Contour Integral

The contour integral of $f$ over a contour $C$ is

provided the limit exists and is independent of the choice of partition and points $c_k$.

Riemann Sum

Let $C$ be a contour with initial point $A$ and terminal point $B$.

A partition of $C$ is a finite set of points $P = \{z_0, z_1, \ldots, z_n\}$, occurring in order along $C$, with $z_0 = A$ and $z_n = B$.

$c_k$ is any point on the part of $C$ between $z_{k-1}$ and $z_k$.

The Riemann sum for this partition, for a function $f$ is

$\|P\|$ is the maximum length of the subarcs determined by $P$.

Evaluation via Parametrization

Let $f(z)$ be continuous on a set containing the contour $C$, and let $z = z(t)$, $a \le t \le b$, be a piecewise smooth parametrization of $C$.

When $z(t)$ is only piecewise smooth, the right side is evaluated piecewise.

Properties

Piecewise Contours

If $C$ is made of smooth curves $C_1, C_2, \ldots, C_n$, the integral is the sum over the pieces.

Reversed Orientation

Integral Triangle Inequality

Let $f(t) = u(t) + iv(t)$ be continuous on $[a,b]$.

ML Inequality

Let $f$ be continuous on a contour $C$, with $|f(z)| \le M$ for all $z$ on $C$, and let $L$ be the length of $C$.

To bound $|f(z)|$ from above, bound the numerator from above and the denominator from below using the reverse triangle inequality.

Cauchy-Goursat Theorem

Let $f$ be analytic in a simply connected domain $D$. If $C$ is a simple closed contour lying in $D$,

Deformation Invariance Theorem

Let $C_1$ and $C_2$ be 2 positively oriented simple closed contours, with $C_2$ lying inside the interior of $C_1$. Let $f$ be analytic in a domain $D$ that contains both $C_1$ and $C_2$, together with the region between them.

The integral over a closed contour depends only on which singularities the contour encloses, not on its shape.

To evaluate $\int_C f(z)\,\text{d}z$, replace $C$ by a small circle $|z-a| = \varepsilon$ around each singular point $a$ inside $C$.

Corollary of DIT

Let $z_0$ be a fixed complex number, and let $C$ be a simple closed contour with positive orientation such that $z_0$ lies inside $C$.

For $n \ge 2$, $\dfrac{-1}{(n-1)(z-z_0)^{n-1}}$ is a single-valued antiderivative on $\mathbb{C}\setminus\{z_0\}$. Every closed integral is $0$.

For $n=1$ the antiderivative is $\ln(z-z_0)$. Its argument increases by $2\pi$ per loop around $z_0$. This jump gives $2\pi i$.

When to Use DIT

Consider deformation invariance when all of these hold:

• $C$ is a closed contour
• The integrand is analytic on and inside $C$ except at finitely many points
• Those exceptional points lie inside $C$

Singularities Inside and Outside

Split the integrand by partial fractions into terms with 1 singular point each.

A singular point on $C$ is not covered and needs separate treatment.

Singularities outside $C$ never obstruct the deformation, since shrinking $C$ only sweeps its interior.

Here:

• $c_a$: coefficient of $\dfrac{1}{z-a}$, summed over the singular points $a$ inside $C$

Logarithm Shortcut

$\displaystyle\int_C \frac{g'(z)}{g(z)}\,\text{d}z$ cannot be evaluated directly as $\ln g$ at the endpoints.

The shortcut $\int_C F'\,\text{d}z = F(\text{end}) - F(\text{start})$ needs $F$ single-valued and analytic on a domain containing $C$. Here $\ln g = \ln|g| + i\arg g$, and $\arg g$ is multivalued when $g$ has zeros inside $C$.

A valid but laborious version uses the change in argument. The $\ln|g|$ part cancels on a closed curve.

Partial fractions with deformation is the preferred route.

Examples

Line Integral of $z$ from $0$ to $1+i$

Let $C$ be the line segment from $0$ to $1+i$, parametrized by $z(t) = t(1+i)$, $0 \le t \le 1$. $z'(t) = 1+i$.

$(1+i)^2 = 2i$ and $\displaystyle\int_0^1 t\,\text{d}t = \dfrac{1}{2}$

ML Bound on a Large Circle

Bound $\displaystyle\int_C \frac{z^2}{(z^2+9)^2}\,\text{d}z$ for $C: |z| = R > 3$.

$L = 2\pi R$. On $C$, $|z^2| = R^2$ and $|z^2+9| \ge R^2 - 9 > 0$.

The bound behaves like $\dfrac{2\pi}{R}$, so it tends to $0$ as $R \to \infty$.

Integral Using Partial Fractions

Show $\displaystyle\int_C \frac{2z}{z^2+2}\,\text{d}z = 4\pi i$ for $C: |z|=2$, positively oriented.

$z^2+2 = (z-i\sqrt{2})(z+i\sqrt{2})$ and $|\pm i\sqrt{2}| = \sqrt{2} < 2$, so both poles lie inside $C$.

Each term integrates to $2\pi i$ by the key integral.

Using the argument change instead, $g(2e^{i\theta}) = 4e^{2i\theta}\left(1+\tfrac{1}{2}e^{-2i\theta}\right)$. The first factor changes its argument by $4\pi$ and the second by $0$, so the integral is $i \cdot 4\pi = 4\pi i$.