MOMENTS OF THE LOGGAMMA FUNCTION BETWEEN 0 AND 1/2

In this post we will prove the beautiful results below related to the moments of the Loggamma function in the interval between zero and 1/2.

$\int_0^{1/2} x \ln\left(\Gamma(x)\right)\,dx=\frac{\ln \left(4 \pi^3 A^{12}\right)}{48} -\frac{7\,\zeta(3)}{32 \pi^2}$

$\int_0^{1/2} x^2 \ln\left(\Gamma(x)\right)\,dx=\frac{\ln \left(\pi^{60} 2^{31}A^{360} \right)}{2880} -\frac{3\zeta(3)}{32\pi^ 2}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8}$

First recall Kummer´s fourier series for the Loggamma function

$\log(\Gamma(x))=\frac{\log(2 \pi)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos(2 \pi k x) + \sum_{k=1}^{\infty}\frac{\gamma+\log(2 \pi )+\ln(k)}{ \pi k}\sin(2 \pi k x)$ (1)

Then multiply both sides of (1) by and integrate from zero and 1/2:

$\begin{aligned} \int_0^{1/2} x \ln\left(\Gamma(x)\right)\,dx&=\frac{\ln 2 \pi}{2}\int_0^{1/2}x\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^{1/2} x \cos\left(2 \pi k x\right)\,dx + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\int_0^{1/2} x \sin\left(2 \pi k x \right)\,dx\\ &=\frac{\ln 2 \pi}{16}-\sum_{k=1}^{\infty}\frac{1}{2k}\left( \frac{1}{4 \pi^2 k^2}+\frac{(-1)^{k-1}}{4 \pi^2 k^2}\right)+ \sum_{k=1}^{\infty}\frac{\gamma+\ln 2 \pi +\ln k}{ \pi k}\left(\frac{(-1)^{k-1}}{4 \pi k}\right)\\ &=\frac{\ln 2 \pi}{16}-\frac{1}{8 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^3}-\frac{1}{8 \pi^2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}+\frac{\gamma}{4 \pi^2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}+\frac{\ln 2 \pi}{4 \pi^2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}+\frac{1}{4 \pi^2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\ln k}{k^2}\\ &=\frac{\ln 2 \pi}{16}-\frac{1}{8 \pi^2}\zeta(3)-\frac{1}{8 \pi^2}\eta(3)+\frac{\gamma}{4 \pi^2}\eta(2)+\frac{\ln 2 \pi}{4 \pi^2}\eta(2)-\frac{1}{4 \pi^2}\eta^\prime(2)\\ &=\frac{\ln 2 \pi}{16}-\frac{1}{8 \pi^2}\zeta(3)-\frac{1}{8 \pi^2}\left(1-\frac14\right)\zeta(3)+\frac{\gamma+\ln 2 \pi}{4 \pi^2} \cdot\frac{\pi^2}{12}-\frac{\eta^\prime(2)}{4 \pi^2}\\ &=\frac{\ln 2 \pi}{16}-\frac{7 \zeta(3)}{32 \pi^2}+\frac{\gamma+\ln 2 \pi}{48} -\frac{1}{4 \pi^2}\left(\frac{\ln 2}{2}\zeta(2)+ \left(1-\frac12 \right)\zeta^\prime(2)\right)\\ &=\frac{ \ln 2 \pi}{16}-\frac{7 \zeta(3)}{32 \pi^2}+\frac{\gamma}{48} +\frac{\ln 2 \pi}{48}-\frac{\ln 2}{48}- \frac{1}{8 \pi^2}\zeta(2)\left(-12 \ln A + \gamma + \ln2 \pi \right)\\ &=\frac{ \ln 2 \pi}{16}-\frac{7 \zeta(3)}{32 \pi^2}+\frac{\gamma}{48} +\frac{\ln 2 \pi}{48}-\frac{\ln 2}{48}- \frac{1}{48 }\left(-12 \ln A + \gamma + \ln2 \pi \right)\\ &=\frac{ \ln 2 \pi}{16}-\frac{7 \zeta(3)}{32 \pi^2}-\frac{\ln 2}{48} +\frac{12 \ln A}{48 } \\ &=\frac{ \ln \left( 4 \pi^3 A^{12}\right)}{48 }-\frac{7 \zeta(3)}{32 \pi^2} \qquad \blacksquare \\ \end{aligned}$

