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 byabdo mathsup 649 cc last updated on 21/May/18 | ||
$${I}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}\pi} ^{\left({n}+\mathrm{1}\right)\pi} \:{e}^{−{tx}} \mid{sint}\mid{dt}\:\:{changement} \\ $$ $${t}\:={n}\pi\:\:+\:{u}\:{give} \\ $$ $${I}\:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:{e}^{−{n}\pi{x}} \:{e}^{−{xu}} \:\mid{sinu}\mid{du} \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:{e}^{−{n}\pi{x}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{xu}} \:{sinu}\:{du}\:\: \\ $$ $${but}\:\:{A}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:{e}^{−{xu}} \:{sin}\left({u}\right){du}\:={Im}\left(\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{xu}\:+{iu}} {du}\right) \\ $$ $$={Im}\left(\:\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(−{x}+{i}\right){u}} {du}\right)\:{but} \\ $$ $$\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(−{x}+{i}\right){u}} {du}\:\:=\left[\:\frac{\mathrm{1}}{−{x}+{i}}\:{e}^{\left(−{x}+{i}\right){u}} \right]_{\mathrm{0}} ^{\pi} \\ $$ $$=\frac{−\mathrm{1}}{{x}−{i}}\left\{\:{e}^{−{x}\pi\:\:+{i}\pi} \:−\mathrm{1}\right\}=\:\frac{\mathrm{1}\:+{e}^{−\pi{x}} }{{x}−{i}} \\ $$ $$=\frac{{x}+{i}}{{x}^{\mathrm{2}} +\mathrm{1}}\left(\:\mathrm{1}+{e}^{−\pi{x}} \right)\Rightarrow\:{A}\left({x}\right)=\:\frac{\mathrm{1}+{e}^{−\pi{x}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ $$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:{e}^{−{n}\pi{x}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−\pi{x}} \right)^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\pi{x}} }\:\:{so} \\ $$ $${I}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\pi{x}} }\:\frac{\mathrm{1}\:+{e}^{−\pi{x}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\:\:\frac{\mathrm{1}+{e}^{−\pi{x}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{e}^{−\pi{x}} \right)}\:. \\ $$ | ||