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I have used sympy to evaluate this for large $a$ and I think that this converges to a value of $\frac{\pi^2}{4}$ but I am not able to arrive at this analytically. So I guess I want to prove analytically: $$ \lim_{a\to\infty}\int_0^a\frac{\tan^{-1}x\tan^{-1}(a-x)}{a}\,\mathrm{d}x = \frac{\pi^2}{4}$$

I have tried Integrating by parts, and Feynman's integral trick but I arrive at integrals which I don't know how to deal.

pii_ke
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    Put $x=at$, then the integral becomes $$ \int_0^1 {\arctan (at)\arctan (a(1 - t))dt} . $$ Now use dominated convergence. – Gary Jun 23 '21 at 05:29
  • @Gary Thank you for helping. From https://math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral#253697, I understand that dominated convergence theorem helps to take the limit operation inside the integral. The results follows. Is this correct? – pii_ke Jun 23 '21 at 06:01
  • Yes, you can dominate the integrand on $(0,1)$ by $\frac{\pi}{2} \times \frac{\pi}{2}$, which is integrable on $(0,1)$. Hence you can move the limit operator inside the integral. See https://en.wikipedia.org/wiki/Dominated_convergence_theorem – Gary Jun 23 '21 at 06:03

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