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Question Number 73331 by mathmax by abdo last updated on 10/Nov/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{ln(1+e^{-3x^2})}{3+x^2}dx$$

Commented by mathmax by abdo last updated on 11/Nov/19

$$letA={\int }_0^{\mathrm{\infty }}\frac{ln(1+e^{-3x^2})}{x^2+3}dx\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{ln(1+e^{-3x^2})}{x^2+3}dx$$$$letW\left(z\right)=\frac{ln(1+e^{-3z^2})}{z^2+3}\Rightarrow W\left(z\right)=\frac{ln(1+e^{-3z^2})}{(z-i\sqrt{3})(z+i\sqrt{3})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,i\sqrt{3})$$$$Res(W,i\sqrt{3})={lim}_{z\to i\sqrt{3}}(z-i\sqrt{3})W\left(z\right)=\frac{ln(1+e^{-3{\left(i\sqrt{3}\right)}^2})}{2i\sqrt{3}}$$$$=\frac{ln(1+e^9)}{2i\sqrt{3}}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi \times \frac{ln(1+e^9)}{2i\sqrt{3}}=\frac{\pi }{\sqrt{3}}ln(1+e^9)\Rightarrow$$$$A=\frac{\pi }{2\sqrt{3}}ln(1+e^9).$$

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