Question

The value of the integral $\underset{\frac{1}{\mathrm{\pi }}}{\overset{\frac{2}{\mathrm{\pi }}}{\int }}\frac{\sin \left({\displaystyle \frac{1}{\mathrm{\pi }}}\right)}{x^2}dx$ is _______________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{\frac{1}{\mathrm{\pi }}}{\overset{\frac{2}{\mathrm{\pi }}}{\int }}\frac{\sin \left({\displaystyle \frac{1}{x}}\right)}{x^2}dx \\ \mathrm{Let}, \\ \frac{1}{x}=t \\ \Rightarrow d\left(\frac{1}{x}\right)=dt \\ \Rightarrow -\frac{1}{x^2}dx=dt \\ \mathrm{Also},\mathrm{if}x=\frac{1}{\mathrm{\pi }},t=\mathrm{\pi } \\ \mathrm{if}x=\frac{2}{\mathrm{\pi }},t=\frac{\mathrm{\pi }}{2} \\ \mathrm{Thus}, \\ I=\underset{\mathrm{\pi }}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}-\sin \left(t\right)dt \\ ={\left[\cos t\right]}_{\mathrm{\pi }}^{\frac{\mathrm{\pi }}{2}} \\ =\left[\cos \frac{\mathrm{\pi }}{2}-\mathrm{cos\pi }\right] \\ =\left[0-\left(-1\right)\right] \\ =1 \end{gathered}$$

​
Hence, the value of the integral $\underset{\frac{1}{\mathrm{\pi }}}{\overset{\frac{2}{\mathrm{\pi }}}{\int }}\frac{\sin \left({\displaystyle \frac{1}{x}}\right)}{x^2}dx$ is 1.