Integrals

Real and Imaginary parts

If f(z) be a complex function, then there exists u(x),v(x) such that f(z) = u(x) + iv(x).

Definition

Let f(z) be a complex function. There exists u(x),v(x) such that f(z) = u(x) + iv(x). Then

\int f(z) = \int u(x) + i \int v(x)

Example

For f(z) = (t + i)^2 we have

\begin{align*} \int_0^1 (t - i)^2 dt &= \int_0^1 t^2 -1 dt - \int_0^1 2ti dt \end{align*}

Example

For f(z) = e^t\cos(t), let u = e^t and v = \cos(t). Then

\begin{align*} \int_0^{\pi/2} e^t\cos(t)\, dt &= \frac{e^{\pi/2 - 1}}{2} \end{align*}

Parametrizing contours

Recall e^{it} = \cos t + i \sin t. Then

e^{t+it} = e^t\cos t + i e^t \sin t

We use eulers method and then separate into both real and complex parts.

Recall from lecture 7 that a parameterized curve in the complex plane is

C \colon z(t) = x(t) + i y(t)

for some continuous real valued functions. We say z(t) is differentiable if both x(t) and y(t) are.

We will integrate complex functions over contours, unions of smooth curves.

In order to do Riemann sums, we split our contour into many different intervals and then take a test under each contour.

\sum_{i=1}^n f(c_i)(z_i - z_{i-1}) = \sum_{i=1}f(c_i)\Delta z

Then we can define the integral.

\int_C f(z) dz = \lim_{n \to \infty} \sum_{i=1}f(c_i)\Delta z

Then we can do some coolio

\begin{align*} \int_C f(z) dz &= \lim_{n \to \infty} \sum_{i=1}f(c_i)\Delta z\\ &= \int_a^b f(z(t))z^\prime(t) dt \end{align*}

Example

For f(z) = z^2, and a contour line from 0 to 3i we need to intgerate. Here we choose the parameterization z(t) = 3ti for 0 \leq t \leq 1. This is clearly a smooth contour. Thus

\begin{align*} \int_C f(z) dz &= \int_0^1 (z(t))^2z^\prime(t) dt\\ &= \int_0^1 (3ti)^2(3i)\\ &= -9i \end{align*}

If we instead parametrize z(t) = t^2i where 0 \leq t \leq \sqrt{3}. Then z^\prime(t) = 2ti.

\begin{align*} \int_C f(z) dz &= \int_0^1 (z(t))^2z^\prime(t) dt\\ &= \int_0^{\sqrt{3}} (t^2i)^2(2ti)\\ &= -9i \end{align*}

Cauchys Contours

Theorem (Deformation of contours)

Let C be a simple closed positively oriented contour, and f(z) be analytic on some simple connected domain D that contains C. Then

\oint_C f(z) = 0

Theorem (Deformation of contours)

Let C_1,C_2 be two simple closed positively oriented contours such that

• C_1 is in the interioir of C_2, and

• f(z) is analytic on some domain D that contains C_1 and C_2 and the area inbetween.

Then \int_{C_1} f(z) = \int_{C_2} f(z).

Proof

C_2 = A +B, and C_1 = E + F. Then

\int_{C_2} - \int_{C_1} = \int_A f + \int_B f - \int_E f - \int_F f

Corollary

Let C be a simple closed positively oriented contour containing some point z_0 \in \mathbb{C} in its interioir. Then

• \displaystyle\int_C \frac{1}{z - z_0} dz = 2\pi i

• \displaystyle\int_C (z - z_0)^n dz = 0 for all integers n \not = -1.

Proof

The function f(z) = 1/(z - z_0) is always analytic except for z = z_0, so we can deform the contour to some circle containing z_0.

\int_C \frac{1}{z - z_0} dz = \int_{C^+_r(z_0)}\frac{1}{z - z_0} dz = 2\pi i.

The function f(z) = (z- z_0)^n is entire for every nonnegative integer n, so it is always 0. When n is negative and not 1, we choose a paramitization and obtain 0.

Theorem (Extended Cauchy-Goursat)

Let C and C_1,\dots,C_k be simple closed positively oriented contours such that

• C_i are all in the interioir of C

• The interiors of C_i are all disjoint.

• f(z) is analytic on some domain D that contains all these contours and the region between C and C_i.

Then \int_{C_1} f(z) = \sum_{i=1}^k \int_{C_i} f(z).

Fundamental Theorems of Integration

So far we have only done definite integrals over contours. Lets expand our purview.

Theorem

If f(z) is an analytic function in a simply connected domain D, such that z_0 is in D and C is any contour in D with an initial point z_0 and terminal point z, then

• \displaystyle F(z) = \int_C f(w) dw

• F(z) is a well defined function of z which is analytic on D such that F^\prime(z) = f(z).

We will often write F(z) = \int^z_{z_0} f(w) dw to mean that we are free to choose any contour C. This integral is called the antiderivative.

Proof

We first need to show that using any two contours will lead us to the same integral.

Then we must prove that the derivative is equivalent to the function we began with.

