# Chaitanya's Random Pages

## March 27, 2012

### A cool integral based on the Gaussian

Filed under: mathematics — ckrao @ 11:45 am

In this post I want to show this nice integral I saw once during my graduate studies. $\displaystyle \int_0^{\infty} \exp \left(-(x-a/x)^2 \right)\ dx = \int_0^{\infty} \exp \left(-x^2 \right)\ dx = \frac{\sqrt{\pi}}{2}, \quad \text{Re}(a) > 0.$

The cool thing about it is the integral’s value is independent of $a$. The shape of the integrand changes as $a$ is varied but the area under the curve remains the same.

To prove this, we denote the left side as $f(a)$ (a continuous function of $a$) and differentiate under the integral sign. $\begin{array}{lcl} \frac{df}{da} &=& \frac{d}{da} \int_0^{\infty} \exp \left(-(x-a/x)^2 \right)\ dx \\&=& \int_0^{\infty} \frac{\partial}{\partial a} \exp \left(-(x-a/x)^2 \right)\ dx\\&=& \int_0^{\infty} \exp \left(-(x-a/x)^2 \right) \left( 2-\frac{2a}{x^2}\right) \ dx. \quad \quad (1)\end{array}$

The interchange of integral and derivative is valid here as the integrand and its derivative are continuous in the open interval of interest, plus the integral exists. Next, note that if we make the substitution $u = a/x$ then $dx = -a/u^2\ du$ and $\displaystyle f(a) = \int_{\infty}^0 \exp \left(-(a/u-u)^2 \right) \frac{(-a)}{u^2}\ du = \int_0^{\infty} \exp \left(-(x - a/x)^2 \right)\frac{a}{x^2}\ dx. \quad \quad (2)$

Note that this is not valid when $\text{Re}(a) \leq 0$. Combining (1) and (2), $\begin{array}{lcl} \frac{\partial f}{\partial a} &=& 2 \left(\int_0^{\infty} \exp \left(-(x-a/x)^2 \right) \ dx - \int_0^{\infty} \exp \left(-(x-a/x)^2 \right)\frac{a}{x^2}\right)\ dx\\ &=& 2\left( f(a) - f(a)\right) \\&=& 0.\end{array}$

Hence the integral is independent of $a$ in the region where it is differentiable as a function of $a$. Hence it is equal to its constant value in the limit $\text{Re}(a) \rightarrow 0+$, which is the Gaussian integral $\int_0^{\infty} \exp \left(-x^2 \right)\ dx = \frac{\sqrt{\pi}}{2},$

as we wished to show. Note that when $\text{Re}(a) < 0$ we can replace $a$ with $-a$ in the above and find that for $\text{Re}(a) > 0$ $\displaystyle \int_0^{\infty} \exp \left(-(x+a/x)^2 \right)\ dx = \exp (-4a) \int_0^{\infty} \exp \left(-(x-a/x)^2 \right)\ dx = \exp(-4a)f(a)$,

which though no longer constant has the same limit $f(0) = \sqrt{\pi}/2$ as $\text{Re}(a) \rightarrow 0+$.

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## 1 Comment »

1. […] Over at math.stackexchange I found an integral very similar to one I had blogged about earlier. […]

Pingback by Another integral independent of a parameter « Chaitanya's Random Pages — July 31, 2012 @ 10:25 am

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