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Question Number 93061 by Ar Brandon last updated on 10/May/20

$${\int }_1^{+\mathrm{\infty }}\frac{\sin \mathrm{u}}{\mathrm{u}}\mathrm{du}$$

Commented by mathmax by abdo last updated on 10/May/20

$$wehave\frac{\pi }{2}={\int }_0^{\mathrm{\infty }}\frac{sinu}{u}du={\int }_0^1\frac{sinu}{u}du+{\int }_1^{+\mathrm{\infty }}\frac{sinu}{u}du\Rightarrow$$$${\int }_1^{+\mathrm{\infty }}\frac{sinu}{u}du=\frac{\pi }{2}-{\int }_0^1\frac{sinu}{u}dubutsinu=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!}{u}^{2n+1}\Rightarrow$$$$\frac{sinu}{u}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!}{u}^{2n}\Rightarrow {\int }_0^1\frac{sinu}{u}du=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!}\times \frac{1}{2n+1}\Rightarrow$$$${\int }_1^{+\mathrm{\infty }}\frac{sinu}{u}du=\frac{\pi }{2}-\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!(2n+1)}$$$$approximatevalueforn=4\Rightarrow$$$${\int }_1^{+\mathrm{\infty }}\frac{sinu}{u}du\sim \frac{\pi }{2}-(1-\frac{1}{3.3!}+\frac{1}{5.5!}-\frac{1}{7.7!}+\frac{1}{9.9!})\Rightarrow$$$${\int }_1^{+\mathrm{\infty }}\frac{sinu}{u}du\sim \frac{\pi }{2}-1+\frac{1}{3.3!}-\frac{1}{5.5!}+\frac{1}{7.7!}-\frac{1}{9.9!}$$

Commented by Ar Brandon last updated on 11/May/20

thank you

Commented by abdomathmax last updated on 14/May/20

$$youarewelcome.$$

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