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On a log Gamma integral using Riemann sums

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \log \Gamma (x) \, \mathrm{d}x\]

using Riemann sums.

Solution

Partition the interval [0, 1] into n subintervals of length \frac{1}{n}. This produces,

(1)   \begin{equation*}  \int_{0}^{1} \log \Gamma(x) \, \mathrm{d}x = \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) \end{equation*}

On the other hand, assuming n is even:

    \begin{align*} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) &= \frac{1}{n} \log \prod_{k=1}^{n} \Gamma \left ( \frac{k}{n} \right ) \\ &= \frac{1}{n} \log \prod_{k=1}^{n/2} \Gamma \left ( \frac{k}{n} \right ) \Gamma \left ( 1 - \frac{k}{n} \right )\\ &=\frac{1}{n} \log \prod_{k=1}^{n/2} \frac{\pi}{\sin \frac{\pi k}{n}} \\ &= \log \sqrt{\pi} - \log \left ( \prod_{k=1}^{n} \sin \frac{\pi k}{n} \right )^{1/n}\\ &= \log \frac{\sqrt{2 \pi}}{\left ( 2n \right )^{1/2n}} \end{align*}

since it holds that

    \[\prod_{k=1}^{n} \sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}\]

Euler’s Gamma reflection formula was used at line (3). Letting n \rightarrow +\infty we get that

    \[\int_0^1 \log \Gamma(x) \, \mathrm{d}x = \log \sqrt{2\pi}\]

If n is odd we work similarly.

 

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