# Double integral to single by magic substitution

1. Jul 9, 2010

### rubenvb

double integral to single by "magic" substitution

Hi,

I have a double (actually quadruple, but the other dimensions don't matter here) integral which looks like this:

$$\iint_0^\infty \frac{d^2 k}{k^2}$$

Now, someone here told me to replace that with

$$\int_0^\infty \frac{1}{2} 2\pi \frac{d k^2}{k^2}$$

How and why is this ok? Thanks!

PS: For anyone wondering where this comes up: it's related to Elementary Particle physics and structure function calculation.

2. Jul 9, 2010

### Anthony

Re: double integral to single by "magic" substitution

If your first integral is meant to be:

$$I= \int_{-\infty}^\infty \int_0^\infty \frac{\mathrm{d}k_1\, \mathrm{d} k_2}{k_1^2 + k_2^2}$$

then you *could* use polar coordinates to give:

$$I = \int_0^\pi \int_0^\infty \frac{ r\mathrm{d} r\, \mathrm{d} \theta}{r^2} = \pi \int_0^\infty \frac{ \mathrm{d} r}{r}$$

which seems to be the same as what you have written. However, the integral doesn't make a lot of sense - it fails to converge (logarithmic singularity). Although perhaps if this is from particle physics, you don't mind this divergence.

3. Jul 10, 2010

### rubenvb

Re: double integral to single by "magic" substitution

Thanks, that looks good. The integrand isn't complete here (waaaay too big to put here).