Application of Complex Variables Considering the integral :
$$\int_{-\infty}^{+\infty}\left(\dfrac{\sin\alpha z}{\alpha z}\right)^2 \dfrac{\pi}{\sin{\pi} z}dz,\quad \alpha \lt \dfrac{\pi}{2}$$
around a circle of large radius, prove that
$$\sum\limits_{m=1}^{\infty} (-1)^{m-1}\dfrac{\sin^2m\alpha}{(m\alpha)^2}=\dfrac{1}{2}$$
If anyone can help me in going about doing this?
 A: The function written above has poles of the first order at points $z=n$, where $n \in \mathbb{Z}$. All because of the $\sin{\pi z}$ in the denominator : The function itself has a linear behaviour in the neighbourhood of all its zeros, so they are of the first order. The $\alpha$ part does not count in pole-behaviour, albeit it looks loke it has a singularity, it has not. (Because the apparent singularity is removable)
Now, consider a contour infinitely large (or just limitingly huge), it containts all poles exept the one at infinity. As you probably know, you can compute the integral on that contour by using the Residue formula, the information is extracted from poles which this contour encloses - i.e. the integral is sum of the residues times a constant. Now, if you know a little bit from projection, you can imagine that the whole complex plain is just so the Riemann sphere. With little bit of imagination you can see now that there are two ways of defining region which that contour encloses, so the poles IN the region could be :
$(1)$ All the integer value poles on the real line (positively oriented contour)
or
$(2)$ Just point on the top of the sphere (infinite point) (negatively or.c.)
Let us compute the residue at each integer point $z=n$, unsing the definition of the residue, the by the L'Hopital's rule :
$$\mathrm{Res}_{z=n}{f(z)}:=\lim_{z \to n} f(z)(z-n) = \lim_{z \to n} \left(\frac{\sin{\alpha z}}{\alpha z}\right)^2 \pi \frac{z-n}{\sin{\pi z}} = \lim_{z \to n} \left(\frac{\sin{\alpha z}}{\alpha z}\right)^2\frac{1}{\cos{\pi z}} = \left(\frac{\sin{\alpha n}}{\alpha n}\right)^2 (-1)^n $$
Of course, at $n=0$, using this limit is equal to $1$
Nevertheless, the residuum at infinity simply vanishes, so, by the fact the value of the contour integral is independent on the "wiew in/out", so does the sum of the residue at integer valued points, therefore :
$$\therefore \sum_{n=-\infty}^{\infty}\left(\frac{\sin{\alpha n}}{\alpha n}\right)^2 (-1)^n=0$$
Now, we can separate the sum into two sympetric sums starting at 1 (excluding n=0)
$$\sum_{n=-\infty}^{\infty}\left(\frac{\sin{\alpha n}}{\alpha n}\right)^2 (-1)^n = \sum_{n=-\infty}^{n=-1}\left(\frac{\sin{\alpha n}}{\alpha n}\right)^2 (-1)^n 
 + 1 + \sum_{n=1}^{n=\infty}\left(\frac{\sin{\alpha n}}{\alpha n}\right)^2 (-1)^n$$
By symmetry (just even function) the two mani sums are equal (because every term is equal), regrupping gives the result which have to be showed.
(Albeit I thing there may be far more rigorous aproaches, from the physicist point of view this may be sufficient)
