A kernel trick for integrals

November 10, 2022. I present a simple trick for doing integrals by swapping the argument of a kernel.


Consider an integral transform with kernel $K(x, y)$. In general, this gives two distinct transforms,

\[T_1f(y) = \int_{\Omega_1} f(x) K(x, y) \, \text{d}x, \quad T_2f(x) = \int_{\Omega_2} f(y) K(x, y) \, \text{d}y,\]

where $T_i$ integrates over argument $i$, and $\Omega_i$ denotes the corresponding domain of integration. If everything is smooth enough to swap integrals (i.e. Fubini’s theorem), then

\[\begin{align*} \int_{\Omega_1} f(x)\cdot T_2g(x) \,\text{d}x & = \int_{\Omega_1} f(x)\left[\int_{\Omega_2} g(y) K(x, y) \, \text{d}y\right] \text{d}x \\ & = \int_{\Omega_2} g(y)\left[\int_\Omega f(x) K(x, y) \, \text{d}x\right] \text{d}y \\ & = \int_{\Omega_2} T_1f(y) \cdot g(y)\, \text{d}y. \end{align*}\]

For a symmetric kernel $K(x, y) = K(y, x)$ and $\Omega_1 = \Omega_2 = \Omega$, we have $T_1 = T_2 = T$, and our result simplifies to:

The symmetric kernel trick.
For an integral transform $T$ defined by a symmetric kernel, $$ \int_{\Omega} f(x) \cdot Tg(x)\, \text{d}x = \int_{\Omega} Tf(y) \cdot g(y)\, \text{d}y. $$

From a pure math standpoint, we’ve basically just observed that the integral transforms $T_1$ and $T_2$ are dual,

\[\langle f, T_2 g\rangle = \langle T_1 f, g\rangle,\]

with respect to a suitably defined inner product $\langle \cdot, \cdot\rangle$. But this turns out to be a useful trick for doing real-life integrals!

Full disclosure. I didn’t come up with this hack, but stole it (with some customizations) from Ramanujan. Also, I’m ignoring many mathematical subtleties! The joys of being a physicist.

The Voigt integral

Let’s take everyone’s favourite example, the 1D Fourier transform:

\[T_\text{F} f(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i\omega x} \, \text{d}x.\]

We can consult a table and pick out, for instance, the pairs

\[\begin{align*} f(x) & = e^{-\alpha x^2}, \quad T_\text{F}f(\omega) = \frac{1}{\sqrt{2\alpha}} e^{-\omega^2/4\alpha} \\ g(x) & = e^{-\beta |x|}, \quad T_\text{F}g(\omega) = \sqrt{\frac{2}{\pi}} \cdot \frac{\beta}{\beta^2 + \omega^2}. \end{align*}\]

Then our kernel trick gives

\[\begin{align*} \int_{-\infty}^\infty \sqrt{\frac{2}{\pi}}\frac{\beta e^{-\alpha^2 x^2}}{\beta^2 + x^2} \, \text{d}x & =\frac{1}{\sqrt{2\alpha}} \int_{-\infty}^\infty e^{-x^2/4\alpha^2 + \beta|x|} \, \text{d}x. \end{align*}\]

The RHS is straightforward to express in terms of the complementary error function:

\[\text{erfc}(z) = \frac{2}{\sqrt{\pi}}\int_z^\infty e^{-x^2}\, \text{d}x.\]

We complete the square, defining $2\alpha u = x + 2\alpha^2\beta$ to find

\[\begin{align*} \int_{-\infty}^\infty e^{-x^2/4\alpha^2 - \beta|x|} \, \text{d}x & = 2\int_{0}^\infty e^{-x^2/4\alpha^2 - \beta x} \, \text{d}x\\ & = 4\sqrt{\alpha} e^{(\alpha\beta)^2}\int_{2\alpha^2\beta}^\infty e^{-u^2} \, \text{d}u\\ & = 2\sqrt{\pi\alpha} e^{(\alpha\beta)^2}\text{erfc}(\alpha\beta). \end{align*}\]

We can finally conclude that

\[\int_{-\infty}^\infty \frac{ e^{-\alpha^2 x^2}}{\beta^2 + x^2} \, \text{d}x = \frac{\sqrt{2}\pi \alpha}{\beta} e^{(\alpha\beta)^2}\text{erfc}(\alpha\beta).\]

I call this the “Voigt integral”after the related convolution in spectroscopy.

