Residue of $\dfrac{\mathrm{e}^\frac{2}{z}}{\sin z}$ at $0$

Asked May 16 '21 at 14:08

Active May 16 '21 at 14:08

Viewed 70 times

Is there a nice way to find (and represent) the residue of the function $\dfrac{\mathrm{e}^\frac{2}{z}}{\sin z}$ at $0$?

I tried using the product of the associated Laurent series for $\mathrm{e}^\frac{2}{z}$ and $\dfrac{1}{\sin z}$, but the result is a complicated series, since $\dfrac{1}{\sin z}$ is expressed using the Bernoulli numbers (as shown in this SE post).

asked May 16 '21 at 14:08

digital-Ink

1,986

1

The residue is $\sum\limits_{n\ge0}\frac{(-1)^{n+1}2^{2n+1}(2^{2n-1}-1)B_{2n}}{(2n)!^2}=\sum\limits_{n\ge0}\frac{2(2^{2n}-2)\zeta(2n)}{\pi^{2n}(2n)!}$ which has no closed form. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ May 16 '21 at 14:24
@TheSimpliFire: That is, indeed, the residue I got, too. – digital-Ink May 16 '21 at 14:43

Residue of $\dfrac{\mathrm{e}^\frac{2}{z}}{\sin z}$ at $0$

0 Answers0