Find Fourier Series of $f(x)=e^x+e^{-x}$

Question

Problem:

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that:

1. $f(x)=e^x+e^{-x},\ \ \forall\ \ x \in [-\pi, \pi)$ and
2. $f(x+2\pi)=f(x),\ \ \forall\ \ x \in \mathbb{R}.$

Calculate the Fourier series of $f$.

What I've done:

First, I attempt to calculate the Fourier Coefficients.

1. Trying to find $a_0$:

$$ \begin{align} a_0 & =\frac{1}{2\pi}\int_{-\pi}^{\pi}{f(x)dx} =\frac{1}{\pi}\int_{0}^{\pi}{f(x)dx} =\frac{1}{\pi}\int_{0}^{\pi}{(e^x+e^{-x})dx} =\frac{1}{\pi}\left[e^x-e^{-x}\right]_{0}^{\pi}\\ & =\frac{1}{\pi}(e^{\pi}-1-e^{-\pi}+1) =\frac{e^{\pi}-e^{-\pi}}{\pi} =\frac{2\sinh(\pi)}{\pi} \end{align} $$

2. Trying to find $a_n$:

$$ \begin{align} a_n & =\frac{1}{\pi}\int_{-\pi}^{\pi}{f(x)\cos(nx)dx} =\frac{2}{\pi}\int_{0}^{\pi}{f(x)\cos(nx)dx} =\frac{2}{\pi}\int_{0}^{\pi}{(e^x+e^{-x})\cos(nx)dx}\\ & =\frac{2}{\pi}\int_{0}^{\pi}{e^x\cos(nx)dx} +\frac{2}{\pi}\int_{0}^{\pi}{e^{-x}\cos(nx)dx} =I+J \end{align} $$

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$$ \begin{align} I & =\frac{2}{\pi}\int_{0}^{\pi}{e^x\cos(nx)dx} =\frac{2}{\pi}\left[e^x\cos(nx)\right]_{0}^{\pi} +\frac{2n}{\pi}\int_{0}^{\pi}{e^x\sin(nx)dx}\\ & =\frac{2}{\pi}(e^{\pi}\cos(n\pi)-1) +\frac{2n}{\pi}\left[e^x\sin(nx)\right]_{0}^{\pi} -\frac{2n^2}{\pi}\int_{0}^{\pi}{e^x\cos(nx)dx}\\ & =\frac{2}{\pi}(e^{\pi}(-1)^n-1) +\frac{2n}{\pi}e^{\pi}\sin(n\pi) -n^2I =\frac{2(e^{\pi}(-1)^n-1)}{(1+n^2)\pi} \end{align} $$

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$$ \begin{align} J & =\frac{2}{\pi}\int_{0}^{\pi}{e^{-x}\cos(nx)dx} =\ ...\ =\frac{2(1-e^{-\pi}(-1)^n)}{(1+n^2)\pi} \end{align} $$

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$$ \begin{align} a_n & =I+J =\frac{2(e^{\pi}(-1)^n-1)}{(1+n^2)\pi} +\frac{2(1-e^{-\pi}(-1)^n)}{(1+n^2)\pi} =\frac{2(e^{\pi}(-1)^n-e^{-\pi}(-1)^n)}{(1+n^2)\pi}\\ & =\frac{2(-1)^n(e^{\pi}-e^{-\pi})}{(1+n^2)\pi} =\frac{4(-1)^n\sinh(\pi)}{(1+n^2)\pi} \end{align} $$

3. Since $f$ is an even function, $b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}{f(x)\sin(nx)dx}=0$.

After finding the Fourier Coefficients, I attempt to find the Fourie Series expression:

$$ \begin{align} s(x) & =a_0+\sum_{n=1}^{\infty}(a_n\cos(nx)+b_n\sin(nx)) =a_0+\sum_{n=1}^{\infty}a_n\cos(nx)\\ & =\frac{2\sinh(\pi)}{\pi}+\sum_{n=1}^{\infty}{\frac{4(-1)^n\sinh(\pi)}{(1+n^2)\pi}}\cos(nx)\\ & =\frac{2\sinh(\pi)}{\pi}+\frac{4\sinh(\pi)}{\pi}\sum_{n=1}^{\infty}{\frac{(-1)^n}{(1+n^2)}}\cos(nx) \end{align} $$

Question:

Is the approach I'm following and the result I've found correct? Is there a better way to tackle the above problem?

Edit:

As mentioned in a comment of mine under @Btzzzz's answer, what troubles me is the graph Desmos produces for my series, whose amplitude is stuck at $21.789$ while the graph of $f$ approaches $\infty$.

Answer 1

Your solution is correct, these kind of calculations are often long and tedious so there is no general shortcut. Something i like to use to verify my solution is correct is by plotting the series using Desmos and comparing it to the graph of $f$. (Noting that it won't be exact since this is just an approximation of the function)

Answer 2

This seems right to me but I think that using the complex coefficient will be easier:

$$f(x)=\sum_{n=-\infty}^\infty a_n e^{inx};\quad a_n=\frac1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\;dx$$Where $i$ is the imaginary unit.

Let's calculate $a_n$ shall we:

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$$\int_{-\pi}^\pi (e^x+e^{-x})e^{-inx}\;dx=\int_{-\pi}^\pi 2 e^{-i n x} \cosh(x)\;dx=\cdots= \dfrac{i e^{-(1 + i n) x} ((n - i) e^{2 x} + n + i)}{n^2 + 1}{\LARGE|}_{-\pi}^\pi\\=\dfrac{4 (\sinh(\pi) \cos(\pi n) + n \cosh(\pi) \sin(\pi n))}{n^2 + 1}$$

We can simplify it by remembering that $n\in\Bbb N$ so $\sin(\pi n)=0,\cos(\pi n)=(-1)^n$:$$\dfrac{4 (\sinh(\pi) \color{blue}{\cos(\pi n)} + \color{red}{n \cosh(\pi) \sin(\pi n)})}{n^2 + 1}=\dfrac{4 \sinh(\pi)(-1)^n}{n^2 + 1}$$

Now dividing by $2\pi$ to get: $$a_n=\frac2{\pi}\dfrac{ \sinh(\pi)(-1)^n}{n^2 + 1}$$

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In the end we get the sum $$f(x)=\sum_{n=-\infty}^\infty (-1)^n\frac2\pi\dfrac{ \sinh(\pi)}{n^2 + 1}e^{inx}$$