# Insane integral tricks

1. Nov 3, 2008

### tim_lou

Ingenious integral tricks

In QFT today I learnt some insane tricks in calculating impossible integrals....I figure it'd be a good idea to see if others have similar tricks so we can all learn from each other.

here goes:
to integrate (bounds are assumed to be from negative inf to inf, k^2 means the vector dot product)
edit: missing factors of 2pi

$$\int \frac{d^d k}{(2\pi)^d}\frac{1}{(k^2 + m^2)^n}$$
substitute
$$\frac{1}{(k^2 + m^2)^n}=\frac{1}{\Gamma(n)}\int_0^\infty t^n e^{-t(k^2 + m^2)} dt$$

and get a gaussian in the integral. At the end of the day,
$$\int \frac{d^d k}{(2\pi)^d}\frac{1}{(k^2 + m^2)^n}=\frac{\Gamma(n-d/2)(m^2)^{d/2-n}}{\Gamma(n)(4\pi)^{d/2}}$$

$$\Gamma(n)=(n-1)!$$

Quite crazy eh? I would've never thought of it myself... what about your favorite crazy tricks?

Last edited: Nov 4, 2008
2. Nov 3, 2008

### tim_lou

Another I have to share (I read it in some analysis book...)

take the following integral
$$\binom{m+n+1}{m,n}\int_0^1 u^{m} (1-u)^n du$$
consider the integral as a sum, we see that given m+n+1 particles on (0,1), it sums the probability of finding m particles in (0, u), 1 particle around the point u and n particles in (u, 1), so it must be one. This gives the beta integral for integers m, n!!!
$$\int_0^1 u^{m} (1-u)^n du=\frac{m!n!}{(m+n+1)!}$$

3. Nov 4, 2008

### Gerenuk

Re: Ingenious integral tricks

I wonder if most of these crazy tricks can be found without guessing. This substitution suggests that the initial problem could be simply solved by Laplace transforms?!