evaluate $\int_1^2\ln{(\Gamma{(x)})}dx$

Question

I want to find the value of $\int_1^2\ln{(\Gamma{(x)})}dx$.

Wolfram alpha can compute that $\int_1^2\ln{(\Gamma{(x)})}dx = \frac{\ln(2\pi)-2}{2}$.

http://www.wolframalpha.com/input/?i=int_1%5E2(log(gamma(x)))dx

How can I prove it ?

Thank you.

Answer 1

From Raabe's formula:

$$\int _{a}^{a+1}\ln \Gamma (z)\,\mathrm {d} z={\tfrac {1}{2}}\ln 2\pi +a\ln a-a,\quad a>0$$ Therefore, $$\int _{1}^{2}\ln \Gamma (z)\,\mathrm {d} z={\tfrac {1}{2}}\ln 2\pi -1$$

Answer 2

Begin by noting that

$$\int_k^{k+1} dx \, \log{\Gamma(x)} = \int_0^1 dx \, \log{\Gamma(x+k)} $$

$$\log{\Gamma(x+k)} = \log{\Gamma(x)} + \sum_{m=0}^{k-1} \log{(x+m)} $$

Thus,

$$\begin{align}\int_k^{k+1} dx \, \log{\Gamma(x)} &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} \int_0^1 dx \, \log{(x+m)}\\ &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} [(m+1) \log{(m+1)} - m \log{m} - (m+1)+m] \\ &= \int_0^1 dx \, \log{\Gamma(x)} + k \log{k} - k \end{align}$$

To evaluate the integral on the RHS, use the duplication formula:

$$\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right ) = 2^{1-x} \sqrt{\pi} \Gamma(x) $$

so that

$$\log{\Gamma(x)} = \log{\left [\frac{\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right )}{\sqrt{\pi} 2^{1-x}} \right ]} = -\frac12 \log{\pi} - \log{2} + \log{\Gamma \left ( \frac{x}{2} \right )} + \log{\Gamma \left ( \frac{x+1}{2} \right )} + x \log{2}$$

Thus,

$$\begin{align}\int_0^1 dx \, \log{\Gamma(x)} &= -\frac12 \log{(2 \pi)} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x}{2} \right )} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x+1}{2} \right )}\\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1/2} dx \, \log{\Gamma(x)} + 2 \int_{1/2}^1 dx \, \log{\Gamma(x)} \\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1} dx \, \log{\Gamma(x)} \end{align}$$

$$\implies \int_0^1 dx \, \log{\Gamma(x)} = \frac12 \log{(2 \pi)} $$

and therefore

$$\int_1^2 dx \, \log{\Gamma(x)} = \frac12 \log{(2 \pi)} -1 $$

Answer 3

HINT:

The log gamma function can be defined as:

$$\ln\Gamma\left(\text{z}\right)=-\gamma\cdot\text{z}-\ln\left(\text{z}\right)+\sum_{\text{k}=1}^\infty\left\{\frac{\text{z}}{\text{k}}-\ln\left(1+\frac{\text{z}}{\text{k}}\right)\right\}$$

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So, when we take the integral of both sides:

$$\int\ln\Gamma\left(\text{z}\right)\space\text{d}\text{z}=-\gamma\int\text{z}\space\text{d}\text{z}-\int\ln\left(\text{z}\right)\space\text{d}\text{z}+\sum_{\text{k}=1}^\infty\left\{\frac{1}{\text{k}}\int\text{z}\space\text{d}\text{z}-\int\ln\left(1+\frac{\text{z}}{\text{k}}\right)\space\text{d}\text{z}\right\}$$

Now, use:

