MOMENTS OF THE LOGGAMMA FUNCTION BETWEEN 0 AND 1-PART 2

Following the previous post, We will evaluate the following two integrals today


\int_0^1 x^3\ln\left(\Gamma(x)\right)\,dx=\frac{\ln\left(2 \pi\right)}{8}+\frac{3 \zeta(3)}{8\pi^2}+\frac{11 }{720}+ \zeta^{\prime}(-3) -\ln A


\int_0^1 x^4\ln\left(\Gamma(x)\right)\,dx=\frac{\ln\left(2 \pi\right)}{10}+\frac{\zeta(3)}{2 \pi^2}-\frac{3 \zeta(5)}{4 \pi^4}+\frac{11}{360}+2\zeta^{\prime}(-3)-\ln A

Recall Kummer´s fourier expansion for LogGamma  0<x<1



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

Multiplying (1) by x^3 and integrating form 0 to 1


\begin{aligned} \int_0^1 x^3\ln\left(\Gamma(x)\right)\,dx&=\frac{\ln\left(2 \pi\right)}{2}\int_0^1 x^3 \,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^1 x^3\cos\left(2 \pi k x\right)\,dx + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi \right)+\ln(k)}{ \pi k}\int_0^1 x^3\sin\left(2 \pi k x \right)\,dx\\ &=\frac{\ln\left(2 \pi\right)}{8}+\frac12 \sum_{k=1}^{\infty}\frac{1}{k}\left( \frac{3}{4\pi^2k^2}\right)+\frac{\left(\gamma+\ln(2 \pi) \right)}{\pi}\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{3}{4\pi^3k^3}-\frac{1}{2 \pi k} \right)+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\ln(k)}{k}\left(\frac{3}{4\pi^3k^3}-\frac{1}{2 \pi k} \right)\\ &=\frac{\ln\left(2 \pi\right)}{8}+\frac{3 \zeta(3)}{8\pi^2}+\frac{3\left(\gamma+\ln(2 \pi) \right)}{4 \pi^4}\zeta(4)-\frac{\left(\gamma+\ln(2 \pi) \right)}{2 \pi^2}\zeta(2)+\frac{3}{4 \pi^4}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^4}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}\\ &=\frac{\ln\left(2 \pi\right)}{8}+\frac{3 \zeta(3)}{8\pi^2}+\frac{\left(\gamma+\ln(2 \pi) \right)}{120}-\frac{\left(\gamma+\ln(2 \pi) \right)}{12 }-\frac{3 \zeta^\prime(4)}{4 \pi^4}+\frac{\zeta^\prime(2)}{2 \pi^2}\\ &=\frac{\ln\left(2 \pi\right)}{8}+\frac{3 \zeta(3)}{8\pi^2}+\frac{\left(\gamma+\ln(2 \pi) \right)}{120}-\frac{\left(\gamma+\ln(2 \pi) \right)}{12 }-\frac{3 }{4 \pi^4}\left(\frac{\pi^{4}}{90}(\ln (2 \pi)+\gamma)-\frac{11 \pi^{4}}{540}-\frac{4 \pi^{4}}{3} \zeta^{\prime}(-3) \right)+\frac{1}{2 \pi^2}\left( \frac{\pi^2}{6}\left(\gamma+\ln(2\pi)-12\ln A \right)\right)\\ &=\frac{\ln\left(2 \pi\right)}{8}+\frac{3 \zeta(3)}{8\pi^2}+\frac{\left(\gamma+\ln(2 \pi) \right)}{120}-\frac{\left(\gamma+\ln(2 \pi) \right)}{12 }-\frac{\left(\gamma+\ln(2 \pi) \right)}{120}+\frac{11 }{720}+ \zeta^{\prime}(-3) +\frac{\left(\gamma+\ln(2 \pi) \right)}{12 }-\ln A \\ &=\frac{\ln\left(2 \pi\right)}{8}+\frac{3 \zeta(3)}{8\pi^2}+\frac{11 }{720}+ \zeta^{\prime}(-3) -\ln A \qquad \blacksquare\\ \end{aligned}

where we used

\zeta^\prime(2)=\frac{\pi^2}{6}\left(\ln(\pi)+\ln(2)+\gamma-12\ln A \right)

