Solve $\int_0^1\ln^2\Gamma(x)\,\mathrm{d}x$

Question

I want to solve the following integral but after some work I didn't find a way to go. Could anyone give me a hint? \begin{equation} I=\int_{0}^{1}\ln^2\Gamma(x)\,\mathrm{d}x \end{equation} The answer is \begin{equation} I=\frac{\ln^2 (2\pi)}{3}+\frac{\pi^2}{48}+\frac{\gamma\ln(2\pi)}{6}+\frac{\gamma^2}{12}+\frac{\zeta''(2)}{2\pi^2}-\frac{\zeta'(2)\ln (2\pi)}{\pi^2}-\frac{\gamma\zeta'(2)}{\pi^2} \end{equation} They only give a hint (using the Fourier Series) which I looked up at https://de.wikipedia.org/wiki/Gammafunktion. \begin{equation} \ln\Gamma(x) = \left(\tfrac{1}{2}-x\right) \bigl(\gamma + \ln(2\pi)\bigr) + \frac{1}{2} \ln\frac{\pi}{\sin(\pi x)} + \frac{1}{\pi} \sum_{k=2}^\infty \frac{\ln k}{k} \sin(2\pi k x) \end{equation}

Want I have tried so far:

• squared the series
• integration by parts and the the fourier series

Answer 1

Theorem $6.1$ from the paper A generalized polygamma function by Olivier Espinosa and Victor H. Moll will bring the light over your question. (see the special case $k=k'=1$)

Answer 2

Use Parseval's Theorem as @James Arathoon metioned and use the Fourier Series given here:

Integral that arises from the derivation of Kummer's Fourier expansion of $\ln{\Gamma(x)}$.

Answer 3

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

$$ \begin{align} &\int_{0}^{1}\log(\Gamma(x))\,dx=\frac{\ln(2 \pi)}{2} \tag{1}\\ & \\ &\int_{0}^{1}\log(\Gamma(x))\cos(2 \pi k x)\,dx=\frac{1}{4 k} \tag{2}\\ &\\ &\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)\,dx=\frac{\gamma+\log(2 \pi k)}{2 \pi k} \tag{3} \end{align} $$

Also recall Kummer´s fourier series for the Loggamma function

$$\log(\Gamma(x))=\frac{\log(2 \pi)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos(2 \pi k x) + \sum_{k=1}^{\infty}\frac{\gamma+\log(2 \pi )+\ln(k)}{ \pi k}\sin(2 \pi k x) \tag{4}$$

Multiplying both sides of (4) by $\log(\Gamma(x))$ and integrating from zero to one and considering the results $(1)-(3)$ we obtain

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

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If we now let $x \to 1-x$ in $(4)$ we obtain

$$\ln\left(\Gamma(1-x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) - \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{5}$$

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

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

Answer 4

An almost-answer, but for brevity some of the details have been skipped.

If you square your formula for $\ln\Gamma(x)$, you'll find many of the terms integrate to $0$ on $[0,\,1]$ due to being of the form $o(x-\tfrac12)$ for odd $o$. Let $f(x)\sim g(x)$ denote the equivalence relation $\int_0^1(f(x)-g(x))dx=0$, so $f(x)\sim\int_0^1f(t)dt$. Before we go any further, I'll mention the famous result $\ln\sin(\pi x)\sim-\ln2$, and something with a similar proof,$$\ln^2\sin(\pi x)\sim\pi\ln4+\ln^22-2\ln\sin\frac{\pi x}{2}\ln\cos\frac{\pi x}{2}.$$Oh, and one more thing I'll need in a moment, for integer $k\ne0$: $x\sin(2\pi kx)\sim\frac{-1}{2\pi k}$. So$$\begin{align}\ln^2\Gamma(x)&=\color{red}{(\gamma+\ln(2\pi))^2(\tfrac12-x)^2}+\color{orange}{\frac14\ln^2\frac{\pi}{\sin(\pi x)}}+\color{limegreen}{\frac{1}{2\pi^2}\sum_{k\ge2}\frac{\ln^2k}{k^2}}\\&+\color{blue}{\frac{\gamma+\ln(2\pi)}{\pi}(1-2x)\sum_{k\ge2}\frac{\ln k}{k}\sin(2\pi kx)}\\&\sim\color{red}{\frac{(\gamma+\ln(2\pi))^2}{12}}\\&+\color{orange}{\frac14(\ln^2\pi+\ln4\ln\pi+\pi\ln4+\ln^22-2\ln\sin\tfrac{\pi x}{2}\ln\cos\tfrac{\pi x}{2})}\\&+\color{limegreen}{\frac{\zeta^{\prime\prime}(2)}{2\pi^2}}-\color{blue}{\frac{\gamma+\ln(2\pi)}{\pi^2}\zeta^\prime(2).}\end{align}$$So now we just need to evaluate $\int_0^1\ln\sin\frac{\pi x}{2}\ln\cos\frac{\pi x}{2}dx$.