How to prove this summation and integral equality? It's related to i factorial.

Question

How should I prove this equality?: $$ \sum_{k=1}^{\infty} \frac{\left ( -1 \right )^k\zeta \left ( 2k+1 \right )}{2k+1}= \sum_{k=1}^{\infty}\left ( \tan^{-1}\left ( \frac1k \right )- \frac1k\right )= \int_0^{\infty} \frac{\frac{\sin(x)}{x}-1}{e^x-1} dx $$ This question is related to the evaluation of i factorial. Someone asked "What is i factorial?" on Quora website.

A person Olof Salberger answered the question with the result as: $$ i!=re^{i\phi},\quad r=\sqrt{\frac{\pi}{\sinh \pi}},\quad \phi=-\gamma+\int_0^{\infty} \frac{1-\operatorname{sinc}(x)}{e^x-1} dx $$ He proved the magnitude r in his answer with Euler reflection formula and mentioned that $$ \int_0^{\infty} \frac{1-\operatorname{sinc}(x)}{e^x-1} dx=\sum_{k=1}^{\infty}\left ( \frac1k-\tan^{-1}\left ( \frac1k \right )\right ) $$ He also mentioned the phase $\phi$ can be computed from the logarithm of the Gamma function, which is: $$ \ln \Gamma(1+z)=-\gamma z+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k}(-z)^k $$ but he didn't show the process how to calculate the result of $\phi$. I am curious how to calculate the phase and how to prove the equality.

I have done the calculation of the magnitude and the phase in zeta function series, which are: $$ \begin{align} &r=exp\left ( \sum_{k=1}^{\infty}\frac{(-1)^k\zeta(2k)}{2k} \right )=\sqrt{\frac{\pi}{\sinh \pi}} \\ &\phi=-\gamma-\sum_{k=1}^{\infty} \frac{\left ( -1 \right )^k\zeta \left ( 2k+1 \right )}{2k+1}=-\gamma-\int_0^{\infty} \frac{\frac{\sin(x)}{x}-1}{e^x-1} dx \end{align} $$ How to prove these two equalities?

Answer 1

The Taylor expansion of $\arctan x$ is $$\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}$$ which leads to $\arctan \left(\frac{1}{k}\right)=\sum_{k=0}^\infty (-1)^k\frac{k^{-2k-1}}{2k+1}$.

Since $2k+1>1,\forall k\geqslant 1$, we can write $\zeta(2k+1)$ as $\sum_{n=1}^{\infty} n^{-2k-1}$. Then we can rearrange the notations $$\sum_{k=1}^\infty(-1)^k\frac{\zeta(2k+1)}{2k+1}=\sum_{n=1}^{\infty}\left(\sum_{k=0}^\infty(-1)^k\frac{n^{-2k-1}}{2k+1}-\frac{1}{k}\right)=\sum_{n=1}^\infty \left(\arctan\left(\frac{1}{k}\right)-\frac{1}{k}\right)$$

The Laplace Transform of $f=\int_{0}^x\frac{\sin t}{t}dt-1$, $\mathcal{L}\{f\}(s)=\int_0^\infty \left(\frac{\sin x}{x}-1\right)\frac{e^{-sx}}{s}dx$, is $$\frac{1}{s}\left(\frac{\pi}{2}-\arctan x-\frac{1}{x}\right)$$ we can put the summation inside the integral to get what we want: $$\sum_{n=1}^\infty \left(\arctan\left(\frac{1}{k}\right)-\frac{1}{k}\right)=\sum_{k=1}^\infty\int_0^\infty k\cdot\mathcal{L}\{f\}(k)dx=\int_0^\infty\left[\left(\frac{\sin x}{x}-1\right)\sum_{k=1}^\infty \left(e^{-x}\right)^k\right]dx$$ $\forall x>0, e^{-x}<1$, so the geometric series is convergent and thus $$\int_0^\infty\left[\left(\frac{\sin x}{x}-1\right)\sum_{k=1}^\infty \left(e^{-x}\right)^k\right]dx=\int_{0}^{\infty}\left(\frac{\sin x}{x}-1\right)\left(\frac{1}{1-e^x}-1\right)dx=\int_{0}^{\infty}\frac{\frac{\sin x}{x}-1}{e^x-1}dx$$

And we finally get the aimed equation: $$\sum_{k=1}^\infty(-1)^k\frac{\zeta(2k+1)}{2k+1}=\sum_{n=1}^\infty \left(\arctan\left(\frac{1}{k}\right)-\frac{1}{k}\right)=\int_{0}^{\infty}\left(\frac{\sin x}{x}-1\right)\frac{1}{e^x-1}dx$$

