# Calculating the residues of $f(z)=\frac{e^{az}}{1+e^z}$

Let $$f(z)=\frac{e^{az}}{1+e^z}$$ where $0<a<1$

Can anyone help me find the residues of this function?

So $$e^z+1=0 \Rightarrow z=i\pi(1+2k)$$ where $k\in \mathbb{Z}$, so these are simple poles (if someone could explain a simple way of showing this that'd be great, other than expansion)

$$\lim_{z\rightarrow i\pi(1+2k)}\frac{(z-i\pi(1+2k))e^{az}}{1+e^z}=\lim_{z\rightarrow i\pi(1+2k)}\frac{a(z-i\pi(1+2k))e^{az}+e^{az}}{e^z}=e^a$$

So i'm trying to evaluate $\int_{\infty}^\infty f(z)$ you see, so will I need to pick a contour with fixed height otherwise the integral around the contour will be equal to $2\pi i \sum_{n=0}^\infty e^a$

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Seems like use l'Hospital is straightforward enough: $$\lim_{z\to i\pi(1+2k)}\frac{(z-i\pi(1+2k)) \mathrm{e}^{az}}{1+\mathrm{e}^z} = \mathrm{e}^{a i\pi(1+2k)} \lim_{z\to 0}\frac{z \mathrm{e}^{az}}{1-\mathrm{e}^z} = \mathrm{e}^{a i\pi(1+2k)} \lim_{z\to 0}\frac{\mathrm{e}^{az}\left(1 + z a\right)}{-\mathrm{e}^z} = -\mathrm{e}^{a i\pi(1+2k)}$$

Added: Then the integral is the sum over poles: $$\int_{-\infty}^\infty \frac{\mathrm{e}^{a z}}{1+\mathrm{e}^z} \mathrm{d} z = 2 \pi i \sum_{k=0}^\infty -\mathrm{e}^{a i\pi(1+2k)} = -2 \pi i \left(\frac{e^{i \pi a}}{-1+e^{2 i \pi a}} \right) = \frac{\pi}{\sin(\pi a)}$$

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When $g(z)$ has a zero of order one at $z = z_0$ (true here since all $g''(z_0) \neq 0$), the residue of $f(z)/g(z)$ at $z = z_0$ is $f(z_0)/g'(z_0)$.

So for the pole at $z = (1 + 2\pi)ki$, the residue is $e^{(1 + 2\pi)aki}/e^{(1 + 2\pi)ki} = - e^{(1 + 2\pi)aki}$.

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Ah brilliant, I see what I did wrong –  Freeman Nov 27 '11 at 14:41