$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解 - Eufisky - The lost book

# $\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解

Eufisky posted @ 2014年7月24日 20:24 in 数学分析 with tags 级数 Mellin变换 , 5439 阅读

Proof.What you require here are the Eisenstein series. In particular the evaluation of

$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$

at $\tau = i.$ Rearrange to get

$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$

See Lambert series for additional information.

The function

$$G_ 2(\tau) = \zeta(2) \left(1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)=\zeta(2)E_2(\tau)$$

satisfies the quasimodular transformation

$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$

And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore

$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$

Hence we obtain

$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$

as given in the comment to the question by Slowsolver.

There is a very nice generalisation of the sum in the question.

For odd $m > 1$ we have

$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$

where $B_k$ are the Bernoulli numbers defined by

$\frac{z}{{{e^z} - 1}} = \sum\limits_{k = 0}^\infty {\frac{{{B_k}}}{{k!}}} {z^k}\;\;\; for |z|<2\pi .$

$$\displaystyle\mathcal{M}\Big[\frac{x}{e^{2\pi x}-1}] = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \ dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)$$

So $$\displaystyle \frac{x}{e^{2\pi x}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) x^{-s}\ ds$$

which implies $$\displaystyle \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds$$

The integrand has poles at $s=-1, s=0$, and $s=1$ (and removable singularities at $s= -2, -3, -4, \ldots$)

I'm going to close the contour with a rectangle that has vertices at $$-i \infty, \frac{3}{2} - i \infty, \frac{3}{2} + i \infty, and \;i \infty$$ and is indented at the origin

$\Gamma(s)$ decays rapidly as $\text{Im} (s) \to \pm \infty$. So the integral goes to zero along the top and bottom of the rectangle.

And on the imaginary axis, the integrand is odd.

So $$\displaystyle \int_{\frac{3}{2}-i\infty}^{\frac{3}{2}+i\infty}(2\pi)^{-s}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds-\pi i \text{Res}[f,0] = 2\pi i\text{Res}[f,1]$$

where $$\displaystyle f(s) = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)$$.

$$\displaystyle \text{Res}[f,0] = \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) = \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s)$$

$$= 1\Big(\frac{1}{2 \pi} \Big)(1)\zeta(0)=-\frac{1}{4 \pi}$$

$$\displaystyle\text{Res}[f,1] = \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s)$$

$$= \displaystyle \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)= 1\Big(\frac{1}{4 \pi^{2}}\Big)(1)\Big(\frac{\pi^{2}}{6}\Big) =\frac{1}{24}$$

Therefore,  $$\displaystyle \sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{2 \pi i} \Big( 2 \pi i (\frac{1}{24}) + \pi i (\frac{-1}{4 \pi}) \Big) = \frac{1}{24} - \frac{1}{8 \pi}.$$

Poof.We will use the Mellin transform technique. Recalling the Mellin transform and its inverse

$$F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds.$$

Now, let's consider the function

$$f(x)= \frac{x}{e^{\pi x}+1}.$$

Taking the Mellin transform of $f(x)$, we get

$$F(s)={\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1- {2}^{-s} \right) \zeta \left( s+1 \right),$$

where $\zeta(s)$ is the zeta function. Representing the function in terms of the inverse Mellin Transform, we have

$$\frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left( 1-{2}^{-s} \right) \zeta \left( s+1 \right) x^{-s}ds.$$

Substituting $x=2n+1$ and summing yields

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma \left( s+1 \right)\left(1-{2}^{-s} \right) \zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$

$$= \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$

Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24}$$

Notes: 1)

$$\sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta \left( s \right).$$

2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as

$$r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$

$$= \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1} {\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) = \frac{1}{24}.$$

For calculating the above limit, we used the facts

$$\lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}.$$

3) Here is the technique for computing the Mellin transform of $f(x)$.
Using the change of variables $u=-\ln(x)$ and the identity

$$\int_{0}^{\infty}\frac{u^{s-1}}{e^u -1}=\zeta{(s)}\Gamma{(s)}$$

we reach to the deisred result

$$\int_0^1 \frac{\ln x }{x-1}= \int_{0}^{\infty}\frac{u}{e^u -1}=\zeta{(2)}\Gamma{(2)} =\sum_{n=1}^\infty \frac{1}{n^2}.$$

Note that,

$$\int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$

$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$

If $Re(s)>1,Re(q)>0$,define$\zeta(s,q)=\sum_{n=0}^{\infty}\frac1{(q+n)^s}.$,then the function has an integral representation in terms of the Mellin transform as $\zeta(s,q)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.$

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