# Beasty integral

1. Sep 5, 2007

This is an integral that Mathematica doesn't seem to be able to do. I don't know how to tackle it either.

The general form is

$$\int^\infty_{-\infty} dz \arctan [d \sqrt{p^2 + z^2}]e^{-b z^2 - i c z} (p^2 + z^2)^n$$

I've thought about integration by parts, by substitution, contour integration, but none of these methods seem suitable. Any help would be much appreciated.

2. Sep 6, 2007

### Gib Z

Quite sure that is not analytically possible.

3. Sep 6, 2007

### HallsofIvy

Staff Emeritus
Since you already have "dz" in your integral what do you mean by "$d\sqrt{p^2+ z^2}$?

4. Sep 6, 2007

### genneth

In general if mathematica can't do it, that means there is no simple closed form, for a definition of simple which means a finite number operations involving arithmetic, trig, exp, logs, etc. The algorithm that it uses produces such a closed form or else a proof that it can't be done.

5. Sep 7, 2007

Thanks.

$d\sqrt{p^2+ z^2}$ means the variable "d" multiplied by the square root of (p^2 + z^2).

Also there's something I missed. The power of the (p^2 + z^2) outside the ArcTan should be n/2 not n, where n is any real integer.

Hmmm I'm sure it can be done. I've seen the answer to an integral of which this is one part. I'm just trying to reproduce it. Mathematica doesn't give a proof it can't be done, it simply spits the integral back at me after a long time...

The answer might well involve a gamma function... Mathematica has been fussy over gamma functions before...

6. Sep 8, 2007

### Gib Z

So there is a variable "d"? How are we meant to distinguish between the differential and the variables..