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Question Number 117948 by mnjuly1970 last updated on 14/Oct/20

$$...niceintegral...$$$$pleaseevaluate::$$$$$$$$\mathrm{I}={\int }_0^1(sin\left(x\right)+sin\left(\frac{1}{x}\right))\frac{dx}{x}=??$$$$$$$$m.n.1970$$$$$$

Answered by Lordose last updated on 14/Oct/20

$$$$$$\mathrm{I}={\int }_0^1\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}+{\int }_0^1\frac{\sin \left(\frac{1}{\mathrm{x}}\right)}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}=\mid \mathrm{Si}\left(\mathrm{x}\right){\mid }_0^1-\mid \mathrm{Si}\left(\frac{1}{\mathrm{x}}\right){\mid }_0^1$$$$\mathrm{I}=\mathrm{Si}\left(1\right)-\mathrm{Si}\left(0\right)-\mathrm{Si}\left(1\right)+\mathrm{Si}\left(\mathrm{\infty }\right)$$$$\mathrm{I}=\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 14/Oct/20

$$grateful..$$

Answered by AbduraufKodiriy last updated on 14/Oct/20

Commented by mnjuly1970 last updated on 14/Oct/20

$$thankyou..$$

Answered by 1549442205PVT last updated on 14/Oct/20

$$\mathrm{I}={\int }_0^1\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}+{\int }_0^1\frac{\sin \frac{1}{\mathrm{x}}}{\mathrm{x}}\mathrm{dx}=\mathrm{A}+\mathrm{B}$$$$\mathrm{A}={\int }_0^1\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}$$$$\mathrm{Put}\mathrm{t}=\frac{1}{\mathrm{x}}\Rightarrow \mathrm{dt}=-\frac{1}{{\mathrm{x}}^2}\mathrm{dx}=-{\mathrm{t}}^2\mathrm{dx}$$$$\mathrm{B}=-{\int }_{\mathrm{\infty }}^1\frac{\mathrm{tsint}}{{\mathrm{t}}^2}\mathrm{dt}={\int }_1^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}$$$$\Rightarrow \mathrm{I}=\mathrm{A}+\mathrm{B}={\int }_0^1\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}+{\int }_1^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}=\frac{\pi }{2}\left(\mathrm{this}\mathrm{is}\mathrm{proved}\mathrm{ago}\right)$$

Commented by mnjuly1970 last updated on 14/Oct/20

$$mercey..$$

Answered by mindispower last updated on 14/Oct/20

$$I={\int }_0^1\frac{sin\left(x\right)}{x}dx+{\int }_0^1\frac{sin\left(\frac{1}{x}\right)}{x}dx$$$$let\frac{1}{x}=tin2nd\Rightarrow$$$${\int }_0^1\frac{sin\left(\frac{1}{x}\right)}{x}dx={\int }_1^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt$$$$I={\int }_0^1\frac{sin\left(t\right)}{t}dt+{\int }_1^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt={\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt=\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 14/Oct/20

$$veryniceasalwaysmr$$$$power..$$

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