how to find a complex integral when the singular point is on the given curve

Question

how to evaluate ∮1/(z-2) dz around the square with vertices 2±2i ,-2±2i the function is not analytic on z=2.but z=2 is on the given curve.so that can't apply cauchy,s integral formula.how can i do this?

Answer 1

As Ted Shifrin mentions, the integral does not make sense. One could try to make sense of the integral by making the contour take a semi-circular detour around $2$, and taking a limit as the radius of the arc goes to $0$, but this picture is not very interesting: if the pole is inside, you would get $2\pi i$ from Cauchy's integral theorem (depending on the orientation, you might get its negative). If the pole is outside the contour, you would get $0$ also from Cauchy's theorem (if a function is analytic on a simply connected domain, the integral of that function over a simple closed curve in the domain gets you zero).

Answer 2

The original problem is:

\begin{equation}\label{original}\tag{1} \oint\limits_{\Gamma}\frac{1}{z-2}\,\mathrm{d}z \end{equation}

Where $\Gamma$ is the following contour:

From the figure we see that: $\Gamma = \Gamma_1 + \Gamma_2 + \Gamma_3 + \Gamma_4$. And the contour orientation is the standard one: counterclockwise.

As stated before, strictly speaking, there is no solution to this problem. The fact that the pole lies on $\Gamma$, stops us from modifying it so that the pole is outside.

But let's pretend nothing stops us to modify $\Gamma$ to calculate the integral. So we modify $\Gamma$ as follows:

Where $\Gamma_\epsilon = L_1 + C_\epsilon + L_2 + L_3 + L_4 + L_5$, the and $C_\epsilon$ is the half circle with radius $\epsilon$ and center $z=2$.

In this case, the contour integral over $\Gamma_\epsilon$ is zero:

\begin{equation}\tag{2}\label{2} \oint\limits_{\Gamma_\epsilon}\frac{1}{z-2}\,\mathrm{d}z = 0 \end{equation}

But let's take a closer look using the segments of $\Gamma_\epsilon$ and letting $\epsilon\rightarrow0$

\begin{eqnarray}\label{modified_expanded} \lim_{\epsilon\rightarrow0} \oint\limits_{\Gamma_\epsilon}\frac{1}{z-2}\,\mathrm{d}z &=& \lim_{\epsilon\rightarrow0} \left(\int\limits_{L_1} + \int\limits_{L_2} + \int\limits_{L_3} +\int\limits_{L_4} + \int\limits_{L_5} +\int\limits_{C_\epsilon}\right)\frac{1}{z-2}\,\mathrm{d}z\\ &=& \oint\limits_{\Gamma}\frac{1}{z-2}\,\mathrm{d}z + \lim_{\epsilon\rightarrow0}\int\limits_{C_\epsilon}\frac{1}{z-2}\,\mathrm{d}z = 0 \tag{3}\label{3} \end{eqnarray}

We now parameterize $C_\epsilon : {z = 2 + \epsilon\,e^{i\theta}}$ with $\theta \in [3\pi/2,\pi/2]$, and insert in \ref{3}: \begin{eqnarray} \oint\limits_{\Gamma}\frac{1}{z-2}\,\mathrm{d}z &=& -\lim_{\epsilon\rightarrow0}\int\limits_{C_\epsilon}\frac{1}{z-2}\,\mathrm{d}z \\ &=&-\lim_{\epsilon\rightarrow0}\int\limits_{3\pi/2}^{\pi/2}\frac{i\epsilon e^{i\theta}}{\epsilon e^{i\theta}}\,\mathrm{d}\theta\nonumber\\ &=& -i \int\limits_{3\pi/2}^{\pi/2}\,\mathrm{d}\theta\nonumber\\ &=& -i(\pi/2 - 3\pi/2)\\ &=& i\pi \end{eqnarray}

So, if we allow $\Gamma$ to be modified we have the result: \begin{equation}\label{4}\tag{4} \oint\limits_{\Gamma}\frac{1}{z-2}\,\mathrm{d}z = i\pi \end{equation}

This makes completely geometric sense because if the pole were inside $\Gamma$, e.g. $\oint\limits_{\Gamma}\frac{1}{z-1}\mathrm{d}\,z$ the result would be $2i\pi$, i.e. if $\Gamma$ encloses totally (half) the pole the result is $2i\pi$ ($i\pi$).

If $\Gamma$ were the circle with radius 2, this would still hold, as the tangent at $z=2$ would still subtend an angle of $\pi$.

On the other hand, if $\Gamma$ were, for example, an equilateral triangle, with one corner at $z=2$, the result would be $i\pi/3$.

Finally, using $C_\epsilon$ "the other way round", would give the same result. Here Cauchy's integral theorem and the "the other way round integral" would give $2i\pi - i\pi = i\pi$ which is equal to \ref{4}