Fourier series. Where is logically wrong? $\text{Define }f(x)=e^x,(-\pi\lt x\le\pi),f(x+2\pi)=f(x).$
Using fourier transformation, we get $e^x=\Sigma C_ne^{inx},(-\pi\lt x\lt\pi)$
Differentiate both sides, and we get $e^x=\Sigma inC_ne^{inx},(-\pi\lt x\lt\pi)$
So, $C_n=(in)C_n \text{ for any n}$.
$\therefore C_n=0 \text { if }n\neq0$
So $e^x=C_0, (-\pi\lt x\lt\pi) $

Where is wrong?
can you check my suspicion?


*

*$\Sigma C_ne^{inx}$ doesn't uniformly convergent to $e^x$. So you can't use the differentiating one by one in the $\Sigma$

*$\Sigma C_ne^{inx}=\Sigma inC_ne^{inx}$ doesn't necessarily mean $C_n=(in)C_n \text{ for any n}$
 A: The Fourier coefficients are
\begin{align}
       c_n&=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{x}e^{-inx}dx
  = \left.\frac{1}{2\pi}\frac{e^{x(1-in)}}{1-in}\right|_{x=-\pi}^{\pi} \\
  &= \frac{e^{\pi(1-in)}-e^{-\pi(1-in)}}{2\pi(1-in)}=\frac{(-1)^n\sinh(\pi)}{\pi(1-in)}.
\end{align}
You correctly questioned the convergence of the series obtained by differentiating term-by-term
$$
       \frac{d}{dx}\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}\sinh(\pi)}{\pi(1-in)}e^{inx}
    \not= \sum_{-\infty}^{\infty}\frac{(-1)^{n}\sinh(\pi)}{\pi(1-in)}ine^{inx}
$$
The series on the right diverges for all $x\in\mathbb{R}$ because the general term does not tend to $0$ as $n\rightarrow\pm\infty$. So equality has no meaning.
Part of the problem is that the function you're dealing, which is the periodic extension of $e^{x}$ has a discontinuity at $\pm \pi$. So the derivative of this function has $\delta$ function components with magnitude $e^{\pi}-e^{-\pi}$ at the endpoints, which is how the $\sinh(\pi)$ components come into play. So it would be wrong if the derived series were to converge.
