December 6, 2024

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Gamma Integrals

Integrals and the Gamma Function
Compute Integrals by Taking Derivatives
Robert F Szalapski, PhD
www.CallMeDrRob.com

Some integrals may actually be computed by taking derivatives! I first learned this technique when studying Quantum Mechanics as an undergraduate physics major at the University of Minnesota. I also read about it in the memoirs of the Richard Feynman, the American physicist who earned a Nobel Prize for his work on renormlization in Quantum Field Theory. I initially found the topic to be somewhat mind blowing, and I find that my students of physics tend to share this reaction when they first learn the technique.

To understand this article the reader should have familiarity with integral calculus and especially the u-substitution technique. Many students forget when integration and differentiation may be interchanged, but below we do change the order of these operations when the parameter is not the integration parameter. The formula for the Gaussian integral is used, but its origin is not critically important here.

We are interested in integrals such as

\displaystyle\int _{0}^{\infty}t^{{2}}e^{{-t^{2}}},\int _{0}^{\infty}t^{{4}}e^{{-t^{2}}},\cdots,\int _{0}^{\infty}t^{{2n}}e^{{-t^{2}}}

and

\displaystyle\int _{0}^{\infty}te^{{-t^{2}}},\int _{0}^{\infty}t^{{3}}e^{{-t^{2}}},\cdots,\int _{0}^{\infty}t^{{2n+1}}e^{{-t^{2}}}

To solve the integrals let’s pull out a copy of Abromowitz & Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Turn to Chapter 6, “Gamma Function and Related Functions”. Of special interest is Formula 6.1.8.

\displaystyle\Gamma(\frac{1}{2})=2\int _{0}^{\infty}e^{{-t^{2}}}dt \displaystyle= \displaystyle\pi^{\frac{1}{2}}

or

\displaystyle\int _{0}^{\infty}e^{{-t^{2}}}dt \displaystyle= \displaystyle\frac{1}{2}\pi^{\frac{1}{2}}

This shouldn’t be a surprise since we already now how to solve the symmetric integral with limits (-\infty,\infty), and this is half of that. Next let’s consider the integral

\displaystyle\int _{0}^{\infty}e^{{-\beta t^{2}}}dt

and make the simple substitution

\displaystyle u \displaystyle= \displaystyle(\beta)^{{\frac{1}{2}}}t
\displaystyle(\beta)^{{-\frac{1}{2}}}du \displaystyle= \displaystyle dt

so that

\displaystyle\int _{0}^{\infty}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle(\beta)^{{-\frac{1}{2}}}\int _{0}^{\infty}e^{{-u^{2}}}du
\displaystyle= \displaystyle\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{1}{2}}}
\displaystyle= \displaystyle\frac{1}{2}\sqrt{\frac{\pi}{\beta}}
\displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma\left(\frac{1}{2}\right)}{\left(\beta\right)^{\frac{1}{2}}}

Notice that we can differentiate this equation with respect to \beta, and we can interchange the order of differentiation and integration.

\displaystyle\frac{d}{d\beta}\int _{0}^{\infty}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{d}{d\beta}\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{1}{2}}}
\displaystyle-\int _{0}^{\infty}t^{2}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle-\frac{1}{2}\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{3}{2}}}
\displaystyle\int _{0}^{\infty}t^{2}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{1}{2}\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{3}{2}}}
\displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma\left(\frac{3}{2}\right)}{\left(\beta\right)^{\frac{3}{2}}}

Differentiate once again with respect to \beta to get the next integral,

\displaystyle\frac{d}{d\beta}\int _{0}^{\infty}t^{2}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{d}{d\beta}\frac{1}{2}\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{3}{2}}}
\displaystyle-\int _{0}^{\infty}t^{4}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle-\frac{3}{2}\frac{1}{2}\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{5}{2}}}
\displaystyle\int _{0}^{\infty}t^{4}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{3}{2}\frac{1}{2}\frac{1}{2}\sqrt{\pi}(\beta)^{{-\frac{5}{2}}}
\displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma\left(\frac{5}{2}\right)}{\left(\beta\right)^{\frac{5}{2}}}

In general we find

\displaystyle\int _{0}^{\infty}t^{{2n}}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma\left(\frac{2n+1}{2}\right)}{\left(\beta\right)^{\frac{2n+1}{2}}}

with the gamma function expansion per Abromowitz and Stegun Formula 6.1.12.

\displaystyle\Gamma\left(n+\frac{1}{2}\right)=\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n}}\Gamma\left(\frac{1}{2}\right)

Next let’s consider the integral

\displaystyle\int _{0}^{\infty}t\; e^{{-\beta t^{2}}}dt

and make the simple substitution

\displaystyle u \displaystyle= \displaystyle\beta t^{2}
\displaystyle\frac{1}{2\beta}du \displaystyle= \displaystyle dt

so that

\displaystyle\int _{0}^{\infty}t\; e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{1}{2\beta}\int _{0}^{\infty}e^{{-u}}du
\displaystyle= \displaystyle-\frac{1}{2\beta}e^{{-u}}\mid _{0}^{\infty}
\displaystyle= \displaystyle\frac{1}{2\beta}

Differentiate with respect to \beta to get the next integral.

\displaystyle\frac{d}{d\beta}\int _{0}^{\infty}t\; e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{d}{d\beta}\frac{1}{2\beta}
\displaystyle-\int _{0}^{\infty}t^{3}\; e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle-\frac{1}{2}\frac{1}{\beta^{2}}
\displaystyle\int _{0}^{\infty}t^{3}\; e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{1}{2}\frac{1}{\beta^{2}}
\displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma(2)}{\beta^{2}}

Differentiate again to produce

\displaystyle\frac{d}{d\beta}\int _{0}^{\infty}t^{3}\; e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{d}{d\beta}\frac{1}{2}\frac{1}{\beta^{2}}
\displaystyle-\frac{d}{d\beta}\int _{0}^{\infty}t^{5}\; e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle-2\frac{1}{2}\frac{1}{\beta^{3}}
\displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma(3)}{\beta^{3}}

In general we find

\displaystyle\int _{0}^{\infty}t^{{2n}}e^{{-\beta t^{2}}}dt \displaystyle= \displaystyle\frac{1}{2}\frac{\Gamma\left(\frac{2n+1}{2}\right)}{\left(\beta\right)^{\frac{2n+1}{2}}}

works for both odd and even powers.

 

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