# Exponential Form of Fourier Series

Problem Suppose $f$ is a continuous function on interval $[-\pi,\pi]$ such that $\sum_{n\in\mathbb{Z}} |c_n| < \infty$ where $c_n = \dfrac {1}{2\pi} \int_{-\pi}^\pi f(x)\cdot \exp(-inx)~dx$, the Fourier coefficient. Let $g(x):=\sum_{k\in\mathbb{Z}} c_n\cdot\exp(ikx)$.

Show $g(x) = f(x)$ for all $x$.

What I've tried: I know that if I can show that $c_n=d_n$ where $d_n$ is the Fourier coefficient of $g$ that I can then proceed using uniqueness of the Fourier series, I think. However, proving this has stumped me.

By definition, $d_n = \dfrac {1}{2\pi} \int_{-\pi}^\pi g(x)\cdot \exp(-inx)~dx=\dfrac {1}{2\pi} \int_{-\pi}^\pi c_n [\sum_{k\in\mathbb{Z}} \exp(ikx)] \exp(-inx)~dx$. Bringing the other exponential into the sum, $\dfrac {1}{2\pi} \int_{-\pi}^\pi c_n \sum_{k\in\mathbb{Z}} \dfrac{\exp(ikx-inx)}{k}~dx$. Since $\sum_{n\in\mathbb{Z}} |c_n| < \infty$ and $\sum_{k\in\mathbb{Z}} |\dfrac{\exp(ikx-inx)}{k}|<\infty$, by Fubini/Tonelli Theorem, I interchange integration and summation, $\dfrac {c_n}{2\pi} \sum_{k\in\mathbb{Z}} \int_{-\pi}^\pi\dfrac{\exp(ikx-inx)}{k}~dx$. This is where I feel like I'm way off track. I think there is something wrong in my work, but I also feel that I may just be going in the complete wrong direction.

Somehow, I want this to equal $c_n = \dfrac {1}{2\pi} \int_{-\pi}^\pi f(x)\cdot \exp(-inx)~dx$

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## 1 Answer

Note that

$$\int_{-\pi}^{\pi}e^{i(k-n)x}\;dx = 2\pi\delta_{k-n}$$

i.e. the integral is only non-zero for $k=n$. Another thing you might want to check is the factor $1/k$ in your integral. Where does it come from?

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