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I have been trying to find the residue of $f(\omega) = \frac{e^{i \omega a} e^{\frac{-b \omega}{\omega + ib}}}{i \omega}$ at the essential singularity $\omega = -ib$ for a while, but it is giving me a headache and I would really appreciate some help. For inspiration I have been looking at Residue of $z^2 e^{1/\sin z}$ at $z=\pi$ and How to compute the residue of a complex function with essential singularity

This is my aproach so far: I moved the singularity to 0 by substituting $y = \omega + \beta i$, then I rewrote and simplified the exponent to $A + By +Cy^{-1}$ which gives us the expression: $$\frac{-i}{y-\beta i}e^{A + By + Cy^{-1}}$$ where we want the residue at $y=0$ Then I used that $$\frac{-i}{y-\beta i} = \sum_{n=0}^{\infty}\frac{y^n}{(\beta i)^n\beta }$$ for $|y|<|\beta|$ and wrote the exponential as an infinite sum to get: $$\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{y^n(A+By+Cy{^-1})^k}{\beta (\beta i)^n \ k!}$$

Now unless I'm mistaken, this should be the Laurent series of $f$, and we should be able to find the residue from this. I just can't wrap my head around the double sum and the triple binomial formula.

I would be very grateful for hints on how to proceed, or on a better approach if there is one.

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John Mikael Modin
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  • Isn't it enough to find the limit as $\omega \rightarrow - b i$ of $(\omega + b i) f(\omega)$? – vonbrand Feb 05 '13 at 12:51
  • You can get $y^{-1}$ members exactly if $k=n+1$, and only from the $C$ part.. – Berci Feb 05 '13 at 12:53
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    @vonbrand: in this case I'm afraid no, because it is not a simple pole. – Berci Feb 05 '13 at 12:54
  • @Berci I agree that we will have $y^{-1}$ when $k = n+1$ from the $C$ part, but we will have a whole lot more from combinations of $A,B$ and $C$ http://en.wikipedia.org/wiki/Multinomial_theorem – John Mikael Modin Feb 05 '13 at 13:34
  • Ah yes, indeed. – Berci Feb 05 '13 at 13:56
  • Pull out factors that don't misbehave at the singularity, and work with the rest. That should get rid of much of the cruft that bothers you. – vonbrand Feb 14 '13 at 15:07
  • @vonbrand I think I did what you recommended, here: http://math.stackexchange.com/questions/304025/does-sum-k-0-infty-sum-n-0-infty-fracbk-cnk1ibn

    That sum is the residue (or almost, pulled out some factors)unless I'm mistaken. Just cant compute it :( hehe

     – John Mikael Modin Feb 14 '13 at 16:05

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The expression can also be written as follows: $$ \frac{e^A}{\beta}\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty} \sum_{n = 0}^{\infty}\frac{y^n y^k B^k y^{-j}C^{j}}{(i\beta)^n k! j!}$$ To find the residue we need the coefficient of $y^-1$, that is we need $ j = n+k+1$. Thus we need to solve $$\frac{e^A}{\beta}\sum_{k = 0}^{\infty} \sum_{n = 0}^{\infty}\frac{B^k C^{(n+k+1)}}{(i\beta)^n k! (n+k+1)!}$$ The sum here is solved in Does $ \sum_{k = 0}^{\infty} \sum_{n = 0}^{\infty}\frac{B^k C^{(n+k+1)}}{(ib)^n k! (n+k+1)!}$ converge? So finally the answer is $\frac{e^{A+B}}{\beta}\left(\frac{e^C-e^{\frac{-iC}{\beta}}}{1+\frac{i}{\beta}} \right)$

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John Mikael Modin
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