Step by Step

Observe the integral:

$\int_{0}^{2}$ $\pi$ ( $\sin{\frac{m\pi\theta}{2}}$ )( $\sin{\frac{n\pi\theta}{2}}$ ) $d\theta$

There are two cases:-

1) n=m

2) n different from m

For n=m,

The integral simplifies to-

$\int_{0}^{2}$ $\pi$ ( $\sin^2({\frac{m\pi\theta}{2}})$ ) $d\theta$

Using $\sin^2(a)$ = $\frac{1-\cos(2a)}{2}$ ,

Here:

a= $\frac{m\pi\theta}{2}$

We get:

$\frac{\pi}{2}$ $\int_{0}^{2}$ $(1-\cos{m\pi\theta})$ $d\theta$ = $\pi$

So, if n=m the integral is just $\pi$ .

If n is different from m,

Recall:

$\cos(a+b)$ = $\cos(a)\cos(b)$ - $\sin(a)\sin(b)$

and

$\cos(a-b)$ = $\cos(a)\cos(b)$ + $\sin(a)\sin(b)$

Therefore,

$\sin(a)\sin(b)$ = $\frac{\cos(a-b)- \cos(a+b)}{2}$

Here:

a= $\frac{m\pi\theta}{2}$

b= $\frac{n\pi\theta}{2}$

Note- the $\cos$ function integrates to $\sin$ function and the angle is an integer multiple of $\pi$ for $\theta$ between 0 and 2.

So, the value of integral evaluated at n different from m is zero.

Conclusion: The integral= $\pi$ only for n=m and zero for all other values of n.

Therefore, the answer is $\pi$ .

Bonus question: What will be the answer if I change the 2s' in the question to some other integer, L?

i.e. $\sum_{n=1}^{\infty}$ $\int_{0}^{L}$ $\pi$ ( $\sin{\frac{m\pi\theta}{L}}$ )( $\sin{\frac{n\pi\theta}{L}}$ ) $d\theta$