Where We used that $\zeta^\prime(2) =\zeta(2)\left(-12 \ln A + \gamma + \ln2 \pi \right)$

And $\eta^{\prime}(s)=\sum_{k=1}^{\infty} \frac{(-1)^{k} \ln k}{k^{s}}=2^{1-s} \ln (2) \zeta(s)+\left(1-2^{1-s}\right) \zeta^{\prime}(s)$

For the second integral, multiply both sides of (1) by x^2 and integrate from 0 to 1/2

$\begin{aligned} \int_0^{1/2} x^2 \ln\left(\Gamma(x)\right)\,dx&=\underbrace{\frac{\ln 2 \pi}{2}\int_0^{1/2} x^2\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^{1/2} x^2 \cos\left(2 \pi k x\right)\,dx}_{(1)} + \underbrace{\frac{\left(\gamma+\ln(2 \pi) \right)}{ \pi }\sum_{k=1}^{\infty}\frac{1}{ k}\int_0^{1/2} x^2 \sin\left(2 \pi k x \right)\,dx}_{(2)} + \underbrace{\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\ln\left(k \right)}{ k}\int_0^{1/2} x^2 \sin\left(2 \pi k x \right)\,dx}_{(3)}\\ \end{aligned}$

$\begin{aligned} (1)&=\frac{\ln 2 \pi}{2}\int_0^{1/2} x^2\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^{1/2} x^2 \cos\left(2 \pi k x\right)\,dx\\ &=\frac{\ln 2 \pi}{48}-\frac{1}{8\pi^ 2}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^3}\\ &=\frac{\ln 2 \pi}{48}-\frac{\eta(3)}{8\pi^ 2}\\ &=\frac{\ln 2 \pi}{48}-\frac{3\zeta(3)}{32\pi^ 2}\\ &=\frac{\ln 2 }{48}+\frac{\ln \pi }{48}-\frac{3\zeta(3)}{32\pi^ 2}\\ &=\frac{60\ln 2 }{2880}+\frac{60\ln \pi }{2880}-\frac{3\zeta(3)}{32\pi^ 2}\\ \end{aligned}$

$\begin{aligned} (2)&=\frac{\left(\gamma+\ln(2 \pi) \right)}{ \pi }\sum_{k=1}^{\infty}\frac{1}{ k}\int_0^{1/2} x^2 \sin\left(2 \pi k x \right)\,dx\\ &=\frac{\left(\gamma+\ln(2 \pi) \right)}{ \pi }\sum_{k=1}^{\infty}\frac{1}{ k}\left(-\frac{(-1)^{k}}{8 \pi k}-\frac{1}{4 \pi^3 k^3}+\frac{(-1)^{k}}{4 \pi^3 k^3}\right)\\ &=\left(\gamma+\ln(2 \pi) \right)\left(\frac{1}{8\pi^2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{ k^2}-\frac{1}{4 \pi^4}\sum_{k=1}^{\infty}\frac{1}{ k^4}-\frac{1}{4 \pi^4}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{ k^4} \right)\\ &=\left(\gamma+\ln(2 \pi) \right)\left(\frac{1}{8\pi^2} \eta(2)-\frac{1}{4 \pi^4}\zeta(4)-\frac{1}{4 \pi^4} \eta(4) \right)\\ &=\left(\gamma+\ln(2 \pi) \right)\left(\frac{1}{8\pi^2} \frac{\pi^2}{12}-\frac{1}{4 \pi^4}\frac{\pi^4}{90}-\frac{1}{4 \pi^4} \frac{7 \pi^4}{720} \right)\\ &=\left(\gamma+\ln(2 \pi) \right)\left(\frac{1}{96} -\frac{1}{360}- \frac{7 }{2880} \right)\\ &= \frac{15\left(\gamma+\ln(2 \pi) \right) }{2880}\\ &= \frac{15\gamma }{2880}+\frac{15\ln(2) }{2880}+\frac{15\ln(\pi) }{2880}\\ \end{aligned}$

