An Incredibly Overpowered Integration Technique – Part 1

Today, we look at a beautiful transformation/substitution which is not presented in many textbooks and hence is often not part of the standard arsenal of tools for tackling integration problems. It is called the Cauchy-Schlömilch transformation “(after Cauchy who knew it by 1823, and the German mathematician Oskar Schlömilch (1823–1901) who popularized it in an 1848 textbook)”. The result follows:

Theorem (Cauchy-Schlömilch transformation): Let $a, b > 0$ and assume that $f$ is a continuous function for which the integrals below are convergent and are construed as Cauchy principal values. Then

$\displaystyle \boxed{\int_0^{\infty} f((ax - b/x)^2) \: \mathrm{d}x = \frac{1}{a} \int_0^{\infty} f(x^2) \: \mathrm{d}x}.$

Proof: Start with the integral on the LHS and use the substitution

$\displaystyle az = \frac{b}{x} \implies \mathrm{d}x = -\frac{b}{az^2} \: \mathrm{d}z.$

So,

$\displaystyle I = \int_0^{\infty} f((ax - b/x)^2) \: \mathrm{d}x = \frac{b}{a} \int_0^{\infty} \frac{1}{z^2} f((az - b/z)^2) \: \mathrm{d}z.$

Adding these two representations and duly taking note of the fact that $z$ is just a dummy variable, we get

$\displaystyle 2I = \int_0^{\infty} \left( 1 + \frac{b}{ax^2}\right) f((ax - b/x)^2) \: \mathrm{d}x .$

Now, use the substitution

$\displaystyle ax - \frac{b}{x} = y \implies \mathrm{d}y = \left( a + \frac{b}{x^2}\right) \: \mathrm{d}x = a \left( 1 + \frac{b}{ax^2}\right) \: \mathrm{d}x.$

Thus,

$\displaystyle 2I = \frac{1}{a} \int_{-\infty}^{\infty} f(y^2) \: \mathrm{d}y = \frac{2}{a} \int_0^{\infty} f(x^2) \: \mathrm{d}x,$

and this completes the proof. $\blacksquare$

Let’s see an application of the Cauchy–Schlömilch transformation using an easy example. Say we are asked to evaluate

$\displaystyle I = \int_0^{\infty} \frac{x^2}{x^4+1} \: \mathrm{d}x.$

One can tidy this up by writing

$\begin{aligned} I = \int_0^{\infty} \frac{x^2}{x^4+1} \: \mathrm{d}x &= \int_0^{\infty} \frac{1}{x^2 + \frac{1}{x^2}} \: \mathrm{d}x \\ &=\int_0^{\infty} \frac{1}{(x-1/x)^2+2} \: \mathrm{d}x \\ & \stackrel{(\star)}{=} \int_0^{\infty} \frac{1}{x^2 + 2} \: \mathrm{d}x \\ &= \frac{1}{\sqrt{2}} \arctan \left( \frac{x}{\sqrt{2}} \right) \Bigg\vert_0^{\infty} \\ &= \frac{\pi}{2 \sqrt{2}}, \end{aligned}$

where $(\star)$ is a consequence of the Cauchy-Schlömilch transformation for $a=b=1$ .

Another beautiful integral that yields to this method of attack is

$\displaystyle I = \int_{-\infty}^{\infty} e^{x - c \sinh ^2 x} \: \mathrm{d}x ,$

for an arbitrary parameter $c$ . We can write

$\begin{aligned} I = \int_{-\infty}^{\infty} e^{x - c \sinh ^2 x} \: \mathrm{d}x &= \int_{-\infty}^{\infty} e^{-\left( \frac{\sqrt{c} e^x - \sqrt{c}e^{-x}}{2} \right)^2} e^x \: \mathrm{d}x \\ &\stackrel{u = e^x}{=} \int_0^{\infty} e^{-\left( \frac{\sqrt{c}}{2} u - \frac{\sqrt{c}}{2} u^{-1} \right)^2} \: \mathrm{d}u \\ & \stackrel{(\star)}{=} \frac{2}{\sqrt{c}} \int_0^{\infty} e^{-u^2} \: \mathrm{d}u \\ &\stackrel{(\star \star)}{=} \frac{2}{\sqrt{c}} \cdot \frac{\sqrt{\pi}}{2} \\ &= \sqrt{\frac{\pi}{c}}, \end{aligned}$

where $(\star)$ is a consequence of the Cauchy-Schlömilch transformation for $a=b=\sqrt{c}/2$ and $(\star \star)$ is the classical result that the integral of the Gaussian function from 0 to infinity is $\sqrt{\pi}/2$ .

We will see in the next post how the Cauchy-Schlömilch transformation is actually the special case of a more generalized substitution that proves incredibly useful in the evaluation of complicated definite integrals.

Challenge Problem

Use the Cauchy-Schlömilch transformation to show that

$\displaystyle \int_0^{\infty} \left( \frac{x^2}{x^4 + 2ax^2 + 1} \right)^r \: \mathrm{d}x = \frac{\mathrm{B} \left( r - \frac{1}{2} , \frac{1}{2} \right)}{2^{r+1/2} (1+a)^{r - 1/2}}.$

Here $\mathrm{B}(\cdot, \cdot)$ is the beta function. Try playing around with your choice of values for $a$ and $r$ and see what wonderful results you get!