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I'd like to ask for a little hint on how to proceed with this result.

In an attempt to evaluate the integral in title I introduced the following auxiliary integral: $$I\left(a\right)=\int_0^\infty \frac{\arctan^2\left(x\right)}{\sqrt{x}\left(a^2+x^2\right)}\ dx$$ which for $\mathit a=1$ gives us the value that we seek. I then defined a complex valued function: $$\mathit f\left(z\right)=-\frac{1}{4}{\log^2\left(i-z\over i+z\right)\over\sqrt{z}\left(z^2+a^2\right)}$$ $$\arg\left(i-z\over i+z\right)\in\left(0,2\pi\right)$$ $$\arg\left(z\right)\in\left(0,2\pi\right)$$ $$a>1$$ with the arguments and values of $a$ as such, so that I could use the approach in this thread, using the same contour.

After doing all of the necessary manipulations, I'm left with the following expression: $$2I\left(a\right)+\int_0^1\left[\pi\sqrt2\log\left({1-y\over 1+y}\right)+\pi^2\sqrt2\right]\ {dy\over\sqrt y\left(a^2-y^2\right)}={\pi\sqrt a\over4a^2}\left[\pi^2\sqrt{2}-2\pi\sqrt{2}\log\left({a-1\over a+1}\right)-\sqrt{2}\log^2\left({a-1\over a+1}\right) \right]$$ with the right hand side representing the sum of the residues (multiplied by $2\pi i$) of the function $f\left(z\right)$ at $z=ia$ and $z=-ia$.

As we let $a$ approach 1 from above, it seems to agree numerically, but I'd like to ask for a tip on how to simplify this gnarly looking thing.

For reference, the integral in title should evaluate to: $$\frac{\sqrt{2}}{8}\left(\frac{5\pi^3}{12}-\pi^2\log\left(2\right)-\pi\log^2\left(2\right)\right)$$

Thank you in advance!

PepperLegs
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  • $$\frac{\sqrt{2}\pi}{96}\left(5\pi^2-12\pi\ln(2)-12(\ln2)^2\right)$$ Seems to be the answer, not the one provided by OP. The integral has been evaluated at this AoPS Post – Miracle Invoker Apr 15 '23 at 06:22
  • Woops, I forgot about the 5 in front of $\frac{\pi^3}{12}$. These two results should now be equivalent, I'm terribly sorry for the mistake! – PepperLegs Apr 15 '23 at 08:12
  • Possibly off the topic, but it seems that the formula $\displaystyle\int_0^\pi\cos (p x) \sin ^{s-1}(x) dx=-\frac{2^{1-s}\pi~ \cos \left(\frac{\pi p}{2}\right) \Gamma (s)}{\Gamma \left(\frac{1+s-p}{2}\right) \Gamma \left(\frac{1+s+p}{2}\right)}$ leads to a quick solution to the original integral (via differentiating w.r.t $p$ twice, plugging in $p=s=\dfrac12$ and a little algebra). – Po1ynomial Apr 16 '23 at 08:28

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