Question

Let $f(x)=\frac{\sin x}{x}$, then $\underset{0}{\overset{\pi /2}{\int }}f(x)f\left(\frac{\pi }{2}-x\right)dx=$

• A. $\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)dx$
• B. $\underset{0}{\overset{\pi }{\int }}f(x)dx$
• C. $\pi \underset{0}{\overset{\pi }{\int }}f(x)dx$
• D. $\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)dx$

Answer 1

Answer: (A)

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x}{x}\frac{\cos x}{\left(\frac{\pi }{2}-x\right)}dx$
$\pi I=\underset{0}{\overset{\pi /2}{\int }}\sin 2x\left[\frac{1}{x}+\frac{1}{\pi /2-x}\right]dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 2x}{x}dx+\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 2x}{\pi /2-x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 2x}{x}dx+\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 2x}{x}dx=4\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 2x}{2x}dx$
$\frac{\pi I}{2}=\underset{0}{\overset{\pi }{\int }}\frac{\sin t}{t}dt$
[Put $2x=t$ ]
$I=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx$