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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.

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Is there a reason you expect this to converge, in general? If $x=y$ is irrational, for example, then $\sin^2(nx\pi)$ does not converge to $0$. –  Thomas Andrews Nov 14 '12 at 17:57
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3 Answers 3

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

$$ \begin{align*} \sum_{n=1}^{\infty} \sin(n\pi x)\sin(n\pi y) &= \frac{1}{2} \sum_{n=1}^{\infty} \left\{ \cos n\pi (x-y) - \cos n \pi (x+y) \right\} \\ &= \frac{1}{4} \sum_{n=-\infty}^{\infty} \left\{ \cos n\pi (x-y) - \cos n \pi (x+y) \right\} \\ &= \frac{1}{2} \sum_{n=-\infty}^{\infty} \left\{ \delta(x-y - 2n) - \delta(x+y - 2n) \right\}. \end{align*}$$

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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 ;)

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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

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But the textbook says that it converges to delta function? –  fsdgfghvvb Nov 14 '12 at 18:00
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Well, that's not convegence at all since the delta function is not a funtion :) –  Hagen von Eitzen Nov 14 '12 at 18:02
    
So, how can one prove that the limit is a delta function ? –  fsdgfghvvb Nov 14 '12 at 18:03
    
What is the exact claim in your textbook? –  Hagen von Eitzen Nov 14 '12 at 18:06
    
@fsdgfghvvb Hint: What's the definition of the $\delta$ function? –  Pragabhava Nov 14 '12 at 18:08
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