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?
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}