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  1. Is it true that $\int_{0}^{\infty}\sin(mx)\sin(nx) \, dx = \delta (m-n) $ although using Euler formula I get a linear combination of $ \delta(m-n) $ and $ \delta (m+n)$?

  2. What is the sum $\sum_{n=0}^\infty (-1)^n \sin(nx)$?

Here $\delta (x)$ is the Dirac delta function.

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  1. Clearly, $$\begin{align*} \int_{0}^{\infty} \sin mx \sin nx \; dx &= \frac{1}{2}\int_{-\infty}^{\infty} \sin mx \sin nx \; dx \\ &= \frac{1}{4}\int_{-\infty}^{\infty} (\cos (m-n)x - \cos(m+n)x ) \; dx \\ &= \frac{1}{4}\int_{-\infty}^{\infty} (e^{i(m-n)x} - e^{i(m+n)x}) \; dx \\ &= \frac{\pi}{2}(\delta(m-n) - \delta(m+n)), \end{align*}$$ in distribution sense. But If $m$ and $n$ are positive, no care is needed for the second term.

  2. We have $$ \sum_{n=0}^{\infty} (-1)^{n}\sin nx = -\frac{1}{2}\tan \frac{x}{2}$$ in Abel summability sense. Indeed, $$ \begin{align*} \sum_{n=0}^{\infty} (-1)^{n}\sin nx &= \lim_{s\to 0^{+}} \sum_{n=0}^{\infty} (-1)^{n}\sin nx e^{-sn} \\ &= \lim_{s\to 0^{+}} \Im \sum_{n=0}^{\infty} (-e^{ix-s})^{n} \\ &= \lim_{s\to 0^{+}} \Im \left( \frac{1}{1 + e^{ix-s}} \right) \\ &= \frac{1}{1 + e^{ix}} \\ &= -\frac{1}{2}\tan \frac{x}{2}. \end{align*}$$ In fact, this convergence can be reduced to Cesaro summability sense, though more calculation is required.

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thank you very much :) ... for the cosine i knew how to evaluate it.. but not for the sine :) –  Jose Garcia May 1 '12 at 13:05
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