Question Number 73331 by mathmax by abdo last updated on 10/Nov/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \right)}{\mathrm{3}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$$${let}\:{W}\left({z}\right)=\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{z}^{\mathrm{2}} } \right)}{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow{W}\left({z}\right)=\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{z}^{\mathrm{2}} } \right)}{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left({W},{i}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\left({z}−{i}\sqrt{\mathrm{3}}\right){W}\left({z}\right)=\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \right)}{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{{ln}\left(\mathrm{1}+{e}^{\mathrm{9}} \right)}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{{ln}\left(\mathrm{1}+{e}^{\mathrm{9}} \right)}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}{ln}\left(\mathrm{1}+{e}^{\mathrm{9}} \right)\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\left(\mathrm{1}+{e}^{\mathrm{9}} \right). \\ $$