Question Number 141062 by mnjuly1970 last updated on 15/May/21
$$\:\:\:\:\: \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{cos}\left({x}\right)} {sin}\left({sin}\left({x}\right)\right)}{{x}}\:{dx}=\frac{\pi}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$
Commented by mindispower last updated on 15/May/21
$${always}\:{pleasur}\:{have}\:{nice}\:{day}“\:{A}\overset{..} {{i}d}\:{Mobarak}'' \\ $$
Commented by mnjuly1970 last updated on 15/May/21
$$\:\:{thank}\:{you}\:{so}\:{much}\:{sir} \\ $$$${power}\:{your}\:{Aid}\:{mobarak}\:.. \\ $$
Commented by mindispower last updated on 15/May/21
$${starte}\:{withe}\:{your}\:{previous}\:{result}\:{quation} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}{a}^{{n}} \frac{{sin}\left({nx}\right)}{{n}!}={e}^{{acos}\left({x}\right)} {sin}\left({asin}\left({x}\right)\right. \\ $$$$\left.\phi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left({a}^{{n}} {sin}\left({nx}\right)\right)/\left({n}!\right)\right)_{{a}=\mathrm{1}} {dx} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({nx}\right)}{{x}\left({n}!\right)}{dx}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({nx}\right)}{{nx}}{d}\left({nx}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}!}.\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left(\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}!}−\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$
Commented by mnjuly1970 last updated on 15/May/21
$$\:{grateful}\:{mr}\:{power}…. \\ $$
Answered by Dwaipayan Shikari last updated on 15/May/21
$${e}^{{cos}\left({x}\right)} {sin}\left({sinx}\right)=\Sigma\frac{{sin}\left({nx}\right)}{{n}!} \\ $$$$\int_{\mathrm{0}} ^{\infty} \underset{{n}=\mathrm{1}} {\sum}\frac{{sin}\left({nx}\right)}{{xn}!}=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}=\frac{\pi}{\mathrm{2}}\left({e}−\mathrm{1}\right) \\ $$
Commented by mnjuly1970 last updated on 15/May/21
$$\:\:\:\:\:{mercey}\:{mr}\:{Payan}… \\ $$