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 \(a,c \gt 0\)
\begin{equation}
\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}
\end{equation}

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
\begin{equation}
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}
\end{equation}

Now make the substitution \(u=cx-(a/x)\)
\begin{equation}
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
\end{equation}

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

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{exp}\left(-\alpha x^{2} – \frac{\beta}{x^{2}}\right) \mathrm{d}x
\end{equation}
Notice that if we expand the expression inside of the integrand in equation \eqref{eq:cst1-1} we have
\begin{equation}
\left(cx – \frac{a}{x} \right)^{2} = c^{2} x^{2} – 2ca + \frac{a^{2}}{x^{2}}
\end{equation}
Rearrangement yields
\begin{equation}
-c^{2} x^{2} – \frac{a^{2}}{x^{2}} = -2ca \,- \left(cx – \frac{a}{x} \right)^{2}
\end{equation}
Thus, \(c^{2} = \alpha\), \(a^{2} = \beta\), and
\begin{equation}
-\alpha x^{2} – \frac{\beta}{x^{2}} = -2\sqrt{\alpha \beta} \,- \left(\sqrt{\alpha} x \,- \frac{\sqrt{\beta}}{x}\right)^{2}
\end{equation}
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.

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