Engineering Mathematics: Calculus and Complex Analysis

Contour integral around simple closed path

A contour integral is an integral of a complex function f(z) taken around a closed path C in the complex plane, written \oint_C f(z)\,dz. A simple closed path is one that does not cross itself. For a function that is analytic on and inside C except for isolated poles, the residue theorem states that \oint_C f(z)\,dz = 2\pi j \sum_k \operatorname{Res}(f, z_k), where the sum runs over the poles z_k lying inside C, provided C is traversed counter-clockwise. A clockwise traversal changes the sign.

A practical procedure:

• Locate the singularities — set the denominator to zero; a pole occurs at each such point where the numerator does not also vanish.
• Decide which poles are enclosed — a pole outside the path contributes nothing. For a circle |z-z_0|=R, compare the distance from z_0 to the pole with R; the pole is inside only if that distance is less than R.
• Compute residues — for a simple pole of f=g/h with h(a)=0 and h'(a)\neq 0, \operatorname{Res}(f,a)=g(a)/h'(a). For a double pole such as g(z)/z^2 with g analytic at 0, the residue is g'(0). A factor that cancels the singularity (a removable singularity) gives zero residue.
• Apply the factor 2\pi j and add the contributions of the enclosed poles.

If f has no poles inside C, Cauchy's theorem gives \oint_C f(z)\,dz = 0. Entire functions such as e^z, \cos z and non-negative powers z^n therefore integrate to zero over any closed path. For an integer n\neq -1, \oint_C z^n\,dz = 0 because z^{n+1}/(n+1) is a single-valued antiderivative, whereas \oint_C z^{-1}\,dz = 2\pi j. A function such as \sec z is analytic inside a contour whenever all its poles (2k+1)\pi/2 lie outside it, which must be checked by comparing moduli.

Vocabulary:

• enclose — lie inside the closed path. e.g. the path encloses the origin if the origin lies inside it.
• removable singularity — a point where a factor cancels, so the function has a finite limit and zero residue. e.g. \sin z / z at z=0.
• Example: For |z|=2 and \oint \frac{1}{z(z-3)}\,dz, only z=0 is enclosed, with residue -\frac{1}{3}, so the integral equals -\frac{2\pi j}{3}.

Convergence of infinite series

A series \sum a_n converges when its partial sums approach a finite limit. The condition a_n \to 0 is necessary but not sufficient, as the harmonic series \sum 1/n shows. The choice of test depends on the form of the terms.

• Ratio test — with L=\lim |a_{n+1}/a_n|: L<1 gives convergence, L>1 gives divergence, and L=1 is inconclusive. It suits terms containing c^n or n!. For \sum n^d/c^n the limit is 1/c for any d, so the series converges whenever c>1. When 0<c<1 the terms grow exponentially and the series diverges.
• p-series — \sum 1/n^p converges if and only if p>1. When c=1, the series \sum n^d/c^n becomes \sum n^d, which converges only for d<-1. A series with p=\tfrac12 diverges.
• Telescoping — write each term as a difference, for example \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}. The partial sums then collapse to a simple expression whose limit can be read off.
• Factorial series — \sum 1/n! converges, because successive ratios tend to 0; summed from n=1 it equals e-1.
• Geometric subsequences — when the exponents in the denominators follow a pattern, group the terms into a geometric series \sum r^k, which converges for |r|<1 with sum \frac{a}{1-r}. If the ratio of successive terms stays below a fixed number less than 1, the series converges by comparison.

Example: for \sum_{n\ge 1} n^2/2^n, the ratio \frac{(n+1)^2}{2n^2}\to \frac12<1, so the series converges.

Double integral over triangular region

A double integral \iint_D f(x,y)\,dA over a triangular region D is evaluated as an iterated integral, and the limits come from the boundary lines of the triangle. Sketch the triangle, identify its vertices, and decide whether to integrate in y first (with x-limits fixed) or in x first. If one edge's boundary changes formula partway through, split the region into two pieces.

• Inner integral — treat the other variable as a constant; for example \int x^2\,dy = x^2 y and \int y^2\,dy = \frac{y^3}{3}. Constant factors can be pulled out of the integral.
• Limits — each edge gives a bound expressed in the other variable; the outer limits are the coordinates of the vertices in that variable.
• Symmetry — if the triangle is symmetric about an axis and the integrand is odd in the corresponding variable, the integral is zero. A triangle with an apex at the origin and a base symmetric about the y-axis is a typical case.

Polynomial integrands such as 3(x^2+y^2) are handled term by term. Each term is integrated first over the inner variable and then over the outer one.

Example: for the triangle with vertices (0,0), (1,0) and (0,1), \iint_D x\,dA = \int_0^1\!\int_0^{1-x} x\,dy\,dx = \int_0^1 x(1-x)\,dx = \frac{1}{6}.

Extrema of function on closed interval

The Extreme Value Theorem states that a function continuous on a closed interval [a,b] attains both an absolute maximum and an absolute minimum there. These occur either at critical points inside (a,b), where f'=0 or f' does not exist, or at the two endpoints.