Holomorphic integrals

Question

I am struggling to understand how the center and radius effect a certain circular contour.

e.g. $ \int _{\gamma} \frac {z^{2}+1} {e^{z}(z-1)(z+1)^{2}} dz $

can anyone explain this?

Answer 1

In general, if $\gamma$ is a closed contour, then the Residue Theorem tells us that

$$\int_\gamma f(z)\,dz = 2\pi i\sum_{z_i}\text{Res}\left(f(z),z_i\right)$$

where $z_i$ are the poles of the meromorphic function $f(z)$ that are enclosed within the contour of integration $\gamma$.

In particular, the value of the integral does not actually depend on, say, the radius of the circle or its center as long as the poles enclosed by the contour of integration do not change.

In your example, we have

$$f(z) = \frac{z^2+1}{e^z(z-1)(z+1)^2}$$

which has a simple pole at $z_1 = 1$ and a pole of order $2$ at $z_2 = -1$.

If we choose a circular contour $\gamma$ that enclosed both of these poles, like the circle with radius $2$ centered at the origin for example, then we obtain

\begin{align} \int_\gamma f(z)\,dz &= 2\pi i\left[\text{Res}\left(f(z), 1\right) + \text{Res}\left(f(z), -1\right)\right]\\ &= 2\pi i\left[\lim\limits_{z\rightarrow 1} \frac{z^2+1}{e^{z}(z+1)^2} + \lim\limits_{z\to -1}\frac{d}{dz}\frac{z^2+1}{e^{z}(z-1)}\right]\\ &=2\pi i\left[\frac{1}{2e} + \lim\limits_{z\to -1}\frac{x(x^2-2x+3}{e^x(x-1)^2}\right]\\ &=2\pi i\left[\frac{1}{2e} + \frac{3e}{2}\right]\\ &=\frac{\pi i}{e}\left(1+3e^2\right) \end{align}

As specified earlier, any other circular, or indeed closed contour that contains both of the poles of $f(z)$ in its interior will yield the same result.