How to evaluate this infinite sum? How to calculate the infinite series $\sum_{n=1}^{\infty }\sin(nx\pi)\sin(ny\pi)$ where $x,y$ are real numbers? 
I assume this is related to Fourier series, but I can't calculate this.
 A: Note that
$$ \frac{1}{T} \sum_{n=-\infty}^{\infty} e^{2\pi i n x / T} = \sum_{n=-\infty}^{\infty} \delta(x - n T).$$
the Dirac comb of period $T > 0$. (This is a functional version of the Poisson summation formula, which is valid at least in the sense of distributions in $\mathcal{S}'(\Bbb{R})$.) Then by noting that the right-hand side is real, together with the property of Dirac delta function, we may rewrite as
$$ \sum_{n=-\infty}^{\infty} \cos (\pi n x) = 2 \sum_{n=-\infty}^{\infty} \delta(x - 2n).$$
Thus by writing
$$ \sin(n\pi x)\sin(n\pi y) = \frac{1}{2} \left\{ \cos n\pi (x-y) - \cos n \pi (x+y) \right\}, $$
we have
$$ \begin{align*}
\sum_{n=1}^{\infty} \sin(n\pi x)\sin(n\pi y)
&= \frac{1}{2} \sum_{n=1}^{\infty} \left\{ \cos n\pi (x-y) - \cos n \pi (x+y) \right\} \\
&= \frac{1}{4} \sum_{n=-\infty}^{\infty} \left\{ \cos n\pi (x-y) - \cos n \pi (x+y) \right\} \\
&= \frac{1}{2} \sum_{n=-\infty}^{\infty} \left\{ \delta(x-y - 2n) - \delta(x+y - 2n) \right\}.
\end{align*}$$
A: Let $x,\,y \in (-1,1)$ (why?). The Dirac $\delta$ satisfies
$$
\int_{-1}^1 \delta(x-y) f(y) dy = f(x)
$$
Now, if we want to express such function in its Fourier Series in the interval metioned above, we have
$$
\delta(x-y) = \frac{a_0}{2} + \sum_{n=1}^\infty \left[a_n \cos(n \pi x) + b_n \sin(n \pi x)\right]
$$
where the Fourier coefficients are (why?)
\begin{align}
a_n &= \int_{-1}^1 \delta(x-y) \cos n \pi x dx = \cos n \pi y, &n=0,1,2,\ldots\\
b_n &= \int_{-1}^1 \delta(x-y) \sin n \pi x dx = \sin n \pi y, &n=1,2,3,\ldots
\end{align}
Then
\begin{align}
\delta(x-y) &= 1 + \sum_{n=1}^\infty \left[\cos(n \pi x)\cos(n \pi y) + \sin(n \pi x)\sin(n \pi y)\right]\\
&= 1 + \sum_{n=1}^\infty\cos n\pi(x-y)
\end{align}
From here, it's easy to see that
$$
\delta(x+y) = 1 + \sum_{n=1}^\infty\cos n\pi(x+y),
$$
Using $\frac{1}{2}\cos n\pi(x-y) - \frac{1}{2}\cos n\pi(x+y) = \sin n\pi x \sin n \pi y$, we have
$$
\frac{1}{2} \delta(x-y) - \frac{1}{2}\delta(x+y) = \sum_{n = 1}^\infty \sin n\pi x \sin n \pi y.
$$
Note 1. Fourier sine and cosine basis are orthogonal on defined intervals, i.e. $\{\sin n \pi x\}$ is orthogonal on $(-1,1)$, but not necesarily in $(a,b)$ arbitrary, hence your question is incomplete.
Note 2. The (why's?) are for you to figure out ;)
A: Note that $\sin\alpha\sin\beta=\frac12\cos(\alpha-\beta)-\frac12\cos(\alpha+\beta)$, thus you should have a look at
$$\sum_{n=1}^\infty \left(\cos(n(x- y)\pi) -\cos(n(x+ y)\pi) \right)$$
If $x,y\in\mathbb Q$, the summands will be periodic, hence no convegence (apart from trivial cases).
And in the irrational case, it doesn't look converging either
