Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 33350 by caravan msup abdo. last updated on 14/Apr/18

prove that ∫_0 ^∞  ((cos(αx))/(chx))dx=  2 Σ_(n=0) ^∞ (−1)^n  ((2n+1)/((2n+1)^2  +α^2 )) .

$${prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\alpha{x}\right)}{{chx}}{dx}= \\ $$$$\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:. \\ $$

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 ))  .

$${let}\:{put}\:{I}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\alpha{x}\right)}{{chx}}{dx} \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\alpha{x}\right)}{{e}^{{x}} \:+{e}^{−{x}} }{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} \:{cos}\left(\alpha{x}\right)}{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{nx}} \right)\:.{e}^{−{x}} \:{cos}\left(\alpha{x}\right){dx} \\ $$$$=\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{cos}\left(\alpha{x}\right){dx}\:{but} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{cos}\left(\alpha{x}\right){dx}\:=\:{Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{e}^{−{i}\alpha{x}} {dx}\right) \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}+{i}\alpha\right){x}} {dx}\right)\:{but}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\:+{i}\alpha\right){x}} {dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}\:+{i}\alpha}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\:+{i}\alpha\right)} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}+{i}\alpha}\:=\:\frac{\mathrm{2}{n}+\mathrm{1}−{i}\alpha}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\Rightarrow\:{A}_{{n}} =\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:\:=\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com