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Question Number 62196 by maxmathsup by imad last updated on 17/Jun/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{ln(2+e^{-t^2})}{t^2+3}dt$$

Commented by mathmax by abdo last updated on 21/Jun/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{ln(2+e^{-t^2})}{t^2+3}dt\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{ln(2+e^{-t^2})}{t^2+3}letconsiderthecomplex$$$$functionw\left(z\right)=\frac{ln(2+e^{-z^2})}{z^2+3}\Rightarrow w\left(z\right)=\frac{ln(2+e^{-z^2})}{(z-i\sqrt{3})(z+i\sqrt{3})}$$$$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(z\right)dz=2i\pi Res(w,i\sqrt{3})$$$$Res(w,i\sqrt{3})={lim}_{z\to i\sqrt{3}}(z-i\sqrt{3})w\left(z\right)=\frac{ln(2+e^{-{\left(i\sqrt{3}\right)}^2})}{2i\sqrt{3}}=\frac{ln(2+e^3)}{2i\sqrt{3}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(z\right)dz=2i\pi \frac{ln(2+e^3)}{2i\sqrt{3}}=\pi \frac{ln(2+e^3)}{\sqrt{3}}\Rightarrow I=\frac{\pi }{2\sqrt{3}}ln(2+e^3).$$

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