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

Today we will solve the two following integrals


\int_0^1 x \ln\left(\Gamma(x)\right)\,dx=\frac{1}{4} \ln \frac{2 \pi}{A^4}


\int_0^1 x^2 \ln\left(\Gamma(x)\right)\,dx=\frac{1}{6} \ln \frac{2 \pi}{A^6} +\frac{\zeta(3)}{4\pi^ 2}


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


\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)


   
  Then, we can write the first integral as


\begin{aligned} \int_0^1 x \ln\left(\Gamma(x)\right)\,dx&=\frac{\ln 2 \pi}{2}\int_0^1 x\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^1 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 x \sin\left(2 \pi k x \right)\,dx\\ &=\frac{\ln 2 \pi}{4}+ \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\left(-\frac{1}{2 \pi k} \right)\\ &=\frac{\ln 2 \pi}{4}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln\left(2 \pi k\right)}{k^2}\\ &=\frac{\ln 2 \pi}{4}-\frac{\gamma}{12}-\frac{\ln 2 \pi}{2\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln k}{k^2}\\ &=\frac{\ln 2 \pi}{4}-\frac{\gamma}{12}-\frac{\ln 2 \pi}{12}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln k}{k^2}\\ &=\frac{\ln 2 \pi}{6}-\frac{\gamma}{12}+\frac{1}{2\pi^2}\zeta^\prime(2)\\ &=\frac{\ln 2 \pi}{6}-\frac{\gamma}{12}+\frac{\zeta(2)}{2\pi^2} \left(-12 \ln A + \gamma + \ln2 \pi \right)\\ &=\frac{\ln 2 \pi}{6}- \ln A +\frac{\ln2 \pi}{12} \\ &=\frac{\ln 2 \pi}{4}- \ln A \\ &=\frac{1}{4} \ln \frac{2 \pi}{A^4} \qquad \blacksquare\\ \end{aligned}


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

  
 For the second integral we follow the same approach:


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

Appendix

\begin{aligned} \int_0^1 x \cos(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{2 \pi k} x \cos(x)\,dx \qquad (2 \pi kx \to x)\\ &=\frac{1}{(2 \pi k)^2}\left(x\sin(x)\Big|_0^{2 \pi k} -\int_0^{2 \pi k}\sin(x)\,dx\right)\\ &=\frac{1}{(2 \pi k)^2}\left(-\cos(x)\Big|_0^{2 \pi k}\right)\\ &=\frac{1}{(2 \pi k)^2}\left(-1+1\right)\\ &=0 \qquad \blacksquare \end{aligned}

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

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

\begin{aligned} \int_0^1 x^2 \sin(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^3}\int_0^{2 \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^{2 \pi k} +2\underbrace{\int_0^{2 \pi k}x\cos(x)\,dx}_{=0}\right)\\ &=\frac{1}{(2 \pi k)^3}\left(-2 \pi k\right)^2\\ &=-\frac{1}{2 \pi k}\qquad \blacksquare \end{aligned}

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