\zeta^{\prime}(4)=\frac{\pi^{4}}{90}(\ln (2 \pi)+\gamma)-\frac{11 \pi^{4}}{540}-\frac{4 \pi^{4}}{3} \zeta^{\prime}(-3)

Multiplying (1) by x^4 and integrating form 0 to 1


\begin{aligned} \int_0^1 x^4\ln\left(\Gamma(x)\right)\,dx&=\frac{\ln\left(2 \pi\right)}{2}\int_0^1 x^4 \,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^1 x^4\cos\left(2 \pi k x\right)\,dx + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi \right)+\ln(k)}{ \pi k}\int_0^1 x^4\sin\left(2 \pi k x \right)\,dx\\ &=\frac{\ln\left(2 \pi\right)}{10}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{ \pi^2 k^2}-\frac{3}{2 \pi^4 k^4} \right)+\frac{\left(\gamma +\ln(2 \pi) \right)}{\pi}\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{3}{2 \pi^3 k^3}-\frac{1}{2 \pi k} \right)+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\ln(k)}{k}\left(\frac{3}{2 \pi^3 k^3}-\frac{1}{2 \pi k} \right)\\ &=\frac{\ln\left(2 \pi\right)}{10}+\frac{\zeta(3)}{2 \pi^2}-\frac{3 \zeta(5)}{4 \pi^4}+\frac{3\left(\gamma +\ln(2 \pi) \right)}{2\pi^4}\zeta(4)-\frac{\left(\gamma +\ln(2 \pi) \right)}{2\pi^2}\zeta(2)+\frac{3}{2 \pi^4}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^4}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}\\ &=\frac{\ln\left(2 \pi\right)}{10}+\frac{\zeta(3)}{2 \pi^2}-\frac{3 \zeta(5)}{4 \pi^4}+\frac{\left(\gamma +\ln(2 \pi) \right)}{60}-\frac{\left(\gamma +\ln(2 \pi) \right)}{12}-\frac{3\zeta^\prime(4)}{2 \pi^4}+\frac{\zeta^\prime(2)}{2 \pi^2}\\ &=\frac{\ln\left(2 \pi\right)}{10}+\frac{\zeta(3)}{2 \pi^2}-\frac{3 \zeta(5)}{4 \pi^4}-\frac{\left(\gamma +\ln(2 \pi) \right)}{15}-\frac{3}{2 \pi^4}\left( \frac{\pi^{4}}{90}(\ln (2 \pi)+\gamma)-\frac{11 \pi^{4}}{540}-\frac{4 \pi^{4}}{3} \zeta^{\prime}(-3)\right)+\frac{1}{2 \pi^2}\left(\frac{\pi^2}{6}\left(-12 \ln A + \gamma + \ln2 \pi \right) \right)\\ &=\frac{\ln\left(2 \pi\right)}{10}+\frac{\zeta(3)}{2 \pi^2}-\frac{3 \zeta(5)}{4 \pi^4}-\frac{\left(\gamma +\ln(2 \pi) \right)}{15}-\frac{\left(\gamma +\ln(2 \pi) \right)}{60}+\frac{11}{360}+2\zeta^{\prime}(-3)-\ln A+\frac{\left(\gamma +\ln(2 \pi) \right)}{12}\\ &=\frac{\ln\left(2 \pi\right)}{10}+\frac{\zeta(3)}{2 \pi^2}-\frac{3 \zeta(5)}{4 \pi^4}+\frac{11}{360}+2\zeta^{\prime}(-3)-\ln A \qquad \blacksquare\\ \end{aligned}

Appendix

\begin{aligned} \int_0^1 x^3\cos(2 \pi k x)\,dx&=\frac{1}{(2 \pi k)^4}\int_0^{2\pi k} x^3\cos(x)\,dx\\ &=\frac{1}{(2 \pi k)^4}\left(x^3\sin(x)\Big|_0^{2\pi k}-3\int_0^{2\pi k} x^2\sin(x)\,dx \right)\\ &=-\frac{3}{(2 \pi k)^4}\left(-x^2\cos(x)\Big|_0^{2\pi k}+2\int_0^{2\pi k} x\cos(x)\,dx \right)\\ &=\frac{3(2 \pi k)^2}{(2 \pi k)^4}-\frac{6}{(2 \pi k)^4}\left(x\sin(x)\Big|_0^{2\pi k}-\int_0^{2\pi k} \sin(x)\,dx \right)\\ &=\frac{3}{4 \pi^2 k^2} \qquad \blacksquare \end{aligned}

