Using DTFT to find the sum of $\sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)$ I am trying to use DTFT (as asked in a problem) to find the following sum 
$$\sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)$$
for real $\alpha_1>0$ and $\alpha_2<1$.
I tried few approaches, but none seems to work. In the following, $x[n]$ is the discrete-time signal (sampled signal) and $X(\omega)$ the continuous-frequency transformed variable (frequency response).

Approach 1
\begin{align}
\sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)&=\sum_{n=-\infty }^{\infty }\frac{1}{\alpha_1\alpha_2 n^2}\sin(n\alpha_1)\sin(n\alpha_2)\\
&=\sum_{n=-\infty }^{\infty }\frac{1}{\alpha_1\alpha_2 n^2}\frac{1}{2}\left [ \cos(\alpha_1-\alpha_2)n-\cos(\alpha_1+\alpha_2)n \right ]\\
&=\frac{1}{2\alpha_1\alpha_2}\sum_{n=-\infty }^{\infty }\frac{1}{n^2}\cos(\alpha_1-\alpha_2)n-\frac{1}{2\alpha_1\alpha_2}\sum_{n=-\infty }^{\infty }\frac{1}{n^2}\cos(\alpha_1+\alpha_2)n 
\end{align}
Then I thought about using this property
$$\sum_{n=-\infty }^{\infty }x[n]\leftrightarrow X(0)$$
And I don't think the DTFT of both terms is easy.
Approach 2
\begin{align}
DTFT\left \{ \sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2) \right \}&=\sum_{m=-\infty}^{\infty}\left \{ \sum_{n=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2) \right \} e^{-i\omega m}\\
&=\sum_{n=-\infty}^{\infty} \sum_{m=-\infty }^{\infty }\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)e^{-i\omega m} 
\end{align}
And then I need to find the DTFT of $\text{sinc}(n\alpha_1)\text{sinc}(n\alpha_2)$ using this property
$$x_1[n]x_2[n]\leftrightarrow \frac{1}{2\pi}X_1(\omega)\ast X_2(\omega)$$
where $\ast$ denotes the convolution operator. The final step would be to sum this result over all $n$, which doesn't seem to help (unless the result has a well-known sum).

Any hints would be appreciated.
 A: I think you already have the answer.
using these two properties:
Property 1.
$$x_1[n]x_2[n]\leftrightarrow \frac{1}{2\pi}X_1(\omega)\ast X_2(\omega)$$
Property 2.
$$ \sum_{n=-\infty }^{\infty }x[n]\leftrightarrow X(0)$$
Step 1. Using property 1.
find the DTFT of 
$y[n]=\text{sinc}[n\alpha_1]\text{sinc}[n\alpha_2]$, which is a periodic convolution of two rectangle function in frequency domain. Now you have a function of $\omega$, $Y(\omega)$
Step2. Using property 2. 
set $\omega$ to $0$ in your previous answer $Y(\omega)$
$$\sum_{n=-\infty }^{\infty }y[n]\leftrightarrow Y(0)$$
Then you have the answer
In detail :
$\text{sinc}[\alpha_1 n]\leftrightarrow X_1(\omega)=\begin{cases}
\frac{1}{\alpha_1} & \text{for} & |\omega|<\pi\alpha_1 \\
0 &\text{for} & \pi\alpha_1<|\omega|\leq \pi\\
\end{cases}$
Similarly 
$\text{sinc}[\alpha_2 n]\leftrightarrow X_2(\omega)=\begin{cases}
\frac{1}{\alpha_2} & \text{for} & |\omega|<\pi\alpha_2 \\
0 &\text{for} & \pi\alpha_2<|\omega|\leq \pi\\
\end{cases}$
and also both $X_1(\omega)$and$X_2(\omega)$ repeats in every $2\pi$ interval.
by property 1, $\text{sinc}[\alpha_1n]\text{sinc}[\alpha_2n]\leftrightarrow \frac{1}{2\pi}X_1(\omega)\ast X_2(\omega)=\frac{1}{2\pi}\int^{\pi}_{-\pi}X_1(\tau)X_2(\omega-\tau)d\tau=Y(\omega)$
Since the final solution is $Y(0)=\frac{1}{2\pi}\int^{\pi}_{-\pi}X_1(\tau)X_2(-\tau)d\tau$ ,the integration is simply calculating the overlapping area of these two rectangular function in $[-\pi,\pi]$, then the answer is $$2*\frac{\pi \text{min}(\alpha_1,\alpha_2)}{2\pi\alpha_1\alpha_2}=\frac{\text{min}(\alpha_1,\alpha_2)}{\alpha_1\alpha_2}$$
(assuming $\alpha_1,\alpha_2>0$ and $\alpha_1,\alpha_2<1$)
A: Assuming $\alpha,\beta> 0$, in order to compute:
$$\sum_{n\in\mathbb{Z}}\text{sinc}(n\alpha)\,\text{sinc}(n\beta)=1+\frac{2}{\alpha \beta}\sum_{n\geq 1}\frac{\sin(n\alpha)\sin(n\beta)}{n^2}\tag{1}$$
it is enough to find a closed formula for:
$$ g(\gamma)=\sum_{n\geq 1}\frac{\cos(n\gamma)}{n^2},\qquad \gamma\geq 0.\tag{2}$$
It is straightforward to notice that $g(\gamma)$ is a continuous, $2\pi$-periodic function, whose value in zero equals $\frac{\pi^2}{6}$ and whose value in $\pi$ equals $-\frac{\pi^2}{12}$. Since $g'(\gamma)$ is a well-known Fourier series (the Fourier series of the sawtooth-wave), it follows that $g(\gamma)$ is the periodic continuation of the function:
$$ h(\gamma)=\frac{\pi^2}{6}-\frac{\gamma(2\pi-\gamma)}{4},\qquad \gamma\in[0,2\pi]\tag{3}$$
hence we have, by further assuming $\alpha+\beta<2\pi$:
$$\begin{eqnarray*}\sum_{n\in\mathbb{Z}}\text{sinc}(n\alpha)\text{sinc}(n\beta)&=&1+\frac{1}{\alpha\beta}\left(g(|\alpha-\beta|)-g(\alpha+\beta)\right)\\&=&1+\frac{1}{4\alpha\beta}\left[(\alpha-\beta)^2-(\alpha+\beta)^2+2\pi\left(\alpha+\beta-|\alpha-\beta|\right)\right]\\&=&1+\frac{1}{4\alpha\beta}\left[-4\alpha\beta+4\pi\min(\alpha,\beta)\right]\\&=&{\frac{\pi}{\alpha\beta}\min(\alpha,\beta)}=\color{red}{\frac{\pi}{\max(\alpha,\beta)}}.\tag{4}\end{eqnarray*}$$
