Prove $\gamma_1\left(\frac34\right)-\gamma_1\left(\frac14\right)=\pi\,\left(\gamma+4\ln2+3\ln\pi-4\ln\Gamma\left(\frac14\right)\right)$

Question

Please help me to prove this identity: $$\gamma_1\left(\frac{3}{4}\right)-\gamma_1\left(\frac{1}{4}\right)=\pi\,\left(\gamma+4\ln2+3\ln\pi-4\ln\Gamma\left(\frac{1}{4}\right)\right),$$ where $\gamma_n(a)$ is a generalized Stieltjes constant and $\gamma$ is the Euler-Mascheroni constant.

Answer 1

By definition, the generalized Stieltjes constant $\gamma_n(a)$ is the coefficient of $(1-s)^n$ in the generalized zeta function $\zeta(s,a)$: $$ \zeta(s,a) = \sum_{n\geq0}(n+a)^{-s}, $$ $$ \gamma_1(a) = -\frac{d}{ds}\Big|_{s=1}\zeta(s,a). $$

Now, use the integral representation (see mathworld) $$ \zeta(s,a) = \int_0^\infty \frac{t^{s-1}}{\Gamma(s)} \frac{e^{-a t}\,dt}{1-e^{-t}}, $$ to get $$ \gamma_1(a) = - \int_0^\infty \frac{\gamma+\log t}{1-e^{-t}}\, e^{-at}\,dt, $$ from which it follows that the desired quantity is $$ Q = \gamma_1(b) - \gamma_1(a) = \int_0^{\infty} \frac{e^{-a t}-e^{-b t}}{1-e^{-t}}(\gamma+\log t)\,dt, $$ with $a=\frac14$, $b=\frac34$.

The first part of that integral is $$ \gamma\int_0^\infty \frac{e^{-at}-e^{-bt}}{1-e^{-t}}\,dt = \gamma(\psi(b)-\psi(a)) = \gamma\pi, $$ by an integral representation of the digamma function, or by computer algebra.

The second part is $$ \int_0^\infty \frac{e^{-at}-e^{-bt}}{1-e^{-t}}\log t\,dt $$ which by substitution of $t=\log u$ turns into $$ 4\int_1^\infty \frac{1/u-1/u^3}{1-1/u^4}(\log 4+\log\log u)\frac{du}{u} \\= 4\int_1^\infty \frac{du}{1+u^2}(\log4+\log\log u) \\= \pi \log4 + 4\int_1^\infty \frac{du}{1+u^2}\log\log u. $$ That last integral can be transformed by $u=\tan\theta$ into $$ \int_1^\infty \frac{du}{1+u^2}\log\log u = \int_{\pi/4}^{\pi/2} \log\log\tan\theta \,d\theta \\= \frac{\pi}{2}\log\left(\frac{\Gamma(\frac34)}{\Gamma (\frac14)} \sqrt{2\pi} \right). $$ This integral is quite hard to evaluate, it is called Vardi's integral, (see mathworld, and the original paper "Integrals: An Introduction to Analytic Number Theory" by Ilan Vardi, at jstor, which is a very interesting read), and it is equal to $$ \frac{d}{ds}\Big|_{s=1} \Gamma(s)L(s), \qquad L(s) = 1-3^{-s}+5^{-s}-7^{-s}+\cdots. $$ In fact, because $L(s) = 4^{-s}(\zeta(s,\frac14)-\zeta(s,\frac34))$, the question is really about computing $$ \frac{d}{ds}\Big|_{s=1} 4^{s}L(s) = 4L(1)\log4 + 4L'(1), $$ which are computed by Vardi in that paper.

To simplify, note that $\Gamma(\frac34) = \pi\sqrt{2}/\Gamma(\frac14)$ and that $\psi(\frac34)-\psi(\frac14) = \pi$.

Finally, putting everything together, $$ Q = \gamma\pi + \pi\log4 + \pi\log \frac{4\pi^3}{\Gamma(\frac14)^4}, $$ as desired.