Prove that the following series converges to $1 - \frac{\pi}{4}$. We got the following identity by solving a problem in two different ways, but we don't know how to prove it. 
$$\sum_{n=0}^{\infty} [Si((4n+1)\pi)-Si((4n+3)\pi)] = 1 - \frac{\pi}{4}$$
where $$Si(x)=\int_0^x \frac{\sin(t)}{t}dt$$
We were able to verify that the first few thousand partial sums are very close to the RHS.
Is there a way to prove this identity?
 A: With the definition
\begin{equation}
 \mathrm{si}\left(z\right)=-\int_{z}^{\infty}\frac{\sin t}{t}\mathrm{d}t=%
\mathrm{Si}\left(z\right)-\tfrac{1}{2}\pi
\end{equation} 
the series can be written as
\begin{align}
 S&=\sum_{n=0}^{\infty} [\mathrm{Si}((4n+1)\pi)-\mathrm{Si}((4n+3)\pi)]\\
 &=\sum_{n=0}^{\infty} [\mathrm{si}((4n+1)\pi)-\mathrm{si}((4n+3)\pi)]
\end{align} 
We use the integral representation (DLMF):
\begin{align}
 \mathrm{si}\left(z\right)&=-\int_{0}^{\pi/2}e^{-z\cos t}\cos\left(z\sin t\right
)\mathrm{d}t\\
&=-\Re\int_{0}^{\pi/2}e^{-z\exp(-it)}\mathrm{d}t
\end{align} 
to express
\begin{align}
 S&=-\Re\sum_{n=0}^{\infty}\int_{0}^{\pi/2}\left[e^{-\left( 4n+1 \right)\pi\exp(-it)}-e^{-\left( 4n+3 \right)\pi\exp(-it)}\right]\mathrm{d}t \\
 &=-2\Re\sum_{n=0}^{\infty}\int_{0}^{\pi/2}e^{-2\left( 2n+1 \right)\pi\exp(-it)}\sinh\left(\pi\exp(-it)\right)\mathrm{d}t \\
 &=-2\Re\int_{0}^{\pi/2}\frac{e^{-2\pi\exp(-it)}}{1-e^{-4\pi\exp(-it)}}\sinh\left(\pi\exp(-it)\right)\mathrm{d}t \\
 &=-\Re\int_{0}^{\pi/2}\frac{\sinh\left(\pi\exp(-it)\right)}{\sinh\left( 2\pi\exp(-it) \right) }\mathrm{d}t\\
 &=-\frac{1}{2}\Re\int_{0}^{\pi/2}\frac{\mathrm{d}t}{\cosh\left( \pi\exp(-it) \right) }\\
\end{align}
From the symmetry of the real part of $\cosh\left(\pi\exp(-it) \right)$, we have
\begin{align}
 S&=-\frac{1}{8}\Re\int_{0}^{2\pi}\frac{\mathrm{d}t}{\cosh\left( \pi\exp(it) \right) }\\
 &=-\frac{1}{8}\Re\int_C\frac{\mathrm{d}z}{iz\cosh\left( \pi z\right) }\\
\end{align}
where $C$ is the unit circle taken counter clockwise. Poles in the circle lie at $z=0,-i/2,i/2$, with residues respectively $1,-2/\pi,-2/\pi$. Then
\begin{align}
 S&=-\frac{1}{8}\Re\left\lbrace2i\pi\left[-i\left( 1-\frac{2}{\pi}-\frac{2}{\pi} \right)\right]\right\rbrace\\
 &=1-\frac{\pi}{4}
\end{align}
