I know this question has already been asked but I have not found something useful for the way I was looking to calculate this serie (using complex exponentials expansion of $cos(nx)$).
I want to find the Fourier serie of $f(x) = e^x$ for $x \in [-\pi, \pi] $
The coefficient $a_0$ is pretty easy to find : \begin{align*} a_0 & = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^x dx \\ & = \frac{1}{\pi}sinh(\pi) \end{align*}
Next, the coefficient $a_n$, that's were I struggle. I'm trying to integrate by expressing $cos(nx)$ in term of complex exponentials : \begin{align*} a_n & = \frac{1}{\pi}\underbrace{\int_{-\pi}^{\pi}e^x cos(nx)dx}_{ = \textbf{I}} \\ \\ I & = \int_{-\pi}^{\pi}e^x\frac{e^{inx} + e^{-inx}}{2} \\ & = \int_{-\pi}^{\pi}\frac{e^{x(1 + in)} + e^{x(1 - in)}}{2} \\ & = \frac{1}{2(1+in)}(e^{\pi(1+in)} - e^{-\pi(1+in)}) + \frac{1}{2(1-in)}(e^{\pi(1-in)} - e^{-\pi(1-in)}) \\ & = \frac{1}{2(1+in)}(e^{\pi}e^{in\pi} - e^{-\pi}e^{-in\pi}) + \frac{1}{2(1-in)}(e^{\pi}e^{-in\pi} - e^{-\pi}e^{in\pi}) \\ & = \frac{1}{2(1+in)}(e^{\pi}(cos(n\pi)+isin(n\pi)) - e^{-\pi}(cos(n\pi)-isin(n\pi))) + \\ & \frac{1}{2(1-in)}(e^{\pi}(cos(n\pi) - isin(n\pi)) - e^{-\pi}(cos(n\pi)+isin(n\pi))) \end{align*}
I am stuck at this last line, I don't know how to go further. I've been spending a few hours on this, there's definitely something I am missing.
Thanks.
EDIT : Here's the answer.
\begin{align*} a_n & = \frac{1}{\pi}\underbrace{\int_{-\pi}^{\pi}e^x cos(nx)dx}_{ = \textbf{I}} \\ \\ I & = \int_{-\pi}^{\pi}e^x\frac{e^{inx} + e^{-inx}}{2} \\ & = \int_{-\pi}^{\pi}\frac{e^{x(1 + in)} + e^{x(1 - in)}}{2} \\ & = \frac{1}{2(1+in)}(e^{\pi(1+in)} - e^{-\pi(1+in)}) + \frac{1}{2(1-in)}(e^{\pi(1-in)} - e^{-\pi(1-in)}) \\ & = \frac{1}{2(1+in)}(e^{\pi}e^{in\pi} - e^{-\pi}e^{-in\pi}) + \frac{1}{2(1-in)}(e^{\pi}e^{-in\pi} - e^{-\pi}e^{in\pi}) \\ & = \frac{1}{2(1+in)}(e^{\pi}(cos(n\pi)+isin(n\pi)) - e^{-\pi}(cos(n\pi)-isin(n\pi))) + \\ & \frac{1}{2(1-in)}(e^{\pi}(cos(n\pi) - isin(n\pi)) - e^{-\pi}(cos(n\pi)+isin(n\pi))) \\ & = \frac{1}{2(1+in)}(e^{\pi}cos(n\pi) - e^{-\pi}cos(n\pi)) + \frac{1}{2(1-in)}(e^{\pi}cos(n\pi) - e^{-\pi}cos(n\pi)) \\ & = \frac{1}{2(1+in)}(cos(n\pi)(e^{\pi} - e^{-\pi}) + \frac{1}{2(1-in)}(cos(n\pi)(e^{\pi} - e^{-\pi})) \\ & = \frac{1}{(1+in)}cos(n\pi)sinh(\pi) + \frac{1}{(1-in)}cos(n\pi)sinh(\pi) \\ & = cos(n\pi)sinh(\pi)(\frac{1}{1+in} + \frac{1}{1-in}) \\ & = cos(n\pi)sinh(\pi)(\frac{1-in}{1^2 - (in)^2} + \frac{1+in}{1^2 - (in)^2}) \\ & = cos(n\pi)sinh(\pi)(\frac{2}{1 + n^2}) \\ a_n & = \frac{2cos(n\pi) sinh(\pi)}{\pi(n^2+1)} \\ & = \frac{2(-1)^n sinh(\pi)}{\pi(n^2+1)} \end{align*}