## Posts Tagged ‘definite integral’

### Another Definite Integral

August 1, 2009

My students claimed they were doing a calculation that required

$\int_{-\infty}^\infty e^{-x^4}dx$.

I’m not sure what physical situation brought up such a question, but we can find the answer anyway. Let’s kill infinity birds with one stone by evaluating

$\int_{0}^\infty x^{\alpha} e^{-x^\beta} dx$.

and treating my students’ problem as a special case.

First define the gamma function by

$\Gamma(n) \equiv \int_0^\infty t^{n-1}e^{-t}dt$.

I have never understood why $\Gamma(n)$ involves $(n-1)$ as the power of $t$, rather than just $n$. It makes even less sense when you consider $\Gamma(n) = (n-1)!$ for natural numbers $n$.

In the definition of the gamma function, make the substitution

$t = x^\beta, t^{n-1} = x^{\beta n-\beta}, dt = \beta x^{\beta-1}$

$\begin{array}{rcl}\Gamma(n) & = & \int_0^\infty x^{\beta n-\beta}e^{-x^\beta}\beta x^{\beta-1}dx \\ { } & = & \int_0^\infty x^{n\beta-1}e^{-x^\beta}dx\end{array}$

We can choose whatever we want for $n$, as long as we think we can find $\Gamma(n)$. So let’s turn this into the original problem by substituting

$\begin{array}{rcl}n\beta-1 & = & \alpha \\ n & = & \frac{\alpha+1}{\beta} \end{array}$

Putting it all together:

$\frac{1}{\beta}\Gamma\left(\frac{\alpha+1}{\beta}\right) = \int_0^\infty x^\alpha e^{-x^\beta}dx$.

So we understand these seemingly-more-complicated definite integrals equally as well as we understand the gamma function.

For the special case my students were interested in, which has $\alpha = 0, \beta = 4$, the integral goes from $-\infty$ to $\infty$, so we need to multiply by two to get

$\int_{-\infty}^{\infty} e^{-x^4}dx = \frac{1}{2}\Gamma(1/4)$.

A computer tells me this evaluates to about 1.8.

December 10, 2008

The definite integral to evaluate was:

$\int_0^{2\pi} \log\left(1 + x^2 - 2x\cos\theta\right) d\theta$

I didn’t know how to do it, so I posted on the physics forum, and somebody named ‘adriank’ helped me out.
Here’s how it goes:

It would be hard to find a substitution to make this work directly. But if the logarithm were something more congenial, like

$\int \log(1+x\theta)d\theta$

maybe we’d have a better shot.

You might be used to pulling this sort of trick:

$\log(24)=\log(2*2*2*3) = 3\log(2) + \log(3)$

We can do the same sort of thing by factoring that quadratic inside the logarithm. So we want to factor

$x^2 - 2x\cos\theta + 1$

Set it equal to zero and use the quadratic formula to find the roots.

$x^2 - 2x\cos\theta + 1 = 0$
$x = \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4}}{2} = \cos\theta \pm \sqrt{-(1-\cos^2\theta)} = \cos\theta \pm i\sin\theta = e^{\pm i\theta}$

therefore

$\int_0^{2\pi} \log\left(1 + x^2 - 2x\cos\theta\right) d\theta$
$= \int_0^{2\pi} \log\left((x-e^{i\theta})(x-e^{-i\theta}) \right) d\theta$
$= \int_0^{2\pi} \log\left((1-xe^{i\theta})(1-xe^{-i\theta}) \right) d\theta$
$= \int_0^{2\pi} \left(\log(1-xe^{i\theta}) + \log(1-xe^{-i\theta})\right) d\theta$

now, so long as $\|t\|<1$, we can expand $\log(1-t)$ as a Taylor series:
$\log(1-t) =-t - \frac{t^2}{2} - \frac{t^3}{3} - \ldots = \sum_{n=1}^{\infty}-\frac{t^n}{n}$

applying to the integral:
$\int_0^{2\pi} \left(\log(1-xe^{i\theta}) + \log(1-xe^{-i\theta})\right) d\theta$
$=\int_0^{2\pi} \sum_{n=1}^\infty \frac{-x^ne^{in\theta}}{n} + \sum_{n=1}^\infty \frac{-x^ne^{-in\theta}}{n} d\theta$
$= -\sum_{n=1}^\infty \frac{x^n}{n} \int_0^{2*\pi}e^{in\theta}+e^{-in\theta} d\theta$
$= -\sum_{n=1}^\infty \frac{x^n}{n} \int_0^{2*\pi}2*\cos(n\theta) d\theta = 0$