$\begin{aligned} (3)&=\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\ln\left(k \right)}{ k}\int_0^{1/2} x^2 \sin\left(2 \pi k x \right)\,dx\\ &=\frac{1}{\pi}\sum_{k=1}^\infty \frac{\ln(k)}{k}\left(-\frac{(-1)^{k}}{8 \pi k}-\frac{1}{4 \pi^3 k^3}+\frac{(-1)^{k}}{4 \pi^3 k^3}\right)\\ &=-\frac{1}{8\pi^2}\sum_{k=1}^\infty \frac{(-1)^k\ln(k)}{k^2}-\frac{1}{4 \pi^4}\sum_{k=1}^\infty \frac{\ln(k)}{k^4}+\frac{1}{4 \pi^4}\sum_{k=1}^\infty \frac{(-1)^k\ln(k)}{k^4}\\ &=-\frac{1}{8\pi^2}\eta^\prime(2)+\frac{1}{4 \pi^4}\zeta^\prime(4)+\frac{1}{4 \pi^4}\eta^\prime(4)\\ &=-\frac{1}{8\pi^2}\eta^\prime(2)+\frac{1}{4 \pi^4}\zeta^\prime(4)+\frac{1}{4 \pi^4}\left(\frac{\pi^4}{720}\ln(2)+\frac78\zeta^\prime(4) \right)\\ &=-\frac{1}{8\pi^2}\eta^\prime(2)+\frac{\ln(2)}{2880}+\frac{15}{8}\frac{\zeta^\prime(4)}{4 \pi^4} \\ &=-\frac{1}{8\pi^2}\frac{\pi^2}{12}\left(2\ln(2)-12\ln A+\gamma+\ln(\pi) \right)+\frac{\ln(2)}{2880}+\frac{15}{8}\frac{1}{4 \pi^4}\left(\frac{\pi^4}{90}\left(\ln(2 \pi)+\gamma \right)-\frac{11 \pi^4}{540}-\frac{4 \pi^4}{3}\zeta^{\prime}(-3) \right) \\ &=-\frac{\left(2\ln(2)-12\ln A+\gamma+\ln(\pi) \right)}{96}+\frac{\ln(2)}{2880}+\frac{15\left(\ln(2 \pi)+\gamma \right)}{2880}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8} \\ &=-\frac{60\ln(2) }{2880}+\frac{360\ln A }{2880}-\frac{30 \gamma }{2880}-\frac{30 \ln(\pi) }{2880}+\frac{\ln(2)}{2880}+\frac{15\left(\ln(2 \pi)+\gamma \right)}{2880}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8} \\ &=-\frac{60\ln(2) }{2880}+\frac{360\ln A }{2880}-\frac{30 \gamma }{2880}-\frac{30 \ln(\pi) }{2880}+\frac{\ln(2)}{2880}+\frac{15\ln(2) }{2880}+\frac{15\ln(\pi) }{2880}+\frac{15\gamma }{2880}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8} \\ \end{aligned}$

