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A very famous log-gamma integral due to Raabe is $$\int_0^1 \log \Gamma (x) \, dx = \frac{1}{2} \log (2\pi).$$ Several proofs of this result can be found here.

I would like to know about the evaluation of the similar looking log-gamma integral

$$I = \int_0^1 \frac{\log \Gamma (x)}{1 - x} \, dx.$$

I very much doubt a closed-form solution for this integral can be found, though I would really like to see one if it could be found (we live in hope you see). Here is what I have managed to come up with so far. Integrating by parts gives $$I = \int_0^1 \log (1 - x) \psi (x) \, dx.$$ Here $\psi (x)$ is the digamma function. Expanding the log term, after swapping the order of the summation with the integration we have $$I = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^n \psi (x) \, dx.$$ Now amazingly, a closed form solution for $\int_0^1 x^n \psi (x) \, dx$ exists. It is $$\int_0^1 x^n \psi (x) \, dx = \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} \left (\zeta'(-k) - \frac{B_{k + 1} H_k}{k + 1} \right ).$$ Thus $$I = \sum_{n = 1}^\infty \sum_{k = 0}^{n - 1} \frac{(-1)^{k + 1}}{n} \binom{n}{k} \left (\zeta'(-k) - \frac{B_{k + 1} H_k}{k + 1} \right ).$$ Here $H_n$ is the $n$th harmonic number, $B_n$ is the $n$th Bernoulli number, while $\zeta' (x)$ denotes the derivative of the Riemann zeta function. I am guessing not much more can be done with these sums.

Any other alternative approaches one could try?

omegadot
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    In the book "Handbuch der Theorie der Gammafunktion" of Nielsen (1906), you can find on page 203 a number of formulas for what amount to Fourier transforms of $\log \Gamma(x)$. Maybe this can help you towards an alternative approach? Maybe a direct formula is even in the book? Didn't have time to skim through it. Another tip, when you look for sophisticated integrals, always check Gradshteyn and Ryzhik to see if they have it tabulated. – Raskolnikov May 29 '20 at 07:46

2 Answers2

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Developing as series around $x=1$, we have $$\frac{\log (\Gamma (x))}{1-x}=-\sum_{n=0}^\infty \frac{\psi ^{(n)}(1)}{\Gamma (n+2)} (x-1)^n$$ $$\int_0^1\frac{\log (\Gamma (x))}{1-x}\,dx=\sum_{n=0}^\infty (-1)^{n+1}\frac{ \psi ^{(n)}(1)}{(n+1) \Gamma (n+2)}$$

The problem is that the convergence is quite slow. Summing up to $k$, we have $$\left( \begin{array}{cc} k & \text{partial sum} \\ 10 & 1.32631 \\ 100 & 1.40004 \\ 1000 & 1.40798 \\ \cdots & \cdots \\ \infty & 1.41321 \end{array} \right)$$

Claude Leibovici
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By changing $x=1-t$, the integral reads \begin{align} I &= \int_0^1 \frac{\log \Gamma (x)}{1 - x} \, dx\\ &=\int_0^1\frac{\log \Gamma (1-t)}t \, dt \end{align} Then, using the identity (DLMF) \begin{equation} \sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}t^{k}=-\gamma t+\ln\Gamma\left(1-t\right) \end{equation} one obtains, by swapping summation and integral, \begin{align} I&=\int_0^\infty\left[\gamma t+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}t^{k}\right]\frac{dt}{t}\\ &=\gamma+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k^2} \end{align} The series can be made more rapidly convergent by introducing the Basel series: \begin{equation} I=\frac{\pi^2}{6}-1+\gamma+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)-1}{k^2} \end{equation}

Paul Enta
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