Contour Integration — Examples

The midpoint sum for the uniform $N$-partition is $S_N={\sum }_{j=0}^{N-1}\exp ((j+1\mathrm{/}2)w\mathrm{/}N)w\mathrm{/}N$; by [L1] these sums converge componentwise to the contour integral.

By [L2], the same limit is ${\int }_0^1\exp (tw)wdt$. Since $(\exp (tw){)}^{\mathrm{\prime }}=w\exp (tw)$ by [L3], this equals $\exp w-1$.

Alternatively, [L3] makes $\exp$ its own primitive on all of $\mathbb{C}$, and since that derivative is $\exp$ itself it is continuous, so [L4] applies and gives the same endpoint increment $\exp w-\exp 0$ directly.

By [L1] the polynomial $(z-i{)}^3$ is entire, hence continuous, and $(z-i{)}^4\mathrm{/}4$ is a primitive of it whose derivative is that same continuous polynomial; so the hypotheses of [L2] hold and [L2] gives $((1-i{)}^4-(-i{)}^4)\mathrm{/}4=(-4-1)\mathrm{/}4=-5\mathrm{/}4$.

By [L1] the polynomial $z$ is entire with continuous derivative and has primitive $z^2\mathrm{/}2$, again meeting the hypotheses of [L2]; both $\sigma$ and $\rho$ have endpoints $0,1$, so [L2] gives $1\mathrm{/}2$ on each.

Direct substitution into [L3] gives ${\int }_0^1\sigma (t){\sigma }^{\mathrm{\prime }}(t)dt$ and ${\int }_0^1\rho (t){\rho }^{\mathrm{\prime }}(t)dt$; each is the endpoint difference of $\gamma (t{)}^2\mathrm{/}2$, confirming the same values with both orientations explicit.

Along $\gamma$, $\overline{\gamma (t)}{\gamma }^{\mathrm{\prime }}(t)=e^{-i(\pi -t)}(-i)e^{i(\pi -t)}=-i$, so [L1] integrated from $0$ to $\pi$ gives $-\pi i$.

On a segment $z(t)=z_0+td$ with $0\le t\le 1$, direct integration gives $\int \bar{z}dz=\overline{z_0}d+\mathrm{\mid }d{\mathrm{\mid }}^2\mathrm{/}2$. For $-1\to 2i$ we have $z_0=-1$ and $d=1+2i$, so $\mathrm{\mid }d{\mathrm{\mid }}^2=5$ and the value is $(-1)(1+2i)+5\mathrm{/}2=3\mathrm{/}2-2i$; for $2i\to 1$ we have $z_0=2i$ and $d=1-2i$, so $\mathrm{\mid }d{\mathrm{\mid }}^2=5$ and the value is $(-2i)(1-2i)+5\mathrm{/}2=-3\mathrm{/}2-2i$.

Add the two segment values by [L2]: $(3\mathrm{/}2-2i)+(-3\mathrm{/}2-2i)=-4i$. Since $-\pi i\mathrm{\ne }-4i$, the unequal results prove path dependence.

On $\mathrm{\mid }z\mathrm{\mid }=2$, the reverse triangle inequality from [L3] gives $\mathrm{\mid }z-3\mathrm{\mid }\ge 1$, so $\mathrm{\mid }1\mathrm{/}(z-3)\mathrm{\mid }\le 1$; [L2] gives semicircle length $2\pi$, and [L1] gives the first bound.

On $z=2+it$, $0\le t\le 1$, one has $\mathrm{\mid }z-i\mathrm{\mid }\ge 2$ and $\mathrm{\mid }z+i\mathrm{\mid }\ge \sqrt{5}$, so [L3] gives $\mathrm{\mid }1\mathrm{/}(z^2+1)\mathrm{\mid }\le 1\mathrm{/}(2\sqrt{5})$.

The segment length is $1$ by [L2], so [L1] gives the second bound. Both contours stay a positive distance from their poles.

Since $dz=ir{e}^{it}dt$ and $z-a=r{e}^{it}$, their quotient is the constant $i$; integration over $[0,\pi ]$ and $[0,2\pi ]$ gives $i\pi$ and $2\pi i$.

The full-circle value agrees with [L1], and [L2] gives the negative values on reversed paths.

The positive-radius hypothesis ensures that the denominator never vanishes.

On $\mathrm{\mid }z\mathrm{\mid }=1$, the exponential tail is bounded by the convergent numerical series $\sum 1\mathrm{/}n!$, so the partial sums converge uniformly; division by $z$ preserves the bound because $\mathrm{\mid }z\mathrm{\mid }=1$.

By [L2] and [L4], integrating the finite sum ${\sum }_{n=0}^N{z}^{n-1}\mathrm{/}n!$ gives $2\pi i$ from the $n=0$ term and $0$ from every $n\ge 1$ term.

Apply [L3] to the uniform convergence in step 1.1 and pass to the limit in step 1.2. The circle excludes $z=0$, so division is defined.

On $[0,1]$, parametrize the pieces by $r+(R-r)t$, $R{e}^{2\pi it}$, $R-(R-r)t$, and $r{e}^{2\pi i(1-t)}$, respectively. Their endpoints match in this order, so [L1] defines a closed concatenation.

Repeated application of [L2] gives the total integral as the sum of the four oriented integrals, with the reversed radial and inner-circle orientations carrying their signs.

By [L3], the piece lengths are $R-r$, $2\pi R$, $R-r$, and $2\pi r$. This verifies rectifiability and bookkeeping without evaluating the integral by Cauchy's theorem or choosing a logarithm branch.

On each affine segment, the four Stieltjes components in [L1] reduce to ordinary integrals against constant coordinate derivatives. Recombination gives ${\int }_0^1{\gamma }_j(t){\gamma }_j^{\mathrm{\prime }}(t)dt$ on that segment.

Each segment integral is $(z_1^2-z_0^2)\mathrm{/}2$. Adding the three endpoint increments by [L3] telescopes to $(i^2-0^2)\mathrm{/}2=-1\mathrm{/}2$.

Formula [L2] gives the same three parametric integrals. The one-sided derivatives at the two corners need not agree.

By [L1], ${\int }_{\gamma }1dz=1(1-0)=1$, whereas ${\int }_{{\gamma }^{-}}1dz=1(0-1)=-1$.

The values differ, in agreement with [L2], so reversal does not preserve the oriented complex integral even though it preserves the absolute integral.

Since the circle is closed, [L1] gives ${\int }_{\gamma }1dz=0$.

By [L2], its absolute integral is its positive length $2\pi r$.

Thus equality fails: $0<2\pi r$. The values still satisfy the inequality in [L3].

Both traces are the same circle of radius $r$, but their speeds are $r$ and $2r$.

By [L1], $L(\gamma )=2\pi r$ and $L(\eta )=4\pi r$.

Since $r>0$, the lengths differ. This does not contradict [L2], because the double covering is not a bijective reparametrization of the single traversal.

By [L2], the forward segment has integral $1$ and the reversed segment has integral $-1$.

The values differ. This is consistent with [L1], whose exact hypothesis is increasing reparametrization and therefore does not include orientation reversal.