INTEGRAL SQUARE OF LOGGAMMA

Today´s post we will prove the following two Loggamma integrals


\int_{0}^{1}\log^2(\Gamma(x))\,dx=\frac{\gamma^2}{12}+\frac{\pi^2}{48}+\frac{\gamma\log(2 \pi ) }{6}+\frac{\log^2(2 \pi)}{3}-\frac{\left(\gamma+\log(2 \pi ) \right)}{ \pi^2}\zeta^\prime(2)+\frac{\zeta^{\prime \prime}(2)}{2 \pi^2}


\int_0^1\ln\left(\Gamma(x)\right)\ln\left(\Gamma(1-x)\right)\,dx=\frac{\log^2(2 \pi)}{4}+\frac{\pi^2}{48} - \frac{\left(\gamma+\log(2 \pi ) \right)^2}{12 }+\frac{\zeta^\prime(2)\left(\gamma+\log(2 \pi ) \right)}{ \pi^2}-\frac{\zeta^{\prime \prime}(2)}{2 \pi^2}


Recall the integrals below from the coefficients of the fourier series of Loggamma function

\begin{aligned} &\int_{0}^{1}\log(\Gamma(x))\,dx=\frac{\ln(2 \pi)}{2}\\ & \\ &\int_{0}^{1}\log(\Gamma(x))\cos(2 \pi k x)\,dx=\frac{1}{4 k}\\ &\\ &\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)\,dx=\frac{\gamma+\log(2 \pi k)}{2 \pi k} \end{aligned}(1)

And also the fourier series of the Loggama 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)(2)


Multiplying both sides of (2) by  \log(\Gamma(x))  and integrating from zero to one and using (1)


\begin{aligned} \int_{0}^{1}\log^2(\Gamma(x))\,dx&=\frac{\log(2 \pi)}{2}\int_{0}^{1}\log(\Gamma(x)) \,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_{0}^{1}\log(\Gamma(x))\cos(2 \pi k x)\,dx + \sum_{k=1}^{\infty}\frac{\gamma+\log(2 \pi )+\ln(k)}{ \pi k}\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)\,dx\\ &=\frac{\log^2(2 \pi)}{4}+\sum_{k=1}^{\infty}\frac{1}{2k}\frac{1}{4k}+\sum_{k=1}^{\infty}\frac{\gamma+\log(2 \pi )+\ln(k)}{ \pi k}\left(\frac{\gamma+\log(2 \pi )+\ln(k)}{2 \pi k} \right)\\ &=\frac{\log^2(2 \pi)}{4}+\frac18 \zeta(2)+ \frac{\left(\gamma+\log(2 \pi ) \right)^2}{2 \pi^2}\zeta(2)+\frac{\left(\gamma+\log(2 \pi ) \right)}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}+\frac{\left(\gamma+\log(2 \pi ) \right)}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln^2(k)}{k^2}\\ &=\frac{\log^2(2 \pi)}{4}+\frac{\log^2(2 \pi )}{12}+\frac{\pi^2}{48}+ \frac{\gamma^2}{12}+\frac{\gamma\log(2 \pi ) }{6}-\frac{\left(\gamma+\log(2 \pi ) \right)}{2 \pi^2}\zeta^\prime(2)-\frac{\left(\gamma+\log(2 \pi ) \right)}{2 \pi^2}\zeta^\prime(2)+\frac{\zeta^{\prime \prime}(2)}{2 \pi^2}\\ &= \frac{\gamma^2}{12}+\frac{\pi^2}{48}+\frac{\gamma\log(2 \pi ) }{6}+\frac{\log^2(2 \pi)}{3}-\frac{\left(\gamma+\log(2 \pi ) \right)}{ \pi^2}\zeta^\prime(2)+\frac{\zeta^{\prime \prime}(2)}{2 \pi^2} \qquad \blacksquare\\ \end{aligned}

If we now let x \to 1-x in (2)

\ln\left(\Gamma(1-x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k(1- x)\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k (1- x) \right)(3)


Recall the formulas for the sine and cosine of a difference:


\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)(4)

\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)(5)


We therefore obtain


\begin{aligned} \cos(2 \pi k -2 \pi k x)&=\cos(2 \pi k)\cos(2 \pi k x)+\sin(2 \pi k)\sin(2 \pi k x)\\ &=\cos(2 \pi k x) \end{aligned}

And

\begin{aligned} \sin(2 \pi k -2 \pi k x)&=\sin(2 \pi k)\cos(2 \pi k x)-\cos(2 \pi k)\sin(2 \pi k x)\\ &=-\sin(2 \pi k x) \end{aligned}

Then (3) becomes


\ln\left(\Gamma(1-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)(6)


Now we multiply (6) by  \ln\left(\Gamma(x)\right)  and integrate from 0 to 1


\begin{aligned} \int_0^1\ln\left(\Gamma(x)\right)\ln\left(\Gamma(1-x)\right)\,dx&=\frac{\ln\left(2 \pi\right)}{2}\int_0^1\ln\left(\Gamma(x)\right)\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^1\ln\left(\Gamma(x)\right)\cos\left(2 \pi k x\right)\,dx - \sum_{k=1}^{\infty}\frac{\left(\gamma+\ln\left(2 \pi \right)+\ln(k)\right)}{ \pi k}\int_0^1\ln\left(\Gamma(x)\right)\sin\left(2 \pi k x \right)\,dx\\ &=\frac{\log^2(2 \pi)}{4}+\sum_{k=1}^{\infty}\frac{1}{2k}\frac{1}{4k}-\sum_{k=1}^{\infty}\frac{\left(\gamma+\log(2 \pi )+\ln(k)\right)}{ \pi k}\left(\frac{\gamma+\log(2 \pi )+\ln(k)}{2 \pi k} \right)\\ &=\frac{\log^2(2 \pi)}{4}+\frac18 \zeta(2)- \frac{\left(\gamma+\log(2 \pi ) \right)^2}{2 \pi^2}\zeta(2)-\frac{\left(\gamma+\log(2 \pi ) \right)}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}-\frac{\left(\gamma+\log(2 \pi ) \right)}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln(k)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln^2(k)}{k^2}\\ &=\frac{\log^2(2 \pi)}{4}+\frac{\pi^2}{48} - \frac{\left(\gamma+\log(2 \pi ) \right)^2}{12 }+\frac{\zeta^\prime(2)\left(\gamma+\log(2 \pi ) \right)}{ \pi^2}-\frac{\zeta^{\prime \prime}(2)}{2 \pi^2} \qquad \blacksquare\\ \end{aligned}

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