On the beauty of Parametric Integration and the Gamma function

/ 153armstrong

Parametric integration is one such technique that once you are made aware of it, you will never for the love of god forget it. Let me demonstrate :

Now this integral might seem familiar to many of you and to evaluate it is rather simple as well.

\int\limits_0^{\infty} e^{-sx} dx = \frac{1}{s}

Knowing this you can do lots of crazy stuff. Lets differentiate this expression wrt to the parameter in the integral – s (Hence the name parametric integration ). i.e

\frac{d}{ds}\int\limits_0^{\infty} e^{-sx} dx = \frac{d}{ds}\left(\frac{1}{s}\right)

\int\limits_0^{\infty} x e^{-sx} dx = \frac{1}{s^2}

Look at that, by simple differentiation we have obtained the expression for another integral. How cool is that! It gets even better.
Lets differentiate it once more:

\int\limits_0^{\infty} x^2 e^{-sx} dx = \frac{2*1}{s^3}

\int\limits_0^{\infty} x^3 e^{-sx} dx = \frac{3*2*1}{s^4}

\vdots

If you keep on differentiating the expression n times, one gets this :

\int\limits_0^{\infty} x^n e^{-sx} dx = \frac{n!}{s^{n+1}}

Now substituting the value of s to be 1, we obtain the following integral expression for the factorial. This is known as the gamma function.

\int\limits_0^{\infty} x^n e^{-x} dx = n! = \Gamma(n+1)

There are lots of ways to derive the above expression for the gamma function, but parametric integration is in my opinion the most subtle way to arrive at it. 😀

Advertisement

One thought on “On the beauty of Parametric Integration and the Gamma function”

  1. Pingback: Feynman’s trick on Laplace Transform (#1) – Ecstasy Shots

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s