1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Laurent series of an integral

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the first three terms of the Laurent expansion in z of
    [itex]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[/itex]

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

    For the a_0 term this is easy:

    [itex]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)[/itex]

    [itex]= \frac{1}{2\pi i} \int_0^1 .. \oint_\gamma \frac{du}{C^u u} = 1[/itex]

    However for the a_1 term I get this:

    [itex]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[/itex]

    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. jcsd
  3. Dec 4, 2011 #2
    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

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

    [tex]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[/tex]

    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

    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):

    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)

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

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

    then proceed to go on to evaluate the outer real integral. However I think I may have a problem with the function [itex]g(x,y)[/itex] 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
  4. Dec 4, 2011 #3
    This is more or less what I tried but it leads to unsolvable integrals..

    A new idea is the substitution
    [itex]x_i \rightarrow \frac{a_i}{1+a_4}[/itex]
    which leads to
    [itex]\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}}[/itex]

    doing a similar substitution 2 more times I get

    [itex]\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}}[/itex]

    Does anyone see what could be done with that? Expanding z around 0 does again give unsolvable integrals.. I thought maybe of a reduction to Beta functions via the substituion

    [itex]c_i \rightarrow \frac{x}{1-x} const [/itex]

    but Im stuck again :/
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook