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Question Number 136943 by Mathspace last updated on 28/Mar/21
find ∫_0 ^∞  e^(−ax) ∣sin(bx)∣ dx  a>0 and b>0
$${find}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ax}} \mid{sin}\left({bx}\right)\mid\:{dx} \\ $$$${a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 30/Mar/21
Φ=∫_0 ^∞  e^(−ax) ∣sin(bx)∣dx ⇒Φ=_(bx=t)  ∫_0 ^∞  e^(−(a/b)t) ∣sint∣(dt/b)  =(1/b) Σ_(n=0) ^∞  ∫_(nπ) ^((n+1)π)  e^(−((at)/b)) ∣sint∣ dt  =_(t=nπ +y)   (1/b)Σ_(n=0) ^∞  ∫^π _0  e^(−(a/b)(nπ+y)) ∣sin(nπ+y)∣ dy  =(1/b)Σ_(n=0) ^∞ e^(−((naπ)/b))  ∫_0 ^π   e^(−((ay)/b))  siny dy  let (a/b)=λ ⇒  Φ=(1/b) Σ_(n=0) ^∞  e^(−nλπ)  .∫_0 ^π  e^(−λy)  siny dy we have  ∫_0 ^π  e^(−λy)  siny dy =Im(∫_0 ^π  e^(−λy+iy) dy) and  ∫_0 ^π  e^((−λ+i)y)   dy =[(1/(−λ+i))e^((−λ+i)y) ]_0 ^π  =((−1)/(λ−i)){e^((−λ+i)π) −1}  =((−(λ+i))/(1+λ^2 )){− e^(−λπ) −1} =(((λ+i)(e^(−λπ) +1))/(1+λ^2 ))  ⇒∫_0 ^π  e^(−λy)  siny dy =((1+e^(−λπ) )/(1+λ^2 )) ⇒  Φ =((1+e^(−λπ) )/(b(1+λ^2 )))Σ_(n=0) ^∞  (e^(−λπ) )^n  =((1+e^(−λπ) )/(b(1+λ^2 )))×(1/(1−e^(−λπ) )) ⇒  Φ =((1+e^(−((aπ)/b)) )/(b(1−e^(−((aπ)/b)) )(1+(a^2 /b^2 )))) ⇒Φ=(b/(a^2 +b^2 ))×((1+e^(−((aπ)/b)) )/(1−e^(−((aπ)/b)) ))
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{ax}} \mid\mathrm{sin}\left(\mathrm{bx}\right)\mid\mathrm{dx}\:\Rightarrow\Phi=_{\mathrm{bx}=\mathrm{t}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{b}}\mathrm{t}} \mid\mathrm{sint}\mid\frac{\mathrm{dt}}{\mathrm{b}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{b}}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{n}\pi} ^{\left(\mathrm{n}+\mathrm{1}\right)\pi} \:\mathrm{e}^{−\frac{\mathrm{at}}{\mathrm{b}}} \mid\mathrm{sint}\mid\:\mathrm{dt} \\ $$$$=_{\mathrm{t}=\mathrm{n}\pi\:+\mathrm{y}} \:\:\frac{\mathrm{1}}{\mathrm{b}}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\underset{\mathrm{0}} {\int}^{\pi} \:\mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{b}}\left(\mathrm{n}\pi+\mathrm{y}\right)} \mid\mathrm{sin}\left(\mathrm{n}\pi+\mathrm{y}\right)\mid\:\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{b}}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \mathrm{e}^{−\frac{\mathrm{na}\pi}{\mathrm{b}}} \:\int_{\mathrm{0}} ^{\pi} \:\:\mathrm{e}^{−\frac{\mathrm{ay}}{\mathrm{b}}} \:\mathrm{siny}\:\mathrm{dy}\:\:\mathrm{let}\:\frac{\mathrm{a}}{\mathrm{b}}=\lambda\:\Rightarrow \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{b}}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{n}\lambda\pi} \:.\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{−\lambda\mathrm{y}} \:\mathrm{siny}\:\mathrm{dy}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{−\lambda\mathrm{y}} \:\mathrm{siny}\:\mathrm{dy}\:=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{−\lambda\mathrm{y}+\mathrm{iy}} \mathrm{dy}\right)\:\mathrm{and} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{\left(−\lambda+\mathrm{i}\right)\mathrm{y}} \:\:\mathrm{dy}\:=\left[\frac{\mathrm{1}}{−\lambda+\mathrm{i}}\mathrm{e}^{\left(−\lambda+\mathrm{i}\right)\mathrm{y}} \right]_{\mathrm{0}} ^{\pi} \:=\frac{−\mathrm{1}}{\lambda−\mathrm{i}}\left\{\mathrm{e}^{\left(−\lambda+\mathrm{i}\right)\pi} −\mathrm{1}\right\} \\ $$$$=\frac{−\left(\lambda+\mathrm{i}\right)}{\mathrm{1}+\lambda^{\mathrm{2}} }\left\{−\:\mathrm{e}^{−\lambda\pi} −\mathrm{1}\right\}\:=\frac{\left(\lambda+\mathrm{i}\right)\left(\mathrm{e}^{−\lambda\pi} +\mathrm{1}\right)}{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} \:\mathrm{e}^{−\lambda\mathrm{y}} \:\mathrm{siny}\:\mathrm{dy}\:=\frac{\mathrm{1}+\mathrm{e}^{−\lambda\pi} }{\mathrm{1}+\lambda^{\mathrm{2}} }\:\Rightarrow \\ $$$$\Phi\:=\frac{\mathrm{1}+\mathrm{e}^{−\lambda\pi} }{\mathrm{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(\mathrm{e}^{−\lambda\pi} \right)^{\mathrm{n}} \:=\frac{\mathrm{1}+\mathrm{e}^{−\lambda\pi} }{\mathrm{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}×\frac{\mathrm{1}}{\mathrm{1}−\mathrm{e}^{−\lambda\pi} }\:\Rightarrow \\ $$$$\Phi\:=\frac{\mathrm{1}+\mathrm{e}^{−\frac{\mathrm{a}\pi}{\mathrm{b}}} }{\mathrm{b}\left(\mathrm{1}−\mathrm{e}^{−\frac{\mathrm{a}\pi}{\mathrm{b}}} \right)\left(\mathrm{1}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\right)}\:\Rightarrow\Phi=\frac{\mathrm{b}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }×\frac{\mathrm{1}+\mathrm{e}^{−\frac{\mathrm{a}\pi}{\mathrm{b}}} }{\mathrm{1}−\mathrm{e}^{−\frac{\mathrm{a}\pi}{\mathrm{b}}} } \\ $$

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