Residues for Integrals

Some resources:

• For guidelines for what techniques to use, see this very detailed note.

• Some good computations here.

• The reverse triangle shows up here often! \begin{align*} {\left\lvert {z-w} \right\rvert} \geq {\left\lvert { {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert}} \right\rvert} \implies {1\over {\left\lvert {z-w} \right\rvert}}\leq {\left\lvert {1\over {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert}} \right\rvert} \\ {\left\lvert {z+w} \right\rvert} = {\left\lvert {z -(-w)} \right\rvert} \geq {\left\lvert { {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert}} \right\rvert} \implies {1\over {\left\lvert {z+w} \right\rvert}}\leq {\left\lvert {1\over {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert}} \right\rvert} \\ .\end{align*}

• Also often useful: \(e^{-x}\) is monotonically decreasing on \({\mathbf{R}}\), so \begin{align*} a\leq b \implies e^{-a} \geq e^{-b} .\end{align*}

• All residue theorems assume counterclockwise orientation. If your curve is clockwise (e.g. in the lower half-plane), so it traces out \([-R, R]\) on \({\mathbf{R}}\)), then a negative sign is introduced.

• Some standard contours:

  

Let \(\deg\) denote the \(z{\hbox{-}}\)adic valuation, so \(\deg(p(x)/q(x)) = \deg(p) - \deg(q)\).

For \(\int_{\mathbf{R}}f\): Semicircle. Always get \begin{align*} \int_{\mathbf{R}}f = \sum_{z_0\in {\mathbb{H}}} \mathop{\mathrm{Res}}_{z=z_0}f(z) .\end{align*}

For \(\int_{{\mathbf{R}}_{\geq 0}} f\): use symmetry to convert to \(\int_{\mathbf{R}}\) (e.g. if \(f\) is even), or find a symmetric ray and use a sector:

Semicircles. Realize \begin{align*} \int_{\mathbf{R}}f(z)\cos(z) = \Re\qty{\int_{\mathbf{R}}f(z)e^{iz}} .\end{align*}

This works for \(f(z)e^{iz}\) where \(f(z) = p(z)/q(z)\) with \(\deg q \geq \deg p+2\) by the ML estimate, or \(\deg q \geq \deg p +1\) by Jordan’s lemma.

Set \(z=e^{i\theta}\), then \begin{align*} 2\cos(z) = z + z^{-1}, \quad 2i\sin(z) = z - z^{-1}, && \,d\theta= {\,dz\over iz} .\end{align*} Then \(\int_{-\pi}^\pi f(\theta)\,d\theta\to \int_{{\left\lvert {z} \right\rvert} = 1} f(z) \,dz\), which reduces to a residue count in \({\mathbb{D}}\).

If \(f(z) \approx \cos(z), \sin(z), \tan(z), \cdots\): check if \(\displaystyle\int f \approx \Re\displaystyle\int (g(e^{iz}))\).

• \(\int_{{\mathbf{R}}_{\geq 0}} f(x)x^\alpha\) for \(\alpha\) rational with \({\left\lvert {\alpha} \right\rvert} < 1\)

• For \(f\) rational with \(\deg f\leq -2\), take a keyhole

  

  • A useful fact: \begin{align*} \lim_{{\varepsilon}\searrow 0} \ln(x-i{\varepsilon}) = \ln(x) + 2\pi i .\end{align*}

  • Another useful fact: for any constant \(c\), \begin{align*} \lim_{x\to 0} {x\ln(x) \over x^n + c} = 0 .\end{align*} Note that you can’t drop the constant!

• \(\int_{{\mathbf{R}}_{\geq 0}} f(x) \log(x)\): indented semicircular contours

• When a log trick is useful: if \(f(z) = \log(z)g(z)\) where \(\int g(z)\) is easy, try a contour that is a rotation of \({\mathbf{R}}\). The substitution usually yields \(\log(\zeta x) \leadsto \ln{\left\lvert {x} \right\rvert} + i\theta\), so \(\int \log(\zeta x)g(\zeta x) = \int \log(x)g(x) + \int i\theta g(x)\) if \(g\) is \(\zeta{\hbox{-}}\)invariant.

The replication trick: if \(f(z) = f(\zeta_m z)\) for some \(m\), try a sector. If \(f(z) = f(z + ib)\) for some \(b\), a rectangle.

For real integrals: \begin{align*} \operatorname{PV} \int_{-\infty}^{\infty} \frac{f(x)}{x-x_{0}} d x=\lim _{\epsilon \backslash 0}\left(\int_{-\infty}^{x_{0}-\epsilon} f(x) d x+\int_{x_{0}+\epsilon}^{\infty} f(x) d x\right) . .\end{align*} The same principle works for complex contours, e.g. for \(\int_{\mathbf{R}}p(z)/q(z)\) where \(q\) has real roots:

The contours along the real line converge to the principal values of the real integral.

If \(C_R\) is a semicircular contour subtending an angle of \(\theta\), \begin{align*} {\left\lvert {\int_{C_R} f\, } \right\rvert}\leq MR\theta ,\end{align*} so provided \(M = { \mathsf{O}}\qty{1\over R^{1+{\varepsilon}}}\), this goes to zero as \(R\to\infty\). If \(M = { \mathsf{O}}(R)\), use Jordan’s lemma!

Compare to the above computation of \(\int_{\mathbf{R}}{ \cos(x) \over x^2 + 1}\). Taking \(\alpha = 1\), this yields \begin{align*} {\left\lvert { \int_{C_R} { e^{iz} \over z^2 + 1 } \,dz} \right\rvert} \leq \pi \sup_{z\in C_R} {\left\lvert {1\over z^2 +1} \right\rvert} \leq {\pi \over R^2 - 1}\to 0 .\end{align*}