\begin{aligned} \int_0^1 x^3\sin(2 \pi k x)\,dx&=\frac{1}{(2 \pi k)^4}\int_0^{2\pi k} x^3\sin(x)\,dx\\ &=\frac{1}{(2 \pi k)^4}\left(-x^3\cos(x)\Big|_0^{2 \pi k}+3\int_0^{2 \pi k} x^3 \cos(x)\,dx \right)\\ &=-\frac{(2 \pi k)^3}{(2 \pi k)^4}+\frac{3}{(2 \pi k)^4}\left(x^2\sin(x)\Big|_0^{2 \pi k} -2\int_0^{2 \pi k}x \sin(x)\,dx\right)\\ &=-\frac{1}{2 \pi k}-\frac{6}{(2 \pi k)^4}\left(-x\cos(x)\Big|_0^{2 \pi k} +\int_0^{2 \pi k} \cos(x)\,dx\right)\\ &=-\frac{1}{2 \pi k}-\frac{12 \pi k}{16 \pi^4 k^4} \\ &=\frac{3}{4 \pi^3 k^3}-\frac{1}{2 \pi k} \qquad \blacksquare \end{aligned}

\begin{aligned} \int_0^1 x^4\cos(2 \pi k x)\,dx&=\frac{1}{(2 \pi k)^5}\int_0^{2\pi k} x^4\cos(x)\,dx\\ &=\frac{1}{(2 \pi k)^5}\left(x^4\sin(x)\Big|_0^{2 \pi k}-4\int_0^{2 \pi k} x^3 \sin(x)\,dx \right)\\ &=-\frac{4}{(2 \pi k)^5}\left(-x^3 \cos(x)\Big|_0^{2 \pi k}+3\int_0^{2 \pi k}x^2 \cos(x)\,dx \right)\\ &=\frac{4(2 \pi k)^3}{(2 \pi k)^5}-\frac{12}{(2 \pi k)^5}\left(x^2 \sin(x)\Big|_0^{2 \pi k}+2\int_0^{2 \pi k}x \sin(x)\,dx \right)\\ &=\frac{4}{(2 \pi k)^2}-\frac{24}{(2 \pi k)^5}\left(x \cos(x)\Big|_0^{2 \pi k}-\int_0^{2 \pi k} \cos(x)\,dx \right)\\ &=\frac{1}{ \pi^2 k^2}-\frac{24(2 \pi k)}{(2 \pi k)^5}\\ &=\frac{1}{ \pi^2 k^2}-\frac{3}{2 \pi^4 k^4} \qquad \blacksquare\\ \end{aligned}

\begin{aligned} \int_0^1 x^4\sin(2 \pi k x)\,dx&=\frac{1}{(2 \pi k)^5}\int_0^{2\pi k} x^4\sin(x)\,dx\\ &=\frac{1}{(2 \pi k)^5}\left(-x^4\cos(x)\Big|_0^{2 \pi k}+4\int_0^{2 \pi k} x^3 \cos(x)\,dx \right)\\ &=-\frac{(2 \pi k)^4}{(2 \pi k)^5}+\frac{4}{(2 \pi k)^5}\left(x^3\sin(x)\Big|_0^{2 \pi k} -3\int_0^{2 \pi k}x^2 \sin(x)\,dx\right)\\ &=-\frac{1}{2 \pi k}-\frac{12}{(2 \pi k)^5}\left(-x^2\cos(x)\Big|_0^{2 \pi k} +2\int_0^{2 \pi k}x \cos(x)\,dx\right)\\ &=-\frac{1}{2 \pi k}+\frac{12(2 \pi k)^2}{(2 \pi k)^5}-\frac{24}{(2 \pi k)^5}\left(x\sin(x)\Big|_0^{2 \pi k} -\int_0^{2 \pi k} \sin(x)\,dx\right)\\ &=-\frac{1}{2 \pi k}+\frac{12}{8 \pi^3 k^3}\\ &=\frac{3}{2 \pi^3 k^3}-\frac{1}{2 \pi k} \qquad \blacksquare \end{aligned}

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