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July 31, 2012

Another integral independent of a parameter

Filed under: mathematics — ckrao @ 10:25 am

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

In this post I will show that the following integral holds.

\displaystyle \int_0^{\infty} \frac{x^2}{x^2 + (x^2 - a)^2}\ dx = \int_0^{\infty} \frac{1}{1 + \left(x - a/x\right)^2}\ dx = \frac{\pi}{2}, \quad \text{Re}(a) > 0.

In other words, the value does not depend on a. This looks similar to this integral based on the Gaussian:

\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 solution for the second integral is given here.

We prove it through similar means. 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} \frac{1}{1+ (x-a/x)^2 }\ dx \\&=& \int_0^{\infty} \frac{\partial}{\partial a} \frac{1}{1+ (x-a/x)^2}\ dx\\&=& \int_0^{\infty} \frac{1}{\left[1+ (x-a/x)^2 \right]^2} \left( 2-\frac{2a}{x^2}\right) \ dx. \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

\begin{array}{lcl} \frac{1}{2} \frac{df}{da} &=& \int_0^{\infty} \frac{1}{\left[1 + (x-a/x)^2 \right]^2} \left(1 - \frac{a}{x^2} \right)\ dx \\& = & \int_0^{\infty} \frac{1}{\left[1 + (x-a/x)^2 \right]^2}\ dx - \int_0^{\infty} \frac{1}{\left[1 + (x-a/x)^2 \right]^2}. \frac{a}{x^2}\ dx\\&=& \int_0^{\infty} \frac{1}{\left[1 + (x-a/x)^2 \right]^2}\ dx - \int_{\infty}^0 \frac{1}{\left[1 + (a/u-u)^2 \right]^2} \ (-du)\\& =& \int_0^{\infty} \frac{1}{\left[1 + (x-a/x)^2 \right]^2}\ dx - \int_0^{\infty} \frac{1}{\left[1 + (a/u-u)^2 \right]^2} \ du\\&=& 0 \end{array}

(Note that this is not valid when \text{Re}(a) \leq 0.)

It follows that 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

\displaystyle \int_0^{\infty} \frac{1}{1+ x^2}\ dx = \left[\arctan x\right]_0^{\infty} = \frac{\pi}{2},

as we wished to show.

The same idea can be applied to general integrals of the form \displaystyle \int_0^{\infty} g((x-a/x)^2)\ dx, where g(x) is any differentiable function so that the integral exists. We have seen the integral is independent of a for \text{Re}(a) > 0 when g(x) = \exp(-x) and g(x) = 1/(1+x).

\displaystyle \int_0^{\infty} g((x-a/x)^2)\ dx = \int_0^{\infty} g(x^2)\ dx, \text{Re}(a) \leq 0.

It seems remarkable to me that the transformation x \rightarrow |x - a/x| for any a with positive real part somehow preserves the area under the curve of g(x^2) for positive x. 🙂

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