Show that: \begin{equation*} \frac{2}{T}\int_0^T\sin(n\omega t)\sin(m\omega t)dt=\delta_{m,n} \end{equation*} Here the period \(T=2\pi/\omega\), and \(n\) and \(m\) are integers greater than zero. Recall that \(\delta_{m,n}\) (the "Kronecker delta") is given by \[\delta_{m,n}= \begin{cases} 1 &m=n\\ 0 &m\ne n \end{cases} \] You will have to treat the two cases separately. Do not choose specific values of \(m\) and \(n\), prove this relationship in general for ANY integer \(m\) and \(n\).
Hints: Since it is easy to integrate exponentials, even if the exponent is a complex number, use Euler's formula to change the sines into exponentials: \begin{equation*} \sin(n\omega t)=\frac{e^{i n\omega t}-e^{-i n\omega t}}{2i} \end{equation*} Beware of zero in the denominator of fractions!
Please evaluate all integrals analytically by hand.
\[\sum_{n=1}^\infty\delta_{n3}=1 \]
\[\sum_{n=1}^\infty b_n\delta_{n3}=b_3\]
\[\sum_{n=1}^{10}\sum_{m=1}^{10}\delta_{nm}=\,\,\,?\]