# Confusion in Cauchy Integral Formula

I am trying to calculate a simple contour integral in three different ways and am getting three different results.

$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz$$

Method $$1$$:

Write $$\frac{1}{z^2 + 1} = \frac{1}{(z+i)(z-i)} = \frac{1/(z+i)}{z-i}$$. Since $$i \in B_2(0)$$, we can apply the Cauchy Integral Formula to get

$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz = \int_{\vert z \vert = 2} \frac{1/(z+i)}{(z-i)}dz = 2\pi i f(i)$$

where $$f(z) = \frac{1}{z+i}$$. Hence, the integral evaluates to $$\pi$$ since $$f(i) = \frac{1}{2i}$$.

Method $$2$$:

This is almost identical to the above. Write $$\frac{1}{z^2 + 1} = \frac{1}{(z+i)(z-i)} = \frac{1/(z-i)}{z-(-i)}$$. Since $$-i \in B_2(0)$$, we can apply the Cauchy Integral Formula to get

$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz = \int_{\vert z \vert = 2} \frac{1/(z-i)}{(z-(-i))}dz = 2\pi i g(-i)$$

where $$g(z) = \frac{1}{z-i}$$. Hence, the integral evaluates to $$-\pi$$ since $$g(-i) = -\frac{1}{2i}$$.

Method $$3$$:

Using partial fractions, $$\frac{1}{z^2+1} = \frac{1}{(z+i)(z-i)} = -\frac{1}{2i}\left(\frac{1}{z+i} - \frac{1}{z-i}\right)$$. So,

$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz = -\frac{1}{2i} \left(\int_{\vert z \vert = 2} \frac{1}{z+i}dz - \int_{\vert z \vert = 2}\frac{1}{z-i}dz\right) = 0$$

since both integrals are $$2\pi i$$.

Which of these three methods is correct and why are the other two wrong? (By symmetry I feel like the third one is correct, though I also think it may be that the first two are correct but somehow represent integrating in opposite directions along the contour.)

• The function has two poles inside the contour. Commented Sep 6, 2021 at 2:11

The first two are wrong. You are forgetting an important hypothesis is Cauchy's Integral Formula. The functions $$f(z)=\frac 1{z+i}$$ and $$f(z)=\frac 1{z-i}$$ are not analytic in any region containing $$\{z: |z| \leq 2\}$$, so the formula is not applicable for these functions.

• They are not analytic inside the curve $|z|=2.$ They are analytic, where they are defined. Commented Sep 6, 2021 at 0:36
• @ThomasAndrews I am , of course, talking about applicability of Cachy's Integral Formual, so I am only looking at anayticity inside the contour. Commented Sep 6, 2021 at 5:02

There are two poles inside $$|z|=2$$, so Cauchy's integral formula leads to $$\oint_{|z|=2}\frac{dz}{z^2+1}=2\pi i \sum_{a\in\{-i,i\}}\operatorname*{Res}_{z=a}\left(\frac{1}{z^2+1}\right)=0.$$

A simpler way to provide a quick reply is to notice that $$\frac{1}{z^2+1}$$ is holomorphic in the annulus $$2\leq |z|\leq R$$ for any $$R>2$$, so

$$\oint_{|z|=2}\frac{dz}{z^2+1} = \oint_{|z|=R}\frac{dz}{z^2+1}.$$

On the other hand for any $$z$$ such that $$|z|=R$$ we have $$|z^2+1|\geq R^2-1$$, so

$$\left|\oint_{|z|=R}\frac{dz}{z^2+1}\right| \leq \frac{2\pi R}{R^2-1}$$ where the RHS converges to zero as $$R\to +\infty$$. In particular the original integral equals zero.

• (+1) This was the process I used. It does not require the computation of any residues.
– robjohn
Commented Sep 6, 2021 at 22:59