Question Number 26570 by abdo imad last updated on 26/Dec/17 | ||
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{px}} /{sinx}/{dx}\:\:\:{with}\:{p}>\mathrm{0} \\ $$ | ||
Commented byabdo imad last updated on 02/Jan/18 | ||
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\propto} \:{e}^{−{px}} /{sinx}/{dx} \\ $$ $${I}=\:{lim}_{{n}−>\propto} \:\int_{\mathrm{0}} ^{{n}\pi} \:{e}^{−{px}} /{sinx}/{dx}\:\:{but}\: \\ $$ $$\int_{\mathrm{0}} ^{{n}\pi} {e}^{−{px}} /{sinx}/{dx}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)\pi} \:{e}^{−{px}} /{sinx}/{dx} \\ $$ $$=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{p}\left({k}\pi+{t}\right)} /{sin}\left({k}\pi+{t}\right)/{dt}\:\left(\:{we}\:{do}\:{the}\:{changement}\:{x}={k}\pi\:+{t}\right) \\ $$ $$=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{pk}\pi} \:\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{pt}} {sint}\:{dt}\:=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({e}^{−{p}\pi} \right)^{{k}} \:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{pt}} \:{sintdt} \\ $$ $$=\:\frac{\mathrm{1}−\:\:{e}^{−{np}\pi} }{\mathrm{1}−\:{e}^{−{p}\pi} }\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{pt}} {sintdt}\:\:\:{but} \\ $$ $$\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{pt}} {sintdt}\:=\:{Im}\left(\:\int_{\mathrm{0}} ^{\pi} \:{e}^{\left({i}−{p}\right){t}} {dt}\:\right) \\ $$ $$\: \\ $$ $$\:\:\: \\ $$ $$={Im}\left(\:\:\left[\:\frac{\mathrm{1}}{{i}−{p}}\:{e}^{\left({i}−{p}\right){t}} \:\right]_{\mathrm{0}} ^{\pi\:} =\:\frac{\mathrm{1}+{e}^{−{p}\pi} }{\mathrm{1}+{p}^{\mathrm{2}} }\right. \\ $$ $${I}=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{p}\pi} }\:.\:\frac{\mathrm{1}+\:{e}^{−{p}\pi} }{\mathrm{1}+{p}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}+\:{e}^{−{p}\pi} }{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)\left(\mathrm{1}−{e}^{−{p}\pi} \right)}\:. \\ $$ | ||