1

given the following function

$$ f(y)= \int_{0}^{\infty}dx \frac{2}{(x+y)(x+y+1)} $$

i wish to expand $ f(y) $ into a Laurent series

is then valid if i use a numerical approxiamtion for the integral over 'x' so

$$ f(y)= \sum_{j}c_{j}\frac{2}{(x_{j}+y)(x_{j}+y+1)}$$

and then i exapnd each summand of the sum (Which depend only on 'y') to get the Laurent series for $ f(y) $

Jose Garcia
  • 8,506

1 Answers1

1

The improper integral $f(y)$ under consideration equals $ 2\ln(y+1)-2\ln(y) $ for each complex number $y$ which does not belong to $\{y:\Im y=0,\,\Re y <0\}$. Because $y=0$ is not an isolated singularity of $f(y)$, the Laurent expansion of $f(y)$ at $y=0$ does not exist.

user64494
  • 5,811
  • well perhaps this exists at some other point $ y=1$ or soemthing :) – Jose Garcia Oct 11 '13 at 18:23
  • 1
    Yes, of course. For example, $$f(y)=\sum _{k=0}^{\infty } \left( 2,{\frac { \left( -1 \right) ^{k+1}}{k+1 }}-2,{\frac { \left( -1/2 \right) ^{k+1}}{k+1}} \right) \left( y-1 \right) ^{k+1} $$ – user64494 Oct 11 '13 at 18:51