Answer 2

The Taylor series of the logarithm of the Gamma function is: \begin{align} &\ln \Gamma(1+z)=-\gamma z+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k}(-z)^k\\ &i!=\Gamma(1+i)=exp\left( \ln \Gamma(1+i)\right)=exp\left( -\gamma i+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k}(-i)^k\right) \end{align} Let $i!=re^{i\phi}$, then \begin{align} &r=exp\left ( \sum_{k=1}^{\infty}\frac{(-1)^k\zeta(2k)}{2k} \right )\\ &\phi=-\gamma-\sum_{k=1}^{\infty} \frac{\left ( -1 \right )^k\zeta \left ( 2k+1 \right )}{2k+1} \end{align} One needs to prove: \begin{align} &exp\left ( \sum_{k=1}^{\infty}\frac{(-1)^k\zeta(2k)}{2k} \right )=\sqrt{\frac{\pi}{\sinh \pi}}\\ &\sum_{k=1}^{\infty} \frac{\left ( -1 \right )^k\zeta \left ( 2k+1 \right )}{2k+1}= \sum_{k=1}^{\infty}\left ( \tan^{-1}\left ( \frac1k \right )- \frac1k\right )= \int_0^{\infty} \frac{\frac{\sin(x)}{x}-1}{e^x-1} dx \end{align} Proof of the magnitude part:

The Taylor series of $\cot z$ is: \begin{align} &\cot z=\frac1z+\sum_{k=1}^{\infty}\frac{(-1)^k2^{2k}B_{2k}}{(2k)!}z^{2k-1}, \text{ where $B_{2k}$ is the Bernoulli number}\\ &\ln (\sin z)=\int \cot z\,dz=\int \left ( \frac1z+\sum_{k=1}^{\infty}\frac{(-1)^k2^{2k}B_{2k}}{(2k)!}z^{2k-1} \right )dz\\ &\ln (\sin z)=\int \frac1zdz+\sum_{k=1}^{\infty}\frac{(-1)^k2^{2k}B_{2k}}{(2k)!}\int z^{2k-1}dz=\ln z+\sum_{k=1}^{\infty}\frac{(-1)^k2^{2k}B_{2k}}{(2k)!}\cdot \frac{z^{2k}}{2k}\\ &\ln\left ( \frac{\sin z}z \right )=\ln (\sin z)-\ln z=\sum_{k=1}^{\infty}\frac{(-1)^k2^{2k}B_{2k}}{2k(2k)!}\cdot z^{2k} \end{align} Let $z=i\pi$ \begin{align} &\ln\left ( \frac{\sin i\pi}{i\pi} \right )=\sum_{k=1}^{\infty}\frac{(-1)^k2^{2k}B_{2k}}{2k(2k)!}\cdot (i\pi)^{2k},\quad note:\zeta(2k)=\frac{(-1)^{k+1}B_{2k}(2\pi)^{2k}}{2(2k)!}\\ &-\ln\left ( \frac{\sin i\pi}{i\pi} \right )=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(2\pi)^{2k}B_{2k}}{2k(2k)!}\cdot (i)^{2k}=2\sum_{k=1}^{\infty}\frac{\zeta(2k)}{2k}(-1)^k\\ &-\frac12\ln\left ( \frac{\sin i\pi}{i\pi} \right )=\sum_{k=1}^{\infty}\frac{(-1)^k\zeta(2k)}{2k}\quad\Rightarrow \quad \sqrt{\frac{i\pi}{\sin i\pi}}=exp\left ( \sum_{k=1}^{\infty}\frac{(-1)^k\zeta(2k)}{2k} \right )\\ \\ &\sqrt{\frac{i\pi }{\sin i\pi}}=\sqrt{\frac{i\pi}{\frac{e^{i\cdot i\pi}-e^{-i\cdot i\pi}}{2i}}}=\sqrt{\frac{-\pi}{\frac{e^{-\pi}-e^{\pi}}{2}}}=\sqrt{\frac{\pi}{\frac{e^{\pi}-e^{-\pi}}{2}}}=\sqrt{\frac{\pi}{\sinh \pi}}\\ &\sqrt{\frac{\pi}{\sinh \pi}}=exp\left ( \sum_{k=1}^{\infty}\frac{(-1)^k\zeta(2k)}{2k} \right ) \end{align} Proof of the phase part:

