Compute the inner product For $n\in \mathbb{Z}$, let $e_n: [-\pi,\pi] \rightarrow \mathbb{C}$ be defeind by $e_n(t)=e^{int}$. We let $C[-\pi, \pi]$ denote the space of continuous $\mathbb{C}$-valued functions on the interval $[-\pi,\pi]$.
Compute the inner product $\langle f, \frac{1}{\sqrt{2\pi}}e_n \rangle$, for $f$ given by $f(t)= \cosh(t) = \frac{1}{2}(e^t+e^{-t})$, $-\pi \leq t\leq \pi$.
We have $\langle f, \frac{1}{\sqrt{2\pi}}e_n \rangle = \frac{1}{2\sqrt{2\pi}}\int_{-\pi}^{\pi} (e^{t}+e^{-t})e^{int}dt = \frac{1}{2\sqrt{2\pi}}\int_{-\pi}^{\pi} e^{t}e^{int}+e^{-t}e^{int}dt  = \frac{1}{2\sqrt{2\pi}}[(\frac{e^{\pi(in-1)}}{in-1}+\frac{e^{\pi(in+1)}}{in+1})- (\frac{e^{-\pi(in-1)}}{in-1}+\frac{e^{-\pi(in+1)}}{in+1})]$.
From here, is there any way to simplify? I think I need some sort of identities here
Thanks in advance!
 A: I start from the finish of your sentence.
First we apply linearity (on the generalized integral, will do the boundaries later)
\begin{align}
\mathcal{J} = \dfrac{1}{2^{\frac{3}{2}} \sqrt{\pi}} \int e^{i nt+t}dt + \dfrac{1}{2^{\frac{3}{2}} \sqrt{\pi}} \int e^{int-t} dt.
\end{align}
We solve each one separately.
\begin{align}
&\int e^{i nt+t}dt \quad \quad \text{substitute $u=int +t \mapsto \dfrac{du}{dt} = in+1$}\\
= &\dfrac{1}{in+1} \int e^u du\\
= &\dfrac{e^{u}}{in+1}\\
= &\dfrac{e^{int+t}}{in+1} \quad \quad \quad \text{undo original substitution}.
\end{align}
As for the other vector the process is exactly the same barring the plus becoming a minus. Thus the other integral is
\begin{align}
\dfrac{e^{int-t}}{in-1}.
\end{align}
As such we have
\begin{align}
\mathcal{J} = \dfrac{e^{int+t}}{2^{\frac{3}{2}} \sqrt{\pi} (in+1)} + \dfrac{e^{int-t}}{2^{\frac{3}{2}} \sqrt{\pi} (in-1)} + C\\
\implies \boxed{\mathcal{J} = -\dfrac{\left(\left(\mathrm{i}n-1\right)\mathrm{e}^{2t}+\mathrm{i}n+1\right)\mathrm{e}^{\left(\mathrm{i}n-1\right)t}}{2^\frac{3}{2}\sqrt{{\pi}}\left(n^2+1\right)}}.
\end{align}
Putting the integrating parameters in the final expression we have
\begin{align}
\mathcal{J}_{-\pi}^{\pi} &= \dfrac{\left(2\mathrm{e}^{2{\pi}}+2\right)n\sin\left({\pi}n\right)+\left(2\mathrm{e}^{2{\pi}}-2\right)\cos\left({\pi}n\right)}{2^\frac{3}{2}\sqrt{{\pi}}\left(\mathrm{e}^{\pi}n^2+\mathrm{e}^{\pi}\right)}\\
&= \dfrac{\mathrm{e}^{-{\pi}}\cdot\left(\left(\mathrm{e}^{2{\pi}}+1\right)n\sin\left({\pi}n\right)+\left(\mathrm{e}^{2{\pi}}-1\right)\cos\left({\pi}n\right)\right)}{\sqrt{2}\sqrt{{\pi}}\left(n^2+1\right)}
\end{align}
and I believe there can be no further simplification.
A: Note:$$\frac{1}{2\sqrt{2\pi}}\int_{-\pi}^{\pi} e^{t}e^{int}+e^{-t}e^{int}dt=\frac{1}{2\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{t+int}dt+\frac{1}{2\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-t+int}dt $$
You can check that $\int e^{t+int}dt=\frac{e^{t+int}}{1+in}$ and $\int e^{-t+int}dt=\frac{e^{-t+int}}{-1+in}$. So we have: $$\frac{1}{2\sqrt{2\pi}}\int_{-\pi}^{\pi} e^{t}e^{int}+e^{-t}e^{int}dt=\frac{1}{2\sqrt{2\pi}}[\frac{e^{t}e^{int}}{1+in}+\frac{e^{-t}e^{int}}{-1+in}]^{\pi}_{-\pi}=\frac{1}{2\sqrt{2\pi}}[\frac{e^\pi(-1)^n}{1+in}+\frac{e^{-\pi}(-1)^n}{-1+in}-\frac{e^{-\pi}(-1)^n}{1+in}-\frac{e^\pi(-1)^n}{-1+in}]=\frac{(-1)^n}{2\sqrt{2\pi}}[\frac{(e^\pi-e^{-\pi})}{1+in}-\frac{(e^\pi+e^{-\pi})}{-1+in}]=\frac{(-1)^n}{2\sqrt{2\pi}}[\frac{2\sinh(\pi)}{1+in}-\frac{2\cosh(\pi)}{-1+in}]$$
This is as simplified as I could get it.
