For give function prove that $\frac{\pi}{2} \int_{0} ^{\pi/2} f(x) f \bigg(\frac{\pi}{2}-x \bigg)dx=\int _{0}^{\pi} f(x)dx$ If $f(x)=\frac{\sin x}{x}, \forall x \in(0,\pi ]$, then prove that 
$$\frac{\pi}{2} \int_{0} ^{\pi/2} f(x) f \bigg(\frac{\pi}{2}-x \bigg)dx=\int _{0}^{\pi} f(x)dx$$
Could someone give me some hint to proceed in this question. I tried putting $t=2x$ in L.H.S. to get desired limit, but not able to general $\pi$ outside integral.
 A: Let's start with the LHS.  We can plug in to get
$$
\frac{\pi}{2}\int_0^{\pi/2}f(x)f\left(\frac{\pi}{2}-x\right)dx=\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin(x)}{x}\cdot\frac{\sin\left(\frac{\pi}{2}-x\right)}{\frac{\pi}{2}-x}dx
$$
We can now use the trig identity $\sin\left(\frac{\pi}{2}-x\right)=\cos(x)$ to simplify the integral into
$$
\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin(x)\cos(x)}{x\left(\frac{\pi}{2}-x\right)}dx.
$$
Next, let's use the trig identity $\sin(2x)=2\sin(x)\cos(x)$ to simplify the integral even more to 
$$
\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin(2x)}{x\left(\frac{\pi}{2}-x\right)}dx.
$$
At this point, it makes sense to use the substitution $u=2x$ so that the numerator and limits match the RHS.  In this case, $du=2dx$ so we get 
$$
\frac{\pi}{8}\int_0^{\pi}\frac{\sin(u)}{\frac{u}{2}\left(\frac{\pi}{2}-\frac{u}{2}\right)}du.
$$
Bringing some of the factors of $2$ from $8$ inside to simplify the denominator, we get
$$
\frac{\pi}{2}\int_0^{\pi}\frac{\sin(u)}{u(\pi-u)}du.
$$
Now, we go with partial fraction decomposition.  In particular, we can write
$$
\frac{1}{u(\pi-u)}=\frac{A}{u}+\frac{B}{\pi-u}.
$$
Clearing fractions, we get 
$$
1=A(\pi-u)+Bu.
$$
Substituting $u=0$ gives $A=\frac{1}{\pi}$ and substituting $u=\pi$ gives $B=\frac{1}{\pi}$ as well.  Therefore,
\begin{align*}
\frac{\pi}{2}\int_0^{\pi}\frac{\sin(u)}{u(\pi-u)}du&=
\frac{\pi}{2}\int_0^{\pi}\frac{\sin(u)}{\pi u}du+\frac{\pi}{2}\int_0^{\pi}\frac{\sin(u)}{\pi (\pi-u)}du\\
&=\frac{1}{2}\int_0^{\pi}\frac{\sin(u)}{u}du+\frac{1}{2}\int_0^{\pi}\frac{\sin(u)}{\pi-u}du
\end{align*}
The first integral is half the desired integral, so we hope that the second integral is also half the desired integral because then two halves make a whole.  Using the $v$-substitution $v=\pi-u$, we get $dv=-du$ and
$$
\frac{1}{2}\int_0^{\pi}\frac{\sin(u)}{\pi-u}du=-\frac{1}{2}\int_{\pi}^{0}\frac{\sin(\pi-v)}{v}dv=\frac{1}{2}\int_{0}^{\pi}\frac{\sin(\pi-v)}{v}dv.
$$
Finally, using trig identities, 
$$
\sin(\pi-v)=\sin\left(\frac{\pi}{2}-\left(v-\frac{\pi}{2}\right)\right)=\cos\left(v-\frac{\pi}{2}\right).
$$
Since cosine is an even function, this equals
$$
\cos\left(\frac{\pi}{2}-v\right)=\sin(v).
$$
Therefore, the second term is also half the desired integral and the computation is done.
