The question is to compute $$\operatorname{res}_0 \frac{1}{(1-e^{-z})^n}$$ for natural number $n$.

I’ve tried directly calculate the integral by definition and Laurent expansion, but not making any progress. The Laurent series of the function become very nasty for large $n$. Is there any other way to solve this? Any help is much appreciated.


To find the residue, we can integrate $\frac{1}{(1-e^{-z})^{n}}$ around a rectangular contour with vertices at $z=\pm R \pm i \pi$, $R \ge 1$.

Doing so, we get $$\int_{-R}^{R} \frac{1}{\left(1-e^{-(t- i \pi)}\right)^{n}} \, dt + \int_{-\pi}^{\pi} \frac{1}{\left(1-e^{-(R+ it)}\right)^{n}} \, i \, dt - \int_{-R}^{R} \frac{1}{\left(1-e^{-(t+ i \pi)}\right)^{n}} \, dt$$ $$- \int_{-\pi}^{\pi} \frac{1}{\left( 1-e^{-(-R + it)}\right)^{n}} \, i \, dt = 2 \pi i \operatorname{Res}\left[\frac{1}{(1-e^{-z})^{n}},0 \right].$$

Since $e^{-t+i \pi} = -e^{-t} = e^{-t-i \pi}$, the first and third integrals cancel each other.

And as $R \to \infty$, the fourth integral vanishes since $$\left|\int_{-\pi}^{\pi} \frac{1}{\left( 1-e^{-(-R + it)}\right)^{n}} \, i \, dt \right|\le \int_{-\pi}^{\pi} \frac{dt}{\left(e^{R}-1\right)^{n}} = \frac{2 \pi}{(e^{R}-1)^{n}}. $$

(Due to symmetry, it appears that the value of integral might actually be zero for all values of $R >0$.)

That leaves us with $$\operatorname{Res}\left[\frac{1}{(1-e^{-z})^{n}},0 \right] = \lim_{R \to \infty} \frac{1}{2 \pi i} \int_{-\pi}^{\pi} \frac{1}{\left(1-e^{-(R+ it)}\right)^{n}} \, i \, dt. $$

But since $$\left| \frac{i}{\left(1-e^{-(R+it)}\right)^{n}}\right| \le \frac{1}{(1-e^{-R})^{n}} \le 2^{n}$$ for $R \ge 1$, the dominated convergence theorem permits us to move the limit inside the integral and conclude that $$\operatorname{Res}\left[\frac{1}{(1-e^{-z})^{n}},0 \right] = \frac{1}{2\pi i} \int_{-\pi}^{\pi} \, i\, dt = 1.$$


(Too long to be a comment)

Starting with the series $$\frac{z}{1-\mathrm{e}^{-z}}=\sum_{m=0}^\infty B_m^+ \frac{z^m}{m!}.$$

we get $$\frac{1}{(1-\mathrm{e}^{-z})^n}=\left[ \frac{1}{z} \sum_{m=0}^\infty B_m^+ \frac{z^m}{m!} \right]^n=\frac{1}{z^n} \sum_{m=0}^\infty C_m z^m $$ where $$C_m=\sum_{m_{n-1}=0}^{m} \cdots \sum_{m_2=0}^{m_3} \sum_{m_1=0}^{m_2} \frac{B_{m_1}^+}{m_1!} \frac{B_{m_2-m_1}^+}{(m_2-m_1)!} \cdots \frac{B_{m-m_{n-1}}^+}{(m-m_{n-1})!}$$ Numerical evidence suggest that the residue is identically 1, hence from a combinatorical perspective we want to show that

$$\sum_{m_{n-1}=0}^{n-1} \cdots \sum_{m_2=0}^{m_3} \sum_{m_1=0}^{m_2} \frac{B_{m_1}^+}{m_1!} \frac{B_{m_2-m_1}^+}{(m_2-m_1)!} \cdots \frac{B_{n-1-m_{n-1}}^+}{(n-1-m_{n-1})!} \equiv 1$$ for all natural $n$. Perhaps you could show this by induction, using the recursive definition of the Bernoulli numbers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.