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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.

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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*}

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  • $\begingroup$ Beautiful answer. $\endgroup$ – Oria Gruber Sep 20 '17 at 9:03
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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?

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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
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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 $$\int_{\gamma}\pi e^{\pi\overline{z}}dz=\int_{t=0}^1\pi e^{\pi(\overline{P}+(\overline{Q}-\overline{P})t)}(Q-P)dt=\frac{Q-P}{\overline{Q}-\overline{P}}[e^{\pi(\overline{P}+(\overline{Q}-\bar{P})t)}]_0^1=\frac{(Q-P)(e^{\pi \overline{Q}}-e^{\pi \overline{P}})}{\overline{Q-P}}.$$

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