## Evaluation of Definite Integrals Via the Cauchy-Schlomilch Transformation

The Cauchy-Schlomilch transformation is one of many so called master formulas for evaluating definite integrals. The idea is that such formulas can be used directly on a variety of integrals and also indirectly by various additional transformations such as differentiating with respect to certain parameters, changes in variables, etc. Here, I present the basic Cauchy-Schlomilch transformation as it appeared in the references listed below. Note that in [1] there is a corollary to this result, but I will not present it here.

For $$a,c \gt 0$$

\int\limits_{0}^{\infty} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \mathrm{d} x = \frac{1}{c} \int\limits_{0}^{\infty} f(u^{2}) \mathrm{d}u
\tag{1}
\label{eq:cst1-1}

Proof: Let $$y=a/(cx)$$, so that \eqref{eq:cst1-1} becomes
\begin{align}
I &= \int\limits_{0}^{\infty} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \mathrm{d} x \\
&= -\frac{a}{c} \int\limits_{0}^{\infty} \frac{1}{y^{2}} f\Big[\left(\frac{a}{y}-cy \right)^{2} \Big] \mathrm{d} y \\
&= \frac{a}{c} \int\limits_{0}^{\infty} \frac{1}{x^{2}} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \mathrm{d} x
\tag{2}
\label{eq:cst1-2}
\end{align}

Combining equations \eqref{eq:cst1-1} and \eqref{eq:cst1-2} yields

2I =
\frac{1}{c} \int\limits_{0}^{\infty} \frac{1}{x^{2}} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \left(c+\frac{a}{x^{2}} \right) \mathrm{d} x
\tag{3}
\label{eq:cst1-3}

Now make the substitution $$u=cx-(a/x)$$

2I = \frac{1}{c} \int\limits_{-\infty}^{\infty} f(u^{2}) \mathrm{d}u = \frac{2}{c} \int\limits_{0}^{\infty} f(u^{2}) \mathrm{d}u

Solving for I yields our desired result.

Notice that $$a$$ vanishes.

## Example 1

\begin{align}
\int\limits_{0}^{\infty} \mathrm{exp}\left(-\Big[ax-\frac{b}{x}\Big]^{2n}\right) \mathrm{d}x
&= \int\limits_{0}^{\infty} \mathrm{exp}\left(-\Big[(ax-\frac{b}{x})^{n}\Big]^{2}\right) \mathrm{d}x \\
&= \frac{1}{a} \int\limits_{0}^{\infty} \mathrm{exp}\left(-[y^{n}]^{2}\right) \mathrm{d}y \\
&= \frac{1}{2an} \Gamma\left(\frac{1}{2n}\right)
\end{align}

## Example 2

\int\limits_{0}^{\infty} \mathrm{exp}\left(-\alpha x^{2} – \frac{\beta}{x^{2}}\right) \mathrm{d}x

Notice that if we expand the expression inside of the integrand in equation \eqref{eq:cst1-1} we have

\left(cx – \frac{a}{x} \right)^{2} = c^{2} x^{2} – 2ca + \frac{a^{2}}{x^{2}}

Rearrangement yields

-c^{2} x^{2} – \frac{a^{2}}{x^{2}} = -2ca \,- \left(cx – \frac{a}{x} \right)^{2}

Thus, $$c^{2} = \alpha$$, $$a^{2} = \beta$$, and

-\alpha x^{2} – \frac{\beta}{x^{2}} = -2\sqrt{\alpha \beta} \,- \left(\sqrt{\alpha} x \,- \frac{\sqrt{\beta}}{x}\right)^{2}

Now we apply the Cauchy-Schlomilch transformation to obtain our final result
\begin{align}
\int\limits_{0}^{\infty} \mathrm{exp}\left(-\alpha x^{2} – \frac{\beta}{x^{2}}\right) \mathrm{d}x
&= \mathrm{e}^{-2\sqrt{\alpha \beta}} \int\limits_{0}^{\infty} \mathrm{exp}\left(-\Big[\sqrt{\alpha} x – \frac{\sqrt{\beta}}{x}\Big]^{2}\right) \mathrm{d}x \\
&= \frac{\mathrm{e}^{-2\sqrt{\alpha \beta}}}{\sqrt{\alpha}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\
&= \frac{\mathrm{e}^{-2\sqrt{\alpha \beta}}}{\sqrt{\alpha}} \frac{\sqrt{\pi}}{2} \lim_{y \to \infty} \mathrm{erf}(y) \\
&= \frac{1}{2} \sqrt{\frac{\pi}{\alpha}} \mathrm{e}^{-2\sqrt{\alpha \beta}}
\end{align}
Note that this integral was evaluated with a different method in Example 8.4.1 here.

References

1. T. Amdeberhan, M. L. Glasser, M. C. Jones, V. H. Moll, R. Posey, D. Varela. The Cauchy-Schlomilch Transformation. 2010.
2. G. Boros, V. Moll. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. 2004.
3. M. L. Glasser. A remarkable property of definite integrals. Math. Comp., 40(40):561–563, 1983.