Using residue theory, show that $\oint_C\frac{e^{z/2}}{1+e^z}dz = 2\pi$

Question

Using residue theory, show that $$\oint_C\frac{e^{z/2}}{1+e^z}dz = 2\pi$$

I've been attempted this problem using residue theory and the Cauchy integral formula that over a closed contour containing a pole (root at $\pi i$),

We should be able to confirm $2\pi$ using the equation $f(i\pi)\cdot 2\pi i$, but when I do this I am getting $-2\pi$, can anyone assist me?

Answer 1

Assuming that $i\pi$ is the only pole inside your contour, and that the contour is counter clockwise oriented, $$ \operatorname{Res}\limits_{z=i\pi} \frac{e^{z/2}}{1+e^z} = \left. \frac{e^{z/2}}{e^z} \right|_{z=i\pi} = \frac{e^{i\pi/2}}{e^{i\pi}} = -i, $$ so $$ \int_C \frac{e^{z/2}}{1+e^z}\,dz = 2\pi i \cdot (-i) = 2\pi. $$

Answer 2

mrf's answer already shows you the way, but because of what you wrote at the end of your question, you apparently want to use Cauchy's theorem

$$f(z_0)=\frac1{2\pi i}\int\limits_\gamma \frac{f(z)}{z-z_0}dz$$

whenever $\;f(z)\;$ is analytic on the simple, closed path $\,\gamma\,$ and inside the domain it encloses.

In our case, we can take the Taylor series around $\;\pi i\;$ :

$$e^z=-1-\frac{(z-\pi i)}{1!}-\frac{(z-\pi i)^2}{2!}-\ldots=-\sum_{k=0}^\infty\frac{(z-\pi i)^k}{k!}\implies$$

$$1+e^z=-(z-\pi i)-\mathcal O((z-\pi i)^2)\implies\frac1{1+e^z}=-\frac1{z-\pi i}\frac1{1+\mathcal O(z-\pi i)}$$

Notice that for $\,|z-\pi i|<1\;$ , we have that

$$\frac1{1+\mathcal O(z-\pi i)}=1-\mathcal O(z-\pi i)+\left(\mathcal O(z-\pi i)\right)^2-\ldots$$

and thus our integral, when $\,C\,$ encloses only the pole $\,\pi i\,$ , is

$$\oint\limits_C\frac{e^{z/2}}{1+e^z}dz=-\oint\limits_C\frac{e^{z/2}}{z-\pi i}dz=\left.-2\pi i\cdot e^{z/2}\right|_{z=\pi i}=-2\pi i\cdot i=2\pi$$

where $\,f(z):=e^{z/2}\implies f(\pi i)=e^{\pi i/2}=i\;$

Answer 3

To verify it with $f(i\pi)$, you need to modify the integrand as $$ f(i\pi )= \frac{1}{2\pi i} \oint_C\frac{e^{z/2}\frac{(e^z+1)}{(z-i\pi)}}{(z-i\pi)}dz, $$

where $f(z)$ is given by

$$f(z)=\begin{cases} e^{z/2}\frac{(e^z+1)}{(z-i\pi)}, \, z\neq i\pi \\ -i,\,z=i\pi \end{cases}.$$

Note:

$$\lim_{z\to i\pi}\frac{(e^z+1)}{(z-i\pi)}=-1.$$

Answer 4

The singularities of $\dfrac{e^{z/2}}{1+e^z}$ are all simple, so we can compute the residue at $\pi i$ by using L'Hospital: $$ \begin{align} \lim_{z\to\pi i}\frac{(z-\pi i)e^{z/2}}{1+e^z} &=e^{\pi i/2}\lim_{z\to\pi i}\frac{z-\pi i}{1+e^z}\\ &=i\lim_{z\to\pi i}\frac1{e^z}\\[6pt] &=-i \end{align} $$ Thus, the integral along any path circling $\pi i$ once counter-clockwise, and no other singularities, is $2\pi i(-i)=2\pi$. Thus, $$ \oint_C\frac{e^{z/2}}{1+e^z}\,\mathrm{d}z=2\pi $$