# Contour Integrals

### Recall

For $$x\in\mathbb{R}, \int_a^b dxf(x) = \lim_{\Delta x\to 0}\sum f(x_i)\Delta x_i$$, but $$z$$ can go more places!

### Complex Extension

In the complex plane, multiple paths from $$a$$ to $$b$$, i.e. multiple contours. Common symbol for a contour is $$\gamma$$ and also has an orientation, i.e. direction, associated with it (i.e. from $$a$$ to $$b$$).

The integral then is defined as expected, $$\int_\gamma dz f(z) = \lim_{n\to\infty}\sum_k^n f(z_k)(z_k-z_{k-1})$$.

From this, it should be obvious that, $$0\leq |\int_\gamma dzf(z)| \leq L\gamma\cdot\sup_{z\in\gamma} f(z)$$, where $|γ| = Lγ =$ length of $$\gamma$$.

## Darboux Inequality

This is useful for provimg different things or estimating, $$|\int_\gamma dzf(z)| \leq L\gamma\cdot\sup_{z\in\gamma} f(z)$$

## Paramertization

We can parameterize $$\gamma$$ by $$\{z(t):0\leq t\leq 1\}$$, $$dz = \frac{dz}{dt}dt$$. Then $$L\gamma = \int_0^1dt |\dot{z}| = \int_\gamma |dz|$$

## Properties

All of the usual properties:

• Linear
• Integration by Parts (Undoing product rule)
• Piecewise Integration: $$\int_\gamma dzf(z) = \int_{\gamma_1}dzf(z) + \int_{\gamma_2}dzf(z)$$, where $$\gamma = \gamma_1 \# \gamma_2$$. The hash denotes the concatenation of the two contours.
• Reversing orientation brings out a negative sign: $$\int_\gamma dzf(z) = \int_{\overline{\gamma}}dzf(z)$$. $$\overline{\gamma}$$ and $$-\gamma$$ typically refer to the same thing.
• Antiderivatives

### Antiderivatives

$$\int_\gamma dz = \lim_{n\to\infty}\sum_k^n (1)\Delta z_k = (z_1-z_0) + (z_2 - z_1) \cdots = z_n - z_0 = b-a$$.

Suppose $$F(z)$$ is an antiderivative of a function $$f(z)$$ is holomorphic in some domain $$\mathcal{D}$$. If $$f(z) = \frac{dF}{dz}(z); z\in\mathcal{D}$$ then for a $$\gamma$$ that remains in $$\mathcal{D}$$, $$\int_\gamma dz f(z) = \int_{t_0}^{t_1} dt \frac{dz}{dt}(t)f(z(t)) = \int_{t_0}^{t_1} dt \frac{dz}{dt}(t)\frac{dF}{dz}(z(t)) = \int_{t_0}^{t_1}dt\frac{dF}{dt}(z(t)) = F(z(t_1)) - F(z(t_0)) = F(b) - F(z)$$.

So for functions that are holomorphic in a region $$\mathcal{D}$$, we have the Fundamental Theorem of Calculus you would expect. Because derivatives are what we expect, c.f. real counterparts, so are (indefinite) integrals. E.g. $$\int dz z^n = \frac{1}{n+1}z^{n+1}$$, $$\int dz\cos z = \sin z$$, etc.

The indefinite integrals arise from letting the endpoint be arbitrary, i.e. let the integral be a function of $$z$$.

An important implication is that, if $$f(z)$$ is holomorphic and the integral only depends on the end points then in the region $$\mathcal{D}$$ the integral is path independent. Further, because of path independence in this region, we have that a closed contour yields zero in this region $$\mathcal{D}$$, where the function is holomorphic. I.e. the region $$\mathcal{D}$$ needs to be simply connected to allow path deformation.

## Examples

 $$z = x+iy = r\exp(i\theta)$$ $$f(z) = u + iv$$ $$z(t) = x(t)+iy(t) = r(t)\exp(i\theta(t))$$ $$f(z) = f(z(t)) = u(t) + iv(t)$$

Choose whichever represenation is suited to you path $$\gamma$$. $$dz = dx + idy = \dot{z}dt = \left(\frac{dx}{dt} + i\frac{dy}{dt}\right)dt = d(r\exp(i\theta)) = ir\exp(i\theta)d\theta + \exp(i\theta) dr$$

### Example 1

Positively oriented semicircle with the line on the real axis and the arc curving to the positive imaginary axis (counter clockwise). $$z=r\exp(i\theta),dz=ir\exp(i\theta)d\theta+dr\exp(i\theta)$$. $$\int_\gamma dz z^2 = \int_{\gamma_1} dz z^2 + \int_{\gamma_2}dzz^2 = \int_{-r}^rdx (x+i0)^2 + \int_0^\pi (ir\exp(i\theta)d\theta) (r\exp(i\theta))^2 = \int_{-r}^rdx x^2 + r^3\int_0^\pi i\text{d}\theta\exp(i3\theta)=\frac{r^3}{3}-\frac{(-r)^3}{3} + r^3\left[\frac{\exp(3i\pi)}{3}-\frac{\exp(0)}{3}\right]=\frac{2r^3}{3}-\frac{2r^3}{3}=0$$.

Alternatively for one of the integrals: $$\int_{-R}^R dr\exp(i0)r^2\exp(2i0)=\exp(3i0)\frac{2r^3}{3}$$.

### Example 2 - Parameterization

$$f(z)=\exp(z)$$. $$x(t)=a t$$ $$y(t)=b t$$ $$z=at+bti$$ $$dz=(a+bi)dt$$ $$\exp(x(t)+iy(t)) = \exp(x(t)+iy(t))=\exp(t(a+bi))$$ $$\int_\gamma dz f(z) = \int_0^adx f(x+iy) + \int_0^b (idy) f(x+iy) + \int_1^0 [(a+bi)dt] f(x(t)+iy(t)) = \int_0^a dx f(x+i0) + \int_0^b dy f(a+iy) + \int_1^0 dt f(at+ibt) = (\exp(a)-\exp(0)) + \frac{i}{i}(\exp(a+bi)-\exp(a)) + \frac{a+bi}{a+bi}(\exp(0)-\exp(a+bi)) = 0$$

### Example 3 - Singularities

$$f(z)=z^{-1}$$

$$\int_\gamma dz f(z) = \int_0^{2\pi}izd\theta f(z) = i\int_0^{2\pi}d\theta = 2\pi i$$

Created: 2023-06-25 Sun 02:24

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