Question Number 62196 by maxmathsup by imad last updated on 17/Jun/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{2}+{e}^{−{t}^{\mathrm{2}} } \right)}{{t}^{\mathrm{2}} \:+\mathrm{3}}{dt} \\ $$
Commented by mathmax by abdo last updated on 21/Jun/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{2}+{e}^{−{t}^{\mathrm{2}} } \right)}{{t}^{\mathrm{2}} \:+\mathrm{3}}\:{dt}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \frac{{ln}\left(\mathrm{2}+{e}^{−{t}^{\mathrm{2}} } \right)}{{t}^{\mathrm{2}} \:+\mathrm{3}}\:\:\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:{w}\left({z}\right)\:=\frac{{ln}\left(\mathrm{2}+{e}^{−{z}^{\mathrm{2}} } \right)}{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:\:\:\Rightarrow{w}\left({z}\right)\:=\frac{{ln}\left(\mathrm{2}+{e}^{−{z}^{\mathrm{2}} } \right)}{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left({w},{i}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\left({z}−{i}\sqrt{\mathrm{3}}\right){w}\left({z}\right)\:=\frac{{ln}\left(\mathrm{2}+{e}^{−\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \right)}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\frac{{ln}\left(\mathrm{2}+{e}^{\mathrm{3}} \right)}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{ln}\left(\mathrm{2}+{e}^{\mathrm{3}} \right)}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\pi\frac{{ln}\left(\mathrm{2}+{e}^{\mathrm{3}} \right)}{\:\sqrt{\mathrm{3}}}\:\Rightarrow{I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\left(\mathrm{2}+{e}^{\mathrm{3}} \right)\:. \\ $$