Question

If $f\left(x\right)=\frac{\sin x}{x}\forall x\in [0,\pi ]$, then $\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}f\left(x\right)f(\frac{\pi }{2}-x)dx$ is equal to

• A. $\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$
• B. $\pi {\int }_0^{\pi }f\left(x\right)dx$
• C. ${\int }_0^{\pi }f\left(x\right)dx$
• D. ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$
  

Answer 1

The correct option is C ${\int }_0^{\pi }f\left(x\right)dx$

Let, $I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}f\left(x\right)f(\frac{\pi }{2}-x)dx$
$=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\sin x}{x}\frac{\cos x}{\frac{\pi }{2}-x}dx=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\sin 2x}{x(\pi -2x)}dx$
$=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin z}{z(\pi -z)}dz$ $($Let $2x=z\Rightarrow 2dx=dz)$
$=\frac{1}{2}{\int }_0^{\pi }\sin z(\frac{1}{z}+\frac{1}{\pi -z})dz$
$=\frac{1}{2}({\int }_0^{\pi }\frac{\sin z}{z}dz+{\int }_0^{\pi }\frac{\sin z}{\pi -z}dz)={\int }_0^{\pi }\frac{\sin z}{z}dz$

$\because {\int }_0^{a}f\left(x\right)={\int }_0^{a}f(a-x)dx$

$\therefore \frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}f\left(x\right)f(\frac{\pi }{2}-x)dx={\int }_0^{\pi }f\left(x\right)dx$

Hence, option C.