Verify Delta Dirac function

Question

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?

Answer 1

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.