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Question Number 133563 by bemath last updated on 23/Feb/21

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}[\sin \frac{1}{\mathrm{x}}-\frac{1}{\pi }\sin \left(\frac{\pi }{\mathrm{x}}\right)]\mathrm{dx}$$

Answered by EDWIN88 last updated on 23/Feb/21

$$\mathrm{Calculate}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}(\sin \left(\frac{1}{\mathrm{x}}\right)-\frac{1}{\pi }\sin \left(\frac{\pi }{\mathrm{x}}\right))\mathrm{dx}$$$$\mathrm{let}\frac{1}{\mathrm{x}}=\mathrm{t}\wedge \mathrm{x}=\frac{1}{\mathrm{t}};\begin{array}{|cc|}\hline \mathrm{x}\to \mathrm{\infty }& \mathrm{t}\to 0^{+}\\ \mathrm{x}\to 0^{+}& \mathrm{t}\to \mathrm{\infty }\\ \hline\end{array}$$$$\mathrm{I}=\underset{\mathrm{\infty }}{\stackrel{0}{\int }}(\sin \mathrm{t}-\frac{1}{\pi }\sin \pi \mathrm{t})(-\frac{1}{{\mathrm{t}}^2})\mathrm{dt}$$$$\mathrm{I}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\left(\frac{\frac{\sin \mathrm{t}}{\mathrm{t}}-\frac{\sin \pi \mathrm{t}}{\pi \mathrm{t}}}{\mathrm{t}}\right)\mathrm{dt};\mathrm{Frullani}\mathrm{integral}$$$$\mathrm{I}=[\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\left(\frac{\sin \mathrm{t}}{\mathrm{t}}\right)-\underset{\mathrm{t}\to 0}{\lim }\frac{\sin \pi \mathrm{t}}{\pi \mathrm{t}}].\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right);\{\begin{array}{l}\mathrm{b}=1\\ \mathrm{a}=\pi \end{array}$$$$\mathrm{I}=(0-1)\ln \left(\frac{1}{\pi }\right)=\ln \pi$$$$$$

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