Exercise 1. The Hankel transform $$ \mathcal{H}^{(\nu)}f(k) = \int_0^\infty f(r) rJ_\nu(kr) \, \text{d}r $$ is defined by an asymmetric kernel $K(r, k) = rJ_\nu(kr)$, where $J_\nu$ is a Bessel function of the first kind of order $\nu$.
(a) Using the kernel trick, show that $$ \int_0^\infty k \mathcal{H}^{(\nu)}\left[\frac{f(r)}{r}\right](k) g(k) \, \text{d}k = \int_0^\infty f(k)\mathcal{H}^{(\nu)}g (k)\, \text{d}k. \tag{1} \label{hankel} $$ (b) Apply $(\ref{hankel})$ to a judicious choice of Hankel transform pairs to derive the expression $$ \int_0^\infty e^{-\alpha^2 u/2} K_0(\beta\sqrt{u})\, \text{d}u = -\frac{1}{\alpha^2} e^{-\beta^2/2\alpha^2}\text{Ei}\left(-\frac{\beta^2}{2\alpha^2}\right), $$ where $K_0$ is a modified Bessel function of the second kind and $\text{Ei}$ is the exponential integral, a special function defined by $$ \text{Ei}(z) = -\int_{-z} \frac{e^{-t}}{t} \, \text{d}t. $$

Mordell integrals

Here’s a fancier example, again using the Fourier transform. Consider the Mordell integral

\[h(z; \tau) = \int_{-\infty}^\infty \frac{e^{\pi i \tau x^2 - 2\pi zx}}{\cosh(\pi x)} \, \text{d}x,\]

with $\Im(\tau) > 0$ to ensure convergence. Note that this is a product of functions which are self-dual under the Fourier transform, up to a change in their parameters:

\[\begin{align*} f(x) & = e^{i\alpha x^2 - \beta x}, \quad T_\text{F}f(\omega) = \frac{1}{\sqrt{-2i\alpha}} e^{i(\beta-i\omega)^2/4\alpha} \\ g(x) & = \frac{1}{\cosh(\gamma x)}, \quad T_\text{F}g(\omega) = \sqrt{\frac{\pi}{2}} \frac{1}{\gamma\cosh(\pi\omega/2\gamma)}. \end{align*}\]

The kernel trick (and the change of variable $x = 2\pi u$) now gives

\[\begin{align*} h(z; \tau) & = \frac{\sqrt{\pi}}{\sqrt{-i\alpha}}\int_{-\infty}^\infty \frac{e^{i(2\pi z-ix)^2/4\pi\tau}}{\cosh(x/2)} \, \text{d}x \\ & = \frac{\sqrt{\pi}e^{i\pi z^2/\tau}}{\sqrt{-i\alpha}}\int_{-\infty}^\infty \frac{e^{-ix^2/4\pi\tau+zx/\tau}}{\cosh(x/2)} \, \text{d}x \\ & = \frac{2\pi^{3/2}e^{i\pi z^2/\tau}}{\sqrt{-i\alpha}}\int_{-\infty}^\infty \frac{e^{-i\pi u^2/\tau+2\pi zu/\tau}}{\cosh(\pi u)} \, \text{d}u \\ & = \frac{2\pi^{3/2}e^{i\pi z^2/\tau}}{\sqrt{-i\alpha}} h\left(-\frac{z}{\tau}; -\frac{1}{\tau}\right). \tag{2}\label{h} \end{align*}\]

This seems like a neat result!

Exercise 2. Ramanujan defined the related integral $$ F_\omega(z) = \int_{-\infty}^\infty \frac{e^{-\pi\omega x^2 + 2\pi x}\sin(\pi x z)}{e^{2\pi x}-1} \, \text{d} x. $$ We'll end with a few exercises on this theme.
(a) Define $\varphi$ by $$ h(z; \tau) = -\frac{2i}{\tau}e^{-(\pi i\tau/4 + \pi i z)}\varphi\left(z + \frac{\tau-1}{2}, \tau\right). $$ Prove that $F_\omega(z)$ and $h(z; \tau)$ are related by $$ F_{-i\tau}(2iz) = \frac{1}{2i\tau}\left[\varphi(z, t) - \varphi(-z, \tau)\right]. \tag{3} \label{varphi} $$ (b) Using equation $(\ref{h})$ and $(\ref{varphi})$ or otherwise, show that $$ F_\omega(z) = -\frac{i}{\sqrt{\omega}} e^{-\pi z^2/4\omega} F_{1/\omega}\left(\frac{iz}{\omega}\right). $$ (c) Set $\omega = \alpha^2$ and $z \to \alpha z/\sqrt{\pi}$. Deduce from part (b) that, for $\alpha\beta = 1$ and $\alpha, \beta > 0$, $$ \sqrt{\alpha}e^{z^2/8}\int_{-\infty}^\infty \frac{e^{-\pi^2\alpha^2x^2}\sin(\sqrt{\pi}\alpha x z)}{e^{2\pi x}-1} \text{d}x = \sqrt{\beta}e^{-z^2/8}\int_{-\infty}^\infty \frac{e^{-\pi^2\beta^2x^2}\sinh(\sqrt{\pi}\beta x z)}{e^{2\pi x}-1} \text{d}x. $$
Written on November 10, 2022
Math   Hacks