Collecting all the results and putting together we obtain

$\begin{aligned} \int_0^{1/2} x^2 \ln\left(\Gamma(x)\right)\,dx&=\frac{60\ln 2 }{2880}+\frac{60\ln \pi }{2880}-\frac{3\zeta(3)}{32\pi^ 2}\\ &+\frac{15\gamma }{2880}+\frac{15\ln(2) }{2880}+\frac{15\ln(\pi) }{2880}\\ &-\frac{60\ln(2) }{2880}+\frac{360\ln A }{2880}-\frac{30 \gamma }{2880}-\frac{30 \ln(\pi) }{2880}+\frac{\ln(2)}{2880}+\frac{15\ln(2) }{2880}+\frac{15\ln(\pi) }{2880}+\frac{15\gamma }{2880}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8}\\ &=\frac{60 \ln(\pi)}{2880}+\frac{31 \ln(2)}{2880}+\frac{360\ln A }{2880}-\frac{3\zeta(3)}{32\pi^ 2}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8}\\ &=\frac{\ln \left(\pi^{60} 2^{31}A^{360} \right)}{2880} -\frac{3\zeta(3)}{32\pi^ 2}-\frac{11 }{1152}-\frac{5\zeta^{\prime}(-3)}{8}\qquad \blacksquare \end{aligned}$

Appendix

$\begin{aligned} \int_0^{1/2} x \cos(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{\pi k} x \cos(x)\,dx \qquad (2 \pi kx \to x)\\ &=\frac{1}{(2 \pi k)^2}\left(x\sin(x)\Big|_0^{ \pi k} -\int_0^{ \pi k}\sin(x)\,dx\right)\\ &=\frac{1}{(2 \pi k)^2}\left(\cos(x)\Big|_0^{ \pi k}\right)\\ &=\frac{1}{(2 \pi k)^2}\left((-1)^k-1\right)\\ &=-\left(\frac{1}{4 \pi^2 k^2}+\frac{(-1)^{k-1}}{4 \pi^2 k^2} \right) \qquad \blacksquare \end{aligned}$

$\begin{aligned} \int_0^1 x \sin(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{\pi k} x \sin(x)\,dx \qquad (2 \pi kx \to x)\\ &=\frac{1}{(2 \pi k)^2}\left(-x\cos(x)\Big|_0^{ \pi k} +\int_0^{ \pi k}\cos(x)\,dx\right)\\ &=\frac{1}{(2 \pi k)^2}\left(- \pi k(-1)^k\right)\\ &=\frac{(-1)^{k-1}}{4 \pi k}\qquad \blacksquare \end{aligned}$

$\begin{aligned} \int_0^{1/2} x^2 \cos(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^3}\int_0^{\pi k} x^2 \cos(x)\,dx \qquad (2 \pi kx \to x)\\ &=\frac{1}{(2 \pi k)^3}\left(x^2\sin(x)\Big|_0^{ \pi k} -2\int_0^{ \pi k}x\sin(x)\,dx\right)\\ &=\frac{-2}{(2 \pi k)^3}\left(\int_0^{ \pi k}x \sin x \,dx\right)\\ &=\frac{-2}{(2 \pi k)^3}\left(-x\cos x\Big|_0^{ \pi k}+\int_0^{ \pi k}\cos x \,dx\right)\\ &=\frac{2}{(2 \pi k)^3}\left(\pi k (-1)^k \right)\\ &= \frac{ (-1)^k}{4 \pi^2 k^2}\qquad \blacksquare \end{aligned}$

$\begin{aligned} \int_0^{1/2} x^2 \sin(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^3}\int_0^{ \pi k} x^2 \sin(x)\,dx \qquad (2 \pi kx \to x)\\ &=\frac{1}{(2 \pi k)^3}\left(-x^2\cos(x)\Big|_0^{ \pi k} +2\int_0^{ \pi k}x\cos(x)\,dx\right)\\ &=\frac{1}{(2 \pi k)^3}\left(- \pi^2 k^2(-1)^k+2\left((-1)^{k}-1 \right)\right)\\ &=\frac{(-1)^{k-1}}{8 \pi k}-\frac{1}{4 \pi^3 k^3}+\frac{(-1)^{k}}{4 \pi^3 k^3}\qquad \blacksquare \end{aligned}$

