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Question Number 141062 by mnjuly1970 last updated on 15/May/21

$$$$$$provethat::$$$$\varphi :={\int }_0^{\mathrm{\infty }}\frac{e^{cos\left(x\right)}sin\left(sin\left(x\right)\right)}{x}dx=\frac{\pi }{2}(e-1)$$

Commented by mindispower last updated on 15/May/21

$$alwayspleasurhaveniceday‘‘A\stackrel{..}{id}{Mobarak}^{″}$$

Commented by mnjuly1970 last updated on 15/May/21

$$thankyousomuchsir$$$$poweryourAidmobarak..$$

Commented by mindispower last updated on 15/May/21

$$startewitheyourpreviousresultquation$$$$\sum _{n⩾1}{a}^n\frac{sin\left(nx\right)}{n!}=e^{acos\left(x\right)}sin(asin\left(x\right)$$$${\varphi ={\int }_0^{\mathrm{\infty }}\frac{1}{x}\sum _{n⩾1}\left(a^{n}sin\left(nx\right)\right)/(n!))}_{a=1}dx$$$$=\sum _{n⩾1}{\int }_0^{\mathrm{\infty }}\frac{sin\left(nx\right)}{x(n!)}dx=\sum _{n⩾1}\frac{1}{n!}{\int }_0^{\mathrm{\infty }}\frac{sin\left(nx\right)}{nx}d\left(nx\right)$$$$=\sum _{n⩾1}\frac{1}{n!}.\frac{\pi }{2}=\frac{\pi }{2}(\sum _{n⩾0}\frac{1}{n!}-1)=\frac{\pi }{2}(e-1)$$

Commented by mnjuly1970 last updated on 15/May/21

$$gratefulmrpower....$$

Answered by Dwaipayan Shikari last updated on 15/May/21

$$e^{cos\left(x\right)}sin\left(sinx\right)=\mathrm{\Sigma }\frac{sin\left(nx\right)}{n!}$$$${\int }_0^{\mathrm{\infty }}\sum _{n=1}\frac{sin\left(nx\right)}{xn!}=\frac{\pi }{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=\frac{\pi }{2}(e-1)$$

Commented by mnjuly1970 last updated on 15/May/21

$$merceymrPayan...$$

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