The Taylor series of $\tan^{-1} z$ is: $$\tan^{-1}z=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}z^{2n+1}\quad\Rightarrow \quad\tan^{-1}\left (\frac1k \right )=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\cdot \frac{1}{k^{2n+1}}$$ The Taylor series of $\sin z$ is: \begin{align} &\sin z=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}z^{2n+1}\quad\Rightarrow \quad\frac{\sin x}x=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n}\\ &\frac1{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=e^{-x}\sum_{k=0}^{\infty}e^{-kx}=\sum_{k=1}^{\infty}e^{-kx}\\ \end{align} \begin{align} &\int_{0}^{\infty}\frac{\frac{\sin x}x-1}{e^x-1}dx=\int_{0}^{\infty}\left (\frac{\sin x}x-1 \right )\sum_{k=1}^{\infty}e^{-kx}dx=\sum_{k=1}^{\infty}\int_{0}^{\infty}\left (\frac{\sin x}x-1 \right )e^{-kx}dx\\ &\qquad\qquad\qquad\quad\,=\sum_{k=1}^{\infty}\left (\int_{0}^{\infty}\frac{\sin x}x e^{-kx}dx-\int_{0}^{\infty}e^{-kx}dx \right )\\ \\ &\int_{0}^{\infty}e^{-kx}dx=-\frac1ke^{-kx}|{_{x=0}^{x=\infty}}=-\frac1k(e^{-\infty}-e^0)=\frac1k\\ \\ &\int_{0}^{\infty}\frac{\sin x}x e^{-kx}dx=\int_{0}^{\infty}\left (\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n}e^{-kx} \right )dx=\sum_{n=0}^{\infty}\left (\frac{(-1)^n}{(2n+1)!}\int_{0}^{\infty}x^{2n}e^{-kx}dx \right )\\ \\ &\int_{0}^{\infty}x^{2n}e^{-kx}dx=-\frac1k\int_{0}^{\infty}x^{2n}d(e^{-kx})=-\frac1k\left (x^{2n}e^{-kx} |_{x=0}^{x=\infty}-\int_{0}^{\infty}e^{-kx}d(x^{2n}) \right )\\ &=\frac{2n}k\int_{0}^{\infty}x^{2n-1}e^{-kx}dx=-\frac{2n}{k^2}\left (x^{2n-1}e^{-kx} |_{x=0}^{x=\infty}-\int_{0}^{\infty}e^{-kx}d(x^{2n-1}) \right )\\ &=\frac{2n(2n-1))}{k^2}\int_{0}^{\infty}x^{2n-2}e^{-kx}dx=-\frac{2n(2n-1))}{k^3}\left (x^{2n-2}e^{-kx} |_{x=0}^{x=\infty}-\int_{0}^{\infty}e^{-kx}d(x^{2n-2}) \right )\\ &=\cdot\cdot\cdot\cdot\cdot\cdot=\frac{2n!}{k^{2n}}\int_{0}^{\infty}e^{-kx}dx=\frac{2n!}{k^{2n}}\cdot \frac1k=\frac{2n!}{k^{2n+1}}\\ \\ &\int_{0}^{\infty}\frac{\sin x}x e^{-kx}dx=\sum_{n=0}^{\infty}\left (\frac{(-1)^n}{(2n+1)!}\int_{0}^{\infty}x^{2n}e^{-kx}dx \right )=\sum_{n=0}^{\infty}\left (\frac{(-1)^n}{(2n+1)!}\cdot\frac{2n!}{k^{2n+1}} \right )\\ &\qquad\qquad\qquad\quad\;=\sum_{n=0}^{\infty}\left (\frac{(-1)^n}{(2n+1)}\cdot\frac1{k^{2n+1}} \right )=\tan^{-1}\left ( \frac1k\right )\\ \\ &\int_{0}^{\infty}\frac{\frac{\sin x}x-1}{e^x-1}dx=\sum_{k=1}^{\infty}\left (\int_{0}^{\infty}\frac{\sin x}x e^{-kx}dx-\int_{0}^{\infty}e^{-kx}dx \right )=\sum_{k=1}^{\infty}\left ( \tan^{-1}\left ( \frac1k \right )- \frac1k\right )\\ \\ &\sum_{k=1}^{\infty}\left ( \tan^{-1}\left ( \frac1k \right )- \frac1k\right )=\sum_{k=1}^{\infty}\left ( \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\cdot \frac{1}{k^{2n+1}}- \frac1k\right )\\ &\qquad\qquad\qquad\qquad\qquad=\sum_{k=1}^{\infty}\left ( \sum_{n=1}^{\infty}\frac{(-1)^n}{(2n+1)}\cdot \frac{1}{k^{2n+1}} \right )=\sum_{n=1}^{\infty}\left ( \sum_{k=1}^{\infty}\frac{(-1)^n}{(2n+1)}\cdot \frac{1}{k^{2n+1}} \right )\\ &\qquad\qquad\qquad\qquad\qquad=\sum_{n=1}^{\infty}\left ( \frac{(-1)^n}{(2n+1)}\sum_{k=1}^{\infty} \frac{1}{k^{2n+1}} \right )=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)}\zeta(2n+1) \end{align}

Answer 3

If I recall correctly, I just used the integral definition of the zeta function and moved the summation inside the integral, which is where the exp(x) - 1 in the denominator comes from. Then it becomes trivial to match the power series you are left with with that of sinc(x) - 1.

To get to the series, you expand the denominator in the integral to a geometric series and integrate termwise.