Question Number 35589 by abdo mathsup 649 cc last updated on 20/May/18
$${let}\:\:\:{I}\:\:=\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{tx}} \:\mid{sint}\mid{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$${find}\:{the}\:{value}\:{of}\:{I}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 21/May/18
![I = Σ_(n=0) ^∞ ∫_(nπ) ^((n+1)π) e^(−tx) ∣sint∣dt changement t =nπ + u give I = Σ_(n=0) ^∞ ∫_0 ^π e^(−nπx) e^(−xu) ∣sinu∣du = Σ_(n=0) ^∞ e^(−nπx) ∫_0 ^π e^(−xu) sinu du but A(x)=∫_0 ^π e^(−xu) sin(u)du =Im( ∫_0 ^π e^(−xu +iu) du) =Im( ∫_0 ^π e^((−x+i)u) du) but ∫_0 ^π e^((−x+i)u) du =[ (1/(−x+i)) e^((−x+i)u) ]_0 ^π =((−1)/(x−i)){ e^(−xπ +iπ) −1}= ((1 +e^(−πx) )/(x−i)) =((x+i)/(x^2 +1))( 1+e^(−πx) )⇒ A(x)= ((1+e^(−πx) )/(1+x^2 )) Σ_(n=0) ^∞ e^(−nπx) = Σ_(n=0) ^∞ (e^(−πx) )^n = (1/(1−e^(−πx) )) so I = (1/(1−e^(−πx) )) ((1 +e^(−πx) )/(1+x^2 )) ⇒ I = ((1+e^(−πx) )/((1+x^2 )(1−e^(−πx) ))) .](https://www.tinkutara.com/question/Q35661.png)
$${I}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}\pi} ^{\left({n}+\mathrm{1}\right)\pi} \:{e}^{−{tx}} \mid{sint}\mid{dt}\:\:{changement} \\ $$$${t}\:={n}\pi\:\:+\:{u}\:{give} \\ $$$${I}\:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:{e}^{−{n}\pi{x}} \:{e}^{−{xu}} \:\mid{sinu}\mid{du} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:{e}^{−{n}\pi{x}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{xu}} \:{sinu}\:{du}\:\: \\ $$$${but}\:\:{A}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:{e}^{−{xu}} \:{sin}\left({u}\right){du}\:={Im}\left(\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{xu}\:+{iu}} {du}\right) \\ $$$$={Im}\left(\:\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(−{x}+{i}\right){u}} {du}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(−{x}+{i}\right){u}} {du}\:\:=\left[\:\frac{\mathrm{1}}{−{x}+{i}}\:{e}^{\left(−{x}+{i}\right){u}} \right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{−\mathrm{1}}{{x}−{i}}\left\{\:{e}^{−{x}\pi\:\:+{i}\pi} \:−\mathrm{1}\right\}=\:\frac{\mathrm{1}\:+{e}^{−\pi{x}} }{{x}−{i}} \\ $$$$=\frac{{x}+{i}}{{x}^{\mathrm{2}} +\mathrm{1}}\left(\:\mathrm{1}+{e}^{−\pi{x}} \right)\Rightarrow\:{A}\left({x}\right)=\:\frac{\mathrm{1}+{e}^{−\pi{x}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:{e}^{−{n}\pi{x}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−\pi{x}} \right)^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\pi{x}} }\:\:{so} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\pi{x}} }\:\frac{\mathrm{1}\:+{e}^{−\pi{x}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\:\:\frac{\mathrm{1}+{e}^{−\pi{x}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{e}^{−\pi{x}} \right)}\:. \\ $$