# Laurent series of an integral

1. Dec 3, 2011

### monsi23

1. The problem statement, all variables and given/known data
Determine the first three terms of the Laurent expansion in z of
$f(z)=\int_0^1 dx_1..dx_4 \frac{\delta(1-x_1-x_2-x_3-x_4)}{(x_1 x_2 a + x_3 x_4 b)^{2+z}},\quad a,b>0$

2. The attempt at a solution
I tried expanding around z = -2.
$f(z)=\sum_{n=-\infty}^\infty a_n (z-(-2))^n$

For the a_0 term this is easy:

$a_0 = \frac{1}{2\pi i} \oint_\gamma \int_0^1 \frac{dx_1 .. dx_4 dz}{C(\lbrace x_i \rbrace)^{z+2}(z+2)}, \quad C=(x_1 x_2 a + x_3 x_4 b)$

$= \frac{1}{2\pi i} \int_0^1 .. \oint_\gamma \frac{du}{C^u u} = 1$

However for the a_1 term I get this:

$a_1 = -\int_0^1 \delta(1-x_1-x_2-x_3-x_4) \ln(x_1 x_2 a + x_3 x_4 b) dx_1 .. dx_4$

and for a_2 the same with -ln(.) -> (1/2) ln^2(.). These are horrible integrals! Am I doing something wrong or is there any trick or substitution Im missing?
I'm thankful for any suggestion!

2. Dec 4, 2011

### jackmell

Alright I wish to propose an approach which may have some holes but it's a start:

How about we drop it down to two for now and consider
$$f(z)=\iint\frac{\delta(1-x-y)}{(axy)^{z+2}}dA$$

Now, just for now I'm going to assume $f(z)$ is analytic in the neighborhood of the origin so that I can write:

$$f(z)=\sum_{n=0}^{\infty} a_n z^n,\quad a_n=\frac{1}{2\pi i}\oint\frac{f(z)}{z^{n+1}}dz$$

and therefore:
$$a_n=\frac{1}{2\pi i}\oint\left\{\iint \frac{\delta(1-x-y)}{(axy)^2 (axy)^z}dA\right\}\frac{1}{z^{n+1}}dz$$

now let
$$g(x,y)=\frac{\delta(1-x-y)}{(axy)^2}$$

$$a_n=\frac{1}{2\pi i}\oint\left\{\iint g(x,y)\frac{1}{(axy)^z}dA\right\}\frac{1}{z^{n+1}}dz$$

Now, just for now too, let's assume I can switch the order of integration (we can work on that more later if necessary):

\begin{aligned} a_n&=\frac{1}{2\pi i}\iint g(x,y)\oint\frac{dz}{(axy)^z z^{n+1}}dzdA \\ &=\frac{1}{2\pi i} \iint g(x,y)\oint \frac{e^{-kz}}{z^{n+1}}dzdA,\quad k=\log(axy) \end{aligned}

Now we can use Cauchy's Integral formula to evaluate the inner integral:

$$\oint \frac{e^{-kz}}{z^{n+1}}dz=\left(\frac{2\pi i}{n!}\frac{d^n}{dz^n}e^{-kz}\right)_{z=0}$$

then proceed to go on to evaluate the outer real integral. However I think I may have a problem with the function $g(x,y)$ as I've defined it above. Not sure that's even a differentiable function or even a continuous one. That would I think prevent switching the order of integration. I would drop delta for now and just work with nice differentiable functions to see if this approach works then try to fit delta into the analysis if possible.

Last edited: Dec 4, 2011
3. Dec 4, 2011

### monsi23

This is more or less what I tried but it leads to unsolvable integrals..

A new idea is the substitution
$x_i \rightarrow \frac{a_i}{1+a_4}$
$\int_0^\infty da_4 \int_0^1 da_1..da_3 (1+a_4)^{2z} \frac{\delta(1-a_1-a_2-a_3)}{(a a_1 a_2 + b a_3 a_4)^{z+2}}$
$\int_0^\infty dc_4..dc_2 \frac{(1+c_4+c_3 + c_2)^{2z}}{(a c_2 + b c_3 c_4)^{z+2}}$
$c_i \rightarrow \frac{x}{1-x} const$