1. $$\int\text{z}\space\text{d}\text{z}=\frac{\text{z}^2}{2}+\text{C}$$
2. Using integration by parts: $$\int\ln\left(\text{z}\right)\space\text{d}\text{z}=\text{z}\ln\left(\text{z}\right)-\int1\space\text{d}\text{z}=\text{z}\left(\ln\left(\text{z}\right)-1\right)+\text{C}$$
3. Use a substitution $u=1+\frac{\text{z}}{\text{k}}$ and $\text{d}u=\frac{1}{\text{k}}\space\text{d}\text{z}$, after that use integration by parts: $$\int\ln\left(1+\frac{\text{z}}{\text{k}}\right)\space\text{d}\text{z}=\text{k}\int\ln\left(u\right)\space\text{d}u=\left(\text{k}+\text{z}\right)\left(\ln\left(\frac{\text{k}+\text{z}}{\text{k}}\right)-1\right)+\text{C}$$

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When we use definite integrals:

1. $$\int_1^2\text{z}\space\text{d}\text{z}=\frac{3}{2}$$
2. $$\int_1^2\ln\left(\text{z}\right)\space\text{d}\text{z}=\ln\left(4\right)-1$$
3. $$\int_1^2\ln\left(1+\frac{\text{z}}{\text{k}}\right)\space\text{d}\text{z}=(2+\text{k})\ln\left(\frac{2+\text{k}}{\text{k}}\right)-(1+\text{k})\ln\left(1+\frac{1}{\text{k}}\right)-1$$

Answer 4

If we start from the Weierstrass product for the $\Gamma$ function: $$ \Gamma(x+1)=e^{-\gamma x}\prod_{n\geq 1}\left(1+\frac{x}{n}\right)^{-1}e^{x/n}\tag{1} $$ we get $$\psi(x+1)=\frac{d}{dx}\log\Gamma(x+1) = \sum_{n\geq 1}\left(\log\left(1+\frac{1}{n}\right)-\frac{1}{n+x}\right)\tag{2}$$ and by integration by parts: $$ \int_{0}^{1}\log\Gamma(x+1)\,dx = \int_{0}^{1}x\,\psi(x+1)\,dx =\sum_{n\geq 1}\left(\frac{1}{2}\log\left(1+\frac{1}{n}\right)-\int_{0}^{1}\frac{x\,dx}{n+x}\right)\tag{3}$$ hence: $$ \int_{0}^{1}\log\Gamma(x+1)\,dx = \sum_{n\geq 1}\left(-1+\left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)\right)\tag{4}$$ where: $$\sum_{n=1}^{N}\left(-1+\left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)\right)=-N+\left(N+\frac{1}{2}\right)\log(N+1)-\sum_{n=1}^{N-1}\log(n+1)\tag{5}$$ and by Stirling's inequality $$ \sum_{n=1}^{N-1}\log(n+1)=\log\Gamma(N)=\left(N-\frac{1}{2}\right)-N\log(N)+\frac{1}{2}\log(2\pi)+\frac{1}{12N}+O\left(\frac{1}{N^3}\right)\tag{6}$$ so by $(4),(5),(6)$ we have: $$ \int_{0}^{1}\log\Gamma(x+1)\,dx = \color{red}{-1+\log\sqrt{2\pi}}.\tag{7}$$ This is essentially a proof of Raabe's formula.

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An alternative approach: $$\begin{eqnarray*} \int_{0}^{1}\log\Gamma(x+1)\,dx &=& \int_{0}^{1}\log(x)\,dx+\int_{0}^{1/2}\log\left(\Gamma(x)\Gamma(1-x)\right)\,dx\\ &=& -1+\frac{\log\pi}{2}-\int_{0}^{1/2}\log\sin(\pi x)\,dx\\&=&-1+\frac{1}{2}\log\pi-\frac{1}{2}\int_{0}^{1}\log\sin(\pi x)\,dx\\&=&-1+\frac{1}{2}\log\pi+\frac{1}{2}\log 2 \tag{8}\end{eqnarray*}$$ where the last identity follows from the well-known identity $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n}\tag{9} $$ and Riemann sums.