7
$\begingroup$

What is the integral of $$\int_{\Gamma}\pi e^{\pi\bar{z}}dz$$ where $\Gamma$ is the square with vertices at $0,1,1+i,i$ oriented anticlockwise?

I am badly stuck at this problem. I thought of using the Residue theorem by using $\bar{z}=\frac{|z|^2}{z}$, but we get an essential singularity. Using Laurent series about zero, I get $a_{-1}=\pi|z|^2$. Is this correct? how do we proceed? Any hints. Thanks beforehand.

| cite | improve this question | |
$\endgroup$
12
$\begingroup$

In addition to José Carlos Santos' answer, we can also utilize Green's theorem:

\begin{align*} \int_{\gamma} \pi e^{\pi \bar{z}} \, dz &= \int_{\gamma} \left( \pi e^{\pi \bar{z}} \, dx + i\pi e^{\pi \bar{z}} \, dy \right) \\ &= \int_{[0,1]^2} \left( \frac{\partial}{\partial x} i\pi e^{\pi \bar{z}} - \frac{\partial}{\partial y} \pi e^{\pi \bar{z}}\right) \, dxdy \\ &= 2\pi^2 i \int_{[0,1]^2} e^{\pi x}e^{-i\pi y} \, dxdy \\ &= 4(e^{\pi} - 1). \end{align*}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Beautiful answer. $\endgroup$ – Oria Gruber Sep 20 '17 at 9:03
7
$\begingroup$

You can't use the residue theorem in a non-analytic function!

Your integral can be naturally broken on four pieces. One of them is$$\int_\gamma\pi e^{\pi\overline z}\,\mathrm dz,\tag{1}$$with $\gamma\colon[0,1]\longrightarrow\mathbb C$ defined by $\gamma(t)=t$. But then $(1)$ is equal to$$\int_0^1\pi e^{\pi\overline{\gamma(t)}}\gamma'(t)\,\mathrm dt=\int_0^1\pi e^{\pi t}\,\mathrm dt=\left[e^{\pi t}\right]_{t=0}^{t=1}=e^{\pi}-e^0=e^{\pi}-1$$Can you compute the other three integrals?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I thought the function is meromorphic with an essential singularity at $0$, isnt it? Anyways, what are four parts you are breaking the integral to: the integral on the four sides of the square? $\endgroup$ – vidyarthi Sep 20 '17 at 8:44
  • 3
    $\begingroup$ @vidyarthi Your function ($e^{\pi\overline z}$) is differentiable nowhere! Forget about meromorphic. Yes, I was talking about the four sides of the square. $\endgroup$ – José Carlos Santos Sep 20 '17 at 8:45
  • $\begingroup$ why do you use the path $\gamma(t)=it$? and how do you determine the limits? $\endgroup$ – vidyarthi Sep 20 '17 at 8:52
  • $\begingroup$ @vidyarthi Its limits are $\gamma(0)=0$ and $\gamma(1)=i$. Actually, I made a mistake. Since you want to go around the square counterclockwise, I should have defined $\gamma(t)=t$, in order to go from $0$ to $1$. Then you should go from $1$ to $1+i$, then from $1+i$ to $i$ and finally go back to $0$, along the four sides of a square. $\endgroup$ – José Carlos Santos Sep 20 '17 at 8:56
  • 1
    $\begingroup$ @vidyarthi, The contour integral $\int_{\gamma} f(z) \, dz$ depends only on the value of the function $f$ along the curve $\gamma$. So as long as you can identify one analytic function that matches $f$ on $\gamma$, you can initiate the residue theorem. For your function $f(z) = \pi e^{\pi \bar{z}}$ it is simply that the circular contour allows one such choice, and the square contour doesn't. $\endgroup$ – Sangchul Lee Sep 20 '17 at 9:35
3
$\begingroup$

Hint. Note that if $\gamma$ is a segment from $P$ to $Q$ then $\gamma(t)=P+(Q-P)t$ for $t\in [0,1]$ and $$\int_{\gamma}\pi e^{\pi\overline{z}}dz=\int_{t=0}^1\pi e^{\pi(\overline{P}+(\overline{Q}-\overline{P})t)}(Q-P)dt=\frac{Q-P}{\overline{Q}-\overline{P}}[e^{\pi(\overline{P}+(\overline{Q}-\bar{P})t)}]_0^1=\frac{(Q-P)(e^{\pi \overline{Q}}-e^{\pi \overline{P}})}{\overline{Q-P}}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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