Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 68597 by Abdo msup. last updated on 14/Sep/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{arctan\left(e^{x^2}\right)}{x^2+8}dx$$

Commented by mathmax by abdo last updated on 14/Sep/19

$$residusmethodbutneedaproof$$$$letI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(e^{x^2}\right)}{x^2+8}\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(e^{x^2}\right)}{x^2+8}dx$$$$letW\left(z\right)=\frac{arctan\left(e^{z^2}\right)}{z^2+8}\Rightarrow W\left(z\right)=\frac{arctan\left(e^{z^2}\right)}{(z-2i\sqrt{2})(z+2i\sqrt{2})}$$$$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,2i\sqrt{2})$$$$=\frac{arctan\left(e^{{\left(2i\sqrt{2}\right)}^2}\right)}{4i\sqrt{2}}=\frac{arctan\left(e^{-8}\right)}{4i\sqrt{2}}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi \times \frac{arctan\left(e^{-8}\right)}{4i\sqrt{2}}$$$$=\frac{\pi }{\sqrt{2}}arctan\left(e^{-8}\right)\Rightarrow I=\frac{\pi }{2\sqrt{2}}arctan\left(e^{-8}\right).$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com