F(z + \Delta z) - F(z) = \int_{C_1} f(w) dw - \int_{C_2} f(w) dw

Due to continuity, we can keep shrinking \Delta z so that |\Delta z| > \delta. Since |w - z| > \delta, we have that |f(w) - f(z)| < \varepsilon.

So F^\prime(z) = f(z).

If C is a positively oriented curve, then integrating over |dz| gives the length of the curve C.

\int_C |dz| = \text{len } C

Bounding integrals

Theorem (Bounding integrals)

If f(z) is continuous on a contour C with a maximum of M.

\begin{align*} \left|\int_C f(z) dz\right| &\leq \int_C |f(z)| |dz|\\ &\leq M\int_C |dz|\\ &= M \cdot \text{len } C \end{align*}

Theorem (Definite integrals)

Let f be analytic in a simply connected domain D. If z_1 and z_2 are points in D joined by a contour C, then

\int_C f(z) dz = \int_{z_1}^{z_2} f(z) dz

Cuachys integral formula

Theorem (Cauchy integral formula)

Let f be analytic in a simply connected domain D, and let C be a simply closed positively oriented contour. If z_0 is in the interior of C,

f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_0} dz.

Proof

Since z_0 is in the interior of your contour, there exists some r > 0 such that C_r(z_0) is contained in the interior of C. Then

\begin{align*} \frac{1}{2\pi i} \int_{C_{r}^+(z_0)} \frac{f(z)}{z - z_0} dz &=\\ &= \frac{f(z_0)}{2\pi i} \int_{C_{r}^+(z_0)} \frac{1}{z - z_0}dz\\ &= \frac{f(z_0)}{2\pi i} 2\pi i = f(z_0) \end{align*}

It comes down to proving

\int_{C_{r}^+(z_0)} \frac{f(z)}{z - z_0} dz = \int_{C_{r}^+(z_0)} \frac{f(z_0)}{z - z_0} dz

Since f(z) is continuous in D, for every \varepsilon > 0 there exists a \delta > 0 such that

|z - z_0| < \delta \implies |f(z) - f(z_0)| < \varepsilon.

Then

\begin{align*} \left| \int_{C_{\delta/2}^+(z_0)} \frac{f(z)}{z - z_0} dz - \int_{C_{\delta/2}^+(z_0)} \frac{f(z_0)}{z - z_0} dz\right| &=\\ &= \left| \int_{C_{\delta/2}^+(z_0)} \frac{f(z) - f(z_0)}{z - z_0} dz \right|\\ &\leq \int_{C_{\delta/2}^+(z_0)} \frac{|f(z) - f(z_0)|}{|z - z_0|} dz\\ &\leq \frac{\varepsilon}{\delta/2} \int_{C_{\delta/2}^+(z_0)} |dz|\\ &\leq \frac{\varepsilon}{\delta/2}(2\pi (\delta/2)) = 2\pi\varepsilon \end{align*}

Then the difference of the two integrals is 0, and thus

\int_{C_{r}^+(z_0)} \frac{f(z)}{z - z_0} dz = \int_{C_{r}^+(z_0)} \frac{f(z_0)}{z - z_0} dz.

Divide each side by 2\pi i and use deformation of contours to prove

f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_0} dz.

Example

Find

\int_{C_2^+(0)} \frac{e^z}{z - i} dz.

Here z_0 = i and f(z) = e^z. Then by the Cauchy integral theorem

\int_{C_2^+(0)} \frac{e^z}{z - i} dz = 2\pi e^i.

Theorem (Cauchy integral formula for nth derivatives)

Let f be analytic in a simply connected domain D, and let C be a simply closed positively oriented contour. If z_0 is in the interior of C, then for any nonnegative integer n \geq 0,

f^(n)(z_0) = \frac{n!}{2\pi i} \int_C \frac{f(z)}{(z - z_0)^{n+1}} dz.

Review

When is a function analytic?

If f is analytic, then the CR equations hold

u_x = v_Y

u_y = - v_x

A function must be continuous and differentiable.

[Study definitions]

Definition

Let z,w \in \mathbb{C}, and f \colon \mathbb{C} \to \mathbb{C} be a function. The limit of f at z is w if for all D_\varepsilon(w), there exists D_\delta(z) such that f(D_\delta(z)) \subset D_\varepsilon(w).

Equivalently, if |z - z_0| < \delta then |f(z) - w_0| < \varepsilon.

Definition

A complex function is continuous if for all z \in D, we have

\lim_{z \to z_0} = f(z_0).

Definition

Let z_0 \in \mathbb{C}, and f be a complex valued function. Then the derivative of f is defined

f^\prime(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}

if limit exists. If f^\prime(z_0) exists, we say f is differentiable at z_0.

Definition

Let z_0 \in \mathbb{C} a7nd f be a complex valued function. The function f is holomorphic at a point z_0 if it is differentiable at at every point in some neighborhood D_\varepsilon(z_0).

A function is harmonic if u_{xx} + u_{yy} = 0.

If f(z) = u + vi, then both u and v are harmonic.

Given a harmonic function, you can find a harmonic conjugate.