$$\int_0^\infty \frac{\arctan(x^2)}{1+x^2}\,\mathrm{d}x$$ Since it doesn't have a antiderivative in terms of elementary functions I cannot rely on the fundamental theorem of calculus. I tried using the Feynman integration technique by adding an extra parameter inside the inverse tangent but the expression after differentiation under the integral sign got so convoluted that I don't think it would've led to the solution. I would appreciate any help!
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this may help: https://math.stackexchange.com/questions/1413507/solve-this-integral-int-0-infty-frac-arctan-xxx21-mathrm-dx – SK19 May 09 '18 at 15:19
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Also note: https://math.stackexchange.com/a/287724/509159 and – SK19 May 09 '18 at 15:25
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Are you sure that your integral converges? – Dr. Sonnhard Graubner May 09 '18 at 15:41
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2Hint: Use the change of variable $t=1/x$ and the fact that for $u>0$, $arctan)(u)+ arctan(1/u)=\pi/2$. – Kelenner May 09 '18 at 15:44
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2A more general result for $\alpha\in \mathbb {R} $ is: $$\int_0^{\infty} \frac {\arctan(x^{\alpha})}{1+x^2}~dx=\frac {\pi^2}{8} $$ – projectilemotion May 09 '18 at 15:46
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@projectilemotion Could you provide a proof of this result? – Sz. Akos May 09 '18 at 15:48
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@Sz.Akos Using Kelenner's hint, you can prove the more general result. – projectilemotion May 09 '18 at 15:56
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@Kelenner Thank you so much, I got it. – Sz. Akos May 09 '18 at 15:56
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@projectilemotion Yes, I just realized it. Thank you all! – Sz. Akos May 09 '18 at 15:57