Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $2\pi i \sum_{n=0}^\infty e^a$

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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: $$ \int_{-\infty}^\infty \frac{\mathrm{e}^{a z}}{1+\mathrm{e}^z} \mathrm{d} z = 2 \pi i \sum_{k=0}^\infty -\mathrm{e}^{a i\pi(1+2k)} = -2 \pi i \left(\frac{e^{i \pi a}}{-1+e^{2 i \pi a}} \right) = \frac{\pi}{\sin(\pi a)} $$

share|improve this answer

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 $e^{(1 + 2\pi)aki}/e^{(1 + 2\pi)ki} = - e^{(1 + 2\pi)aki}$.

share|improve this answer
    
Ah brilliant, I see what I did wrong –  Freeman Nov 27 '11 at 14:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.