How to prove this summation and integral equality? It's related to i factorial. 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?
 A: 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:
$$\sum_{k=1}^\infty(-1)^k\frac{\zeta(2k+1)}{2k+1}=\sum_{n=1}^\infty \left(\arctan\left(\frac{1}{k}\right)-\frac{1}{k}\right)=\int_{0}^{\infty}\left(\frac{\sin x}{x}-1\right)\frac{1}{e^x-1}dx$$
A: 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.
