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Question Number 76353 by mathmax by abdo last updated on 26/Dec/19

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{arctan\left(sin\left(\pi x^2\right)\right)}{x^2+{\pi }^2}dx$$

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

$$letA={\int }_0^{\mathrm{\infty }}\frac{arctan\left(sin\left(\pi x^2\right)\right)}{x^2+{\pi }^2}dx\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(sin\left(\pi x^2\right)\right)}{x^2+{\pi }^2}$$$$letW\left(z\right)=\frac{arctan\left(sin\left(\pi z^2\right)\right)}{z^2+{\pi }^2}\Rightarrow W\left(z\right)=\frac{arctan(sin\left(\pi z^2\right)}{(z-i\pi )(z+i\pi )}$$$$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,i\pi )$$$$=2i\pi \times \frac{\mid arctan(sin\left(\pi (-{\pi }^2)\right)\mid }{2i\pi }=arctan\left(sin\left({\pi }^3\right)\right)$$$$\Rightarrow A=\frac{arctan\left(sin\left({\pi }^3\right)\right)}{2}$$

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