Verify Delta Dirac function Verify that $\frac{1}{2π} \sum_{r=-∞}^{∞} e^{ir(x-x_0)} $ 
is a Dirac delta function $δ(x − x_0)$ by showing that it satisfies the
definition, $$\int_{-π}^{π} f(x) \frac{1}{2π} \sum_{r=-∞}^{∞} e^{ir(x-x_0)} 
 = f(x_0)$$
Hint: Represent f(x) by an exponential Fourier series.
From some other material from the class, I have that the exponential Fourier series is $f(x) = \sum_{r=-∞}^{∞} c_r e^{\frac{2πrix}{L}}$. However, I don't know how to proceed after that because just plugging that in didn't get me the answer. Using Euler's formula didn't really lead to anything either? 
 A: Let $$f(x) = \sum_{n = -\infty}^\infty c_n e^{i n x}$$
Then, substituting:
\begin{align*}
\int_{-\pi}^\pi f(x) \left(\frac{1}{2\pi} \sum_{k = -\infty}^\infty e^{ik(x - x_0)} \right) dx 
&= \int_{-\pi}^{\pi} \sum_{n = -\infty}^\infty c_n e^{i n x} \left(\frac{1}{2\pi} \sum_{k = -\infty}^\infty e^{ik(x - x_0)} \right) dx \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty c_n e^{i n x}  e^{ik(x - x_0)} dx \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty c_n e^{i (n + k) x} e^{-ik x_0} dx \\
&= \frac{1}{2\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty \int_{-\pi}^{\pi} c_n e^{i (n + k) x} e^{-ik x_0} dx \\
&= \frac{1}{2\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty c_n e^{-ik x_0} \int_{-\pi}^{\pi} e^{i (n + k) x}  dx \\
&= \frac{1}{2\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty c_n e^{-ik x_0} \int_{-\pi}^{\pi} e^{i (n + k) x}  dx \\
\end{align*}
Evaluating the integral in the last line of the above calculation requires special care. If $n + k = 0$, then 
$$
\int_{-\pi}^{\pi} e^{i (n + k) x}  dx = \int_{-\pi}^{\pi} e^{i 0 x}  dx = \int_{-\pi}^{\pi}  dx = 2\pi
$$
If $n + k \neq 0$, then 
\begin{align*}
\int_{-\pi}^{\pi} e^{i (n + k) x}  dx 
&= \left. \left( \frac{e^{i(n + k)x}}{i(n + k)} \right) \right|^{\pi}_{-\pi} \\
&= \frac{2}{(n + k)}\frac{e^{i(n + k)\pi} - e^{-i(n + k)\pi}}{2i} \\
&= \frac{2\sin((n + k)\pi)}{n + k} = \frac{2\cdot 0}{n + k} = 0
\end{align*}
In summary, $\int^{\pi}_{-\pi} e^{i(n + k)x} dx$ is $2\pi$ if $n + k = 0$ and $0$ if $n + k \neq 0$. More compactly, this means 
$$
\int^{\pi}_{-\pi} e^{i(n + k)x} dx = 2\pi\delta_{n + k}
$$
where the $\delta_{n + k}$ on the right hand side is the Kronecker delta (equal to $1$ when $n + k = 0$ and $0$ when $n + k \neq 0$). Substituting, we obtain
\begin{align*}
\int_{-\pi}^\pi f(x) \left(\frac{1}{2\pi} \sum_{k = -\infty}^\infty e^{ik(x - x_0)} \right) dx 
&= \frac{1}{2\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty c_n e^{-ik x_0} \int_{-\pi}^{\pi} e^{i (n + k) x}  dx \\
&= \frac{1}{2\pi} \sum_{n = -\infty}^\infty \sum_{k = -\infty}^\infty c_n e^{-ik x_0} \left( 2\pi \delta_{n + k} \right) \\
&= \sum_{n = -\infty}^\infty c_n \sum_{k = -\infty}^\infty e^{-ik x_0} \delta_{n + k}\\
&= \sum_{n = -\infty}^\infty c_n \left( e^{-i(-n)x_0} \right)\\
&= \sum_{n = -\infty}^\infty c_n e^{in x_0}\\
&= f(x_0)
\end{align*}
Thus, we have show
$$
\int_{-\pi}^\pi f(x) \left(\frac{1}{2\pi} \sum_{k = -\infty}^\infty e^{ik(x - x_0)} \right) dx = f(x_0)
$$
Notice that for this proof to go through, we must assume that $f$ can be written as a Fourier series on $[-\pi, \pi]$ and that exchanging the sum and integral signs is valid in all steps where we do so. 
