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Question Number 60679 by maxmathsup by imad last updated on 24/May/19
calculate ∫_0 ^∞ ((ln(1+e^(−x^2 ) ))/(x^2 +4)) dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 26/May/19
let A =∫_0 ^∞ ((ln(1+e^(−x^2 ) ))/(x^2 +4)) dx ⇒2A = ∫_(−∞) ^(+∞) ((ln(1+e^(−x^2 ) ))/(x^2 +4))dx let ϕ(z) =((ln(1+e^(−z^2 ) ))/(z^2 +4)) we have ϕ(z) =((ln(1+e^(−z^2 ) ))/((z−2i)(z+2i))) so the poles of ϕ are 2i and −2i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,2i) Res(ϕ,2i) =lim_(z→2i) (z−2i)ϕ(z) =((ln(1+e^(−(2i)^2 )) )/(4i)) =((ln(1+e^4 ))/(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((ln(1+e^4 ))/(4i)) =(π/2) ln(1+e^4 ) ⇒ A =(π/4)ln(1+e^4 )
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:{dx}\:\Rightarrow\mathrm{2}{A}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{ln}\left(\mathrm{1}+{e}^{−{z}^{\mathrm{2}} } \right)}{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\:\:{we}\:{have}\:\varphi\left({z}\right)\:=\frac{{ln}\left(\mathrm{1}+{e}^{−{z}^{\mathrm{2}} } \right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\: \\ $$$$\mathrm{2}{i}\:{and}\:−\mathrm{2}{i}\:\:{residus}\:{theorem}\:{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,\mathrm{2}{i}\right) \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}} \left({z}−\mathrm{2}{i}\right)\varphi\left({z}\right)\:=\frac{{ln}\left(\mathrm{1}+{e}^{\left.−\left(\mathrm{2}{i}\right)^{\mathrm{2}} \right)} \right.}{\mathrm{4}{i}}\:=\frac{{ln}\left(\mathrm{1}+{e}^{\mathrm{4}} \right)}{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{ln}\left(\mathrm{1}+{e}^{\mathrm{4}} \right)}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:{ln}\left(\mathrm{1}+{e}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{1}+{e}^{\mathrm{4}} \right) \\ $$

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