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Question Number 110214 by mathdave last updated on 27/Aug/20

show that ∫_(−∞) ^(+∞) (((1+(x/π))sin(πx))/(x^2 +4x+5))dx=(1/e^π )

$${show}\:{that}\: \\ $$$$\int_{−\infty} ^{+\infty} \frac{\left(\mathrm{1}+\frac{{x}}{\pi}\right)\mathrm{sin}\left(\pi{x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}}{dx}=\frac{\mathrm{1}}{{e}^{\pi} } \\ $$

Answered by mathmax by abdo last updated on 27/Aug/20

I =∫_(−∞) ^(+∞) (((1+(x/π))sin(πx))/(x^2 +4x +5))=Im(∫_(−∞) ^(+∞) (((1+(x/π))e^(iπx) )/(x^2 +4x+5))) let condider the complex function ϕ(z) =(((1+(z/π))e^(iπz) )/(z^2 +4z +5)) poles of ϕ? z^2 +4z +5 =0→Δ^′ =4−5 =−1 ⇒z_1 =−2+i and z_2 =−2−i residus theorem ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,−2+i) we have ϕ(z) =(((1+(z/π))e^(iπz) )/((z−z_1 )(z−z_2 ))) ⇒Res(ϕ,−2+i) = =lim_(z→z_1 ) (z−z_1 )ϕ(z) = (((1+(z_1 /π))e^(iπz_1 ) )/(2i)) =(((1+((−2+i)/π))e^(iπ(−2+i)) )/(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((π−2+i)e^(−π) )/(2iπ)) =(π−2 +i)e^(−π) =(π−2)e^(−π) +i e^(−π) ⇒ Im(∫....) = e^(−π) ⇒★ I =(1/e^π )★

$$\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\left(\mathrm{1}+\frac{\mathrm{x}}{\pi}\right)\mathrm{sin}\left(\pi\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4x}\:+\mathrm{5}}=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\left(\mathrm{1}+\frac{\mathrm{x}}{\pi}\right)\mathrm{e}^{\mathrm{i}\pi\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4x}+\mathrm{5}}\right)\:\:\mathrm{let}\:\mathrm{condider}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{function} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\left(\mathrm{1}+\frac{\mathrm{z}}{\pi}\right)\mathrm{e}^{\mathrm{i}\pi\mathrm{z}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{4z}\:+\mathrm{5}}\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{4z}\:+\mathrm{5}\:=\mathrm{0}\rightarrow\Delta^{'} \:=\mathrm{4}−\mathrm{5}\:=−\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =−\mathrm{2}+\mathrm{i}\:\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =−\mathrm{2}−\mathrm{i} \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,−\mathrm{2}+\mathrm{i}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\varphi\left(\mathrm{z}\right)\:=\frac{\left(\mathrm{1}+\frac{\mathrm{z}}{\pi}\right)\mathrm{e}^{\mathrm{i}\pi\mathrm{z}} }{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)}\:\Rightarrow\mathrm{Res}\left(\varphi,−\mathrm{2}+\mathrm{i}\right)\:= \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\varphi\left(\mathrm{z}\right)\:=\:\:\frac{\left(\mathrm{1}+\frac{\mathrm{z}_{\mathrm{1}} }{\pi}\right)\mathrm{e}^{\mathrm{i}\pi\mathrm{z}_{\mathrm{1}} } }{\mathrm{2i}}\:=\frac{\left(\mathrm{1}+\frac{−\mathrm{2}+\mathrm{i}}{\pi}\right)\mathrm{e}^{\mathrm{i}\pi\left(−\mathrm{2}+\mathrm{i}\right)} }{\mathrm{2i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi×\frac{\left(\pi−\mathrm{2}+\mathrm{i}\right)\mathrm{e}^{−\pi} }{\mathrm{2i}\pi}\:=\left(\pi−\mathrm{2}\:+\mathrm{i}\right)\mathrm{e}^{−\pi} \\ $$$$=\left(\pi−\mathrm{2}\right)\mathrm{e}^{−\pi} \:+\mathrm{i}\:\mathrm{e}^{−\pi} \:\Rightarrow\:\mathrm{Im}\left(\int....\right)\:=\:\mathrm{e}^{−\pi} \:\Rightarrow\bigstar\:\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{e}^{\pi} }\bigstar \\ $$$$ \\ $$

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