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Question Number 76350 by mathmax by abdo last updated on 26/Dec/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{arctan\left(e^{x^2}\right)}{x^2+4}dx$$

Commented by mathmax by abdo last updated on 29/Dec/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(e^{x^2}\right)}{x^2+4}dx\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(e^{x^2}\right)}{x^2+4}$$$$let\phi \left(z\right)=\frac{arctan\left(e^{z^2}\right)}{z^2+4}\Rightarrow \phi \left(z\right)=\frac{arctan\left(e^{z^2}\right)}{(z-2i)(z+2i)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,2i)=2i\pi \times \frac{arctan\left(e^{-4}\right)}{4i}=\frac{\pi }{2}arctan\left(e^{-4}\right)$$$$=\frac{\pi }{2}arctan\left(\frac{1}{e^4}\right)=\frac{\pi }{2}(\frac{\pi }{2}-arctan\left(e^4\right))=\frac{{\pi }^2}{4}-\frac{\pi }{2}arctan\left(e^4\right)$$

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