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?
Answer
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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