Question Number 33350 by caravan msup abdo. last updated on 14/Apr/18
$${prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\alpha{x}\right)}{{chx}}{dx}= \\ $$$$\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 17/Apr/18
![let put I =∫_0 ^∞ ((cos(αx))/(chx))dx I = 2 ∫_0 ^∞ ((cos(αx))/(e^x +e^(−x) ))dx =2 ∫_0 ^∞ ((e^(−x) cos(αx))/(1+e^(−2x) ))dx =2 ∫_0 ^∞ (Σ_(n=0) ^∞ (−1)^n e^(−2nx) ) .e^(−x) cos(αx)dx =2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(2n+1)x) cos(αx)dx but A_n =∫_0 ^∞ e^(−(2n+1)x) cos(αx)dx = Re( ∫_0 ^∞ e^(−(2n+1)x) e^(−iαx) dx) =Re( ∫_0 ^∞ e^(−(2n+1+iα)x) dx) but we have ∫_0 ^∞ e^(−(2n+1 +iα)x) dx =[−(1/(2n+1 +iα)) e^(−(2n+1 +iα)) ]_0 ^(+∞) =(1/(2n+1+iα)) = ((2n+1−iα)/((2n+1)^2 +α^2 )) ⇒ A_n =((2n+1)/((2n+1)^2 +α^2 )) ⇒ I =2 Σ_(n=0) ^∞ (((−1)^n (2n+1))/((2n+1)^2 +α^2 )) .](https://www.tinkutara.com/question/Q33490.png)
$${let}\:{put}\:{I}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\alpha{x}\right)}{{chx}}{dx} \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\alpha{x}\right)}{{e}^{{x}} \:+{e}^{−{x}} }{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} \:{cos}\left(\alpha{x}\right)}{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{nx}} \right)\:.{e}^{−{x}} \:{cos}\left(\alpha{x}\right){dx} \\ $$$$=\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{cos}\left(\alpha{x}\right){dx}\:{but} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{cos}\left(\alpha{x}\right){dx}\:=\:{Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{e}^{−{i}\alpha{x}} {dx}\right) \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}+{i}\alpha\right){x}} {dx}\right)\:{but}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\:+{i}\alpha\right){x}} {dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}\:+{i}\alpha}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\:+{i}\alpha\right)} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}+{i}\alpha}\:=\:\frac{\mathrm{2}{n}+\mathrm{1}−{i}\alpha}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\Rightarrow\:{A}_{{n}} =\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:\:=\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\:. \\ $$