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.

Answer 2

The problem discussed here can be easily solved if using explicit expressions for $\gamma_1(1/4)$ and $\gamma_1(3/4)$, see the paper by Donal F. Conon "The difference between two Stieltjes constants" and another paper "Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results" which I wrote two years ago (see p. 100). More general results regarding the first generalized Stieltjes constant at rational argument may be found in the preprint already referenced by Vladimir (thanks). If someone is interested in the journal version of this article, which is considerably enlarged with respect to the arXiv preprint, I can send it privately by e-mail.

Also would like to remark that the integral, which Kirill attributed to Ilan Vardi, was first evaluated by Carl J. Malmsten in 1842, see "Rediscovery of Malmsten's integrals, their evaluation by contour integration methods and some related results" (Wolfram's site contains a lot of errors, especially as concerns the attribution of formulae). By the way, there are numerous ways to evaluate such a kind of integrals (which should be certainly called Malmsten's integrals rather than Vardi's integrals): by contour integration, by using the Hurwitz zeta-function and via polylogarithms, solution proposed by Ilan Vardi in 1988 being one of most complicated (see the same paper).