ I=\Im \left(2\pi i\text{Res}(f(z),z=-i)\right)=\frac{\pi}{2}\log^2(2)$$

where the residue is easy to calculate since the pole is simple.

The vansihing of the contrbutions at infinity follows from the fact that $R|f(Re^{i\phi})|\sim \log^2(R)/R^2$ in sector of $\mathbb{C}$ we are interested in.

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  • $\begingroup$ You also need that the integrand is even (to change the integral into one over $\mathbb{R}$), and that the singularity at zero is removable (so it doesn't contribute to the integral at all). $\endgroup$ – Chappers Jul 14 '18 at 20:17

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