Derivative of the Dirichlet Eta function

$\eta^{\prime}(s)=\sum_{k=1}^{\infty} \frac{(-1)^{k} \ln k}{k^{s}}=2^{1-s} \ln (2) \zeta(s)+\left(1-2^{1-s}\right) \zeta^{\prime}(s)$

Then :

$\begin{aligned} \eta^{\prime}(2)&=\frac12\left(\ln(2)\zeta(2) \right)+\frac12 \zeta^\prime(2)\\ &=\frac12\left(\ln(2)\zeta(2) \right)+\frac12\zeta(2)\left(-12 \ln A + \gamma + \ln2 \pi \right) \\ &=\frac{\pi^2}{12}\left(2\ln(2)-12\ln A+\gamma+\ln(\pi) \right) \qquad \blacksquare \end{aligned}$

We used that:

$\zeta^{\prime}(2)=\zeta(2)\left(-12 \ln A + \gamma + \ln2 \pi \right)$

And

$\begin{aligned} \eta^{\prime}(4)&=\frac18\left(\ln(2)\zeta(4) \right)+\left(1-\frac18 \right)\zeta^\prime(4)\\ &=\frac{\pi^4}{720}\ln(2)+\frac78\zeta^\prime(4)\qquad \blacksquare \end{aligned}$

$\zeta^{\prime}(-3)$ in terms of $\zeta^{\prime}(4)$

Now, recall the functional equation of the Riemann Zeta function

$\zeta(s) = \frac{1}{\pi}(2 \pi)^{s} \sin \left( \frac{\pi s}{s} \right) \Gamma(1-s) \zeta(1-s)$

Differentiating w.r. to we obtain:

$\zeta'(s) = \frac{1}{\pi} \log(2 \pi)(2 \pi)^{s} \sin \left( \frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) + \frac{\pi}{2} (2 \pi)^{s} \cos \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)- \frac{1}{\pi}(2 \pi)^{s} \sin \left(\frac{\pi s}{2} \right)\Gamma^{'}(1-s) \zeta(1-s) - \frac{1}{\pi}(2 \pi)^{s} \sin \left(\frac{\pi s}{2} \right)\Gamma(1-s) \zeta^{\prime}(1-s)$

Let $s \to -3$

$\begin{aligned} \zeta^{\prime}(-3) &= \frac{\ln(2 \pi)}{8\pi^4} \sin \left( \frac{-3\pi }{2} \right) \Gamma(4) \zeta(4) + \frac{1}{16 \pi^3} \cos \left(\frac{-3\pi }{2} \right) \Gamma(4) \zeta(4)- \frac{1}{8\pi^4} \sin \left(\frac{-3\pi }{2} \right)\Gamma^{\prime}(4) \zeta(4) - \frac{1}{8\pi^4} \sin \left(\frac{-3\pi }{2} \right)\Gamma(4) \zeta^{\prime}(4)\\ &= \frac{\ln(2 \pi)}{8\pi^4} \frac{6\pi^4}{90} - \frac{1}{8\pi^4} \Gamma(4)\psi(4) \zeta(4) - \frac{6}{8\pi^4} \zeta^{\prime}(4)\\ &= \frac{\ln(2 \pi)}{120} - \frac{1}{8\pi^4} \frac{6\pi^4}{90}\left(-\gamma+\frac{11}{6} \right) - \frac{3}{4\pi^4} \zeta^{\prime}(4)\\ &= \frac{\ln(2 \pi)}{120} - \frac{1}{120}\left(-\gamma+\frac{11}{6} \right) - \frac{3}{4\pi^4} \zeta^{\prime}(4)\\ &= \frac{\ln(2 \pi)+\gamma}{120} - \frac{11}{720} - \frac{3}{4\pi^4} \zeta^{\prime}(4)\\ \end{aligned}$