I have to find the value of $$\mathcal L^{-1}\left(\frac{1}{s(1+e^{as})}\right)$$ where $a \gt 0$ and $s$ is a complex variable and $\mathcal L^{-1}$ is the inverse Laplace transform.
In the book, this exercise is solved using The Cauchy residue theorem. The function $$F(s)=\frac{1}{s(1+e^{as})}$$ has a simple pole at $s=0$ and at $s_n=\frac{2n-1}{a}\pi i$ for $n=0, \pm1,\pm2,...$. Then the sum of residues becomes $$\sum _{n=-\infty}^{+\infty}\mathcal{Res}(s_n)=\frac12-\sum _{n=-\infty}^{+\infty}\frac1{(2n-1)\pi i}e^{t\frac{2n-1}{a}\pi i}=\frac12-\frac2\pi\sum _{n=1}^{+\infty}\frac1{(2n-1)}\sin{\left(t\frac{2n-1}{a}\pi\right)}$$
This is where I'm stuck. I don't understand why these sums are equal $$\sum _{n=-\infty}^{+\infty}\frac1{(2n-1)\pi i}e^{t\frac{2n-1}{a}\pi i}=\frac2\pi\sum _{n=1}^{+\infty}\frac1{(2n-1)}\sin{\left(\frac{2n-1}{a}t\pi\right)}$$
Hint: The series of interest consists of two parts, one with $n\geq 0$ and the other with $n<0$. Split the series up accordingly, and relabel the sum from $n=0$ to $\infty$ using $n'=1-n$. Now forget the label on $n'$ and combine the series.