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Question Number 133107 by mnjuly1970 last updated on 18/Feb/21

$$...nicecalculus...$$$$find::\mathit{\varphi }\stackrel{???}{=}{\int }_0^1(sin\left(x\right)+sin\left(\frac{1}{x}\right))\frac{dx}{x}$$

Answered by Dwaipayan Shikari last updated on 18/Feb/21

$$\mathrm{\Phi }={\int }_0^1(sin\left(x\right)+sin\left(\frac{1}{x}\right))\frac{dx}{x}\frac{1}{x}=u\Rightarrow -\frac{1}{x^2}=\frac{du}{dx}$$$$={\int }_1^{\mathrm{\infty }}\frac{sin\left(\frac{1}{u}\right)}{u}+\frac{sinu}{u}du={\int }_0^{\mathrm{\infty }}\frac{sin\left(\frac{1}{u}\right)}{u}+\frac{sinu}{u}du-\mathrm{\Phi }$$$$2\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{sinu}{u}+\frac{1}{u}.sin\left(\frac{1}{u}\right)du$$$$\mathrm{\Phi }=\frac{\pi }{4}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}du\frac{1}{u}=t\Rightarrow -\frac{1}{u^2}=\frac{dt}{du}$$$$=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 19/Feb/21

$$gratefulmrpayan...$$

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