Prove $\int_{-\frac{1}{4}}^{\frac{1}{4}} \left(\left(H_{x+\frac{1}{4}}-H_{x-\frac{1}{4}}\right) \cos (2 \pi x)+\pi \right) \, dx=\text{Si}(\pi )$ I recently stumbled upon a curious relationship between harmonic numbers and the Wilbraham-Gibbs constant:
$$\int_{-\frac{1}{4}}^{\frac{1}{4}} \left(\left(H_{x+\frac{1}{4}}-H_{x-\frac{1}{4}}\right) \cos (2 \pi  x)+\pi \right) \, dx=\text{Si}(\pi )$$
Where $H_n$ is the $nth$ harmonic number, and $\mathrm{Si}(\pi)$ is the sine integral of $\pi$ (Wilbraham-Gibbs constant).
Is this an expected result, and if not, is it possible to prove? I have confirmed equivalency up to 40,000 digits.
 A: Using the relation between the Harmonic numbers and digamma function, namely $H_{x} = \gamma + \psi(x +1)$, the relation between the digamma function and the gamma function, 
$$\psi(x) = \frac{d}{dx} \, \ln(\Gamma(x)),$$
and the integrals:
\begin{align}
\int_{0}^{1} \sin\left(\frac{\pi \, t}{2}\right) \, \ln\left(\frac{1+t}{1-t}\right) \, dt &= - \frac{2}{\pi} \, \left[ Si\left(\frac{\pi(1-t)}{2}\right) - 
Si\left(\frac{\pi(1+t)}{2}\right) + \cos\left(\frac{\pi t}{2}\right) \, \ln\left(\frac{1 + t}{1-t}\right) \right]_{0}^{1} \\
&= \frac{2}{\pi} \, Si(\pi) 
\end{align}
\begin{align}
\int_{0}^{1} \sin\left(\frac{\pi \, t}{2}\right) \, \ln\left(cot\left(\frac{\pi(1+t)}{4}\right)\right) \, dt &= (-1) \, \left[ t + \frac{2}{\pi} \, \cos\left(\frac{\pi t}{2}\right) \, \ln\left( 
\frac{\cos\left(\frac{\pi t}{4}\right) - \sin\left(\frac{\pi t}{4}\right)}{\cos\left(\frac{\pi t}{4}\right) + \sin\left(\frac{\pi t}{4}\right)} \right)\right]_{0}^{1} \\
&= -1
\end{align}
the following is obtained.
\begin{align}
I &= \int_{-1/4}^{1/4} \left[ \pi + \left(H_{x+1/4} - H_{x-1/4}\right) \, \cos(2\pi x) \right] \, dx \\
&= \frac{\pi}{2} + \int_{-1/4}^{1/4} \left[ \psi\left(x + \frac{5}{4}\right) - \psi\left(x + \frac{3}{4}\right) \right] \, \cos(2 \pi x) \, dx 
\end{align}
Using integration by parts yields
\begin{align}
I &= \frac{\pi}{2} + 2\pi \, \int_{-1/4}^{1/4} \sin(2\pi t) \, \ln\left( \frac{\Gamma\left(t + \frac{5}{4}\right)}{\Gamma\left(t + \frac{3}{4}\right)}\right) \, dt \\
&= \frac{\pi}{2} + \frac{\pi}{2} \, \int_{-1}^{1} \sin\left(\frac{\pi t}{2}\right) \, \ln\left( \frac{\Gamma\left(\frac{5+t}{4}\right)}{\Gamma\left( \frac{3+t}{4}\right)}\right) \, dt \text{ (obtained by setting $t \to t/4$) } \\
&= \frac{\pi}{2} \, \left[ 1 + \int_{0}^{1} \sin\left(\frac{\pi t}{2}\right) \, 
\ln\left[\frac{\Gamma\left(\frac{5+t}{4}\right) \, \Gamma\left(\frac{3-t}{4}\right)}{\Gamma\left( \frac{3+t}{4}\right) \, \Gamma\left(\frac{5-t}{4}\right)} \right] \, dt \right] \\
&= \frac{\pi}{2} \, \left[ 1 + \int_{0}^{1} \sin\left(\frac{\pi t}{2}\right) \, 
\ln\left[\frac{1+t}{1-t} \, cot\left(\frac{\pi (1+t)}{4}\right) \right] \, dt \right] \\
&= \frac{\pi}{2} \, \left[ 1 + \frac{2}{\pi} \, Si(\pi) -1 \right] \\
&= Si(\pi).
\end{align}
Hence,
$$\int_{-1/4}^{1/4} \left[ \pi + \left(H_{x+1/4} - H_{x-1/4}\right) \, \cos(2\pi x) \right] \, dx = Si(\pi).$$
