Log o' Gamma

$\displaystyle I = \int_{a}^{a+1} \ln \Gamma (x) dx$

Apply integration by parts with $u=\ln \Gamma (x)$ and $dv=dx$ , and use the definition of the Digamma Function :

$\displaystyle I = x\ln \Gamma (x) |_{a}^{a+1} - \int_{a}^{a+1} x\psi (x) dx$

Recall the Harmonic Number - Digamma Function Relation:

$\displaystyle \psi (x) = H_{x-1} - \gamma = H_x - \frac{1}{x} - \gamma$ ,

where $\gamma$ is the Euler-Mascheroni Constant .

$\displaystyle => I = a\ln a + \ln a! -\int_a^{a+1} x(H_x-\frac{1}{x} -\gamma) dx$

$\displaystyle = a\ln a +\ln a! +1+\frac{2a+1}{2}\gamma - \int_a^{a+1} xH_x dx$

$\displaystyle = a\ln a +\ln a! +1+\frac{2a+1}{2}\gamma -\int_{a}^{a+1} x \sum_{k=1}^{\infty} \frac{x }{k(x+k)} dx$

$\displaystyle = a\ln a +\ln a! +1+\frac{2a+1}{2}\gamma -\sum_{k=1}^{\infty} \int_{a}^{a+1} \frac{x^2 dx }{k(x+k)}$

this follows from the definition of the Harmonic Number for all real values. Now let's work out the last term alone: this integral can be evaluated easily using basic integration techniques, to get at last:

$\displaystyle \sum_{k=1}^{\infty} \int_{a}^{a+1} \frac{x^2 dx }{k(x+k)}$

$\displaystyle = \sum_{k=1}^{\infty} \bigg(\frac{2a+1}{2}\frac{1}{k} -1 +k\ln (a+k+1) - k \ln (a+k) \bigg)$

$\displaystyle = \lim_{n\to \infty} \sum_{k=1}^{n} \bigg(\frac{2a+1}{2}{\color{#20A900}{\frac{1}{k} }}- {\color{#3D99F6}{1 }}+{\color{#D61F06}{(k+1) \ln (a+k+1) - k \ln (a+k)}} - {\color{#69047E}{\ln (a+k+1) }}\bigg)$

$\displaystyle \lim_{n\to \infty} \bigg(\frac{2a+1}{2}{\color{#20A900}{H_n}}- {\color{#3D99F6}{n }}+{\color{#D61F06}{(n+1) \ln (a+n+1) - \ln (a+1)}} - {\color{#69047E}{\ln \frac{(a+n+1)!}{(a+1)! }}}\bigg)$

Rearrange the logarithms into one log, with $n=\ln e^n$ :

$\displaystyle = \lim_{n\to \infty} \bigg(\frac{2a+1}{2} H_n +\ln \bigg( \frac{(a+n+1)^{n+1} a!}{(a+n+1)! e^n } \bigg) \bigg)$

Use Stirling's formula for the factorial in the log ( as we have $n \to \infty$ ). After simplification, we get:

$\displaystyle = \lim_{n\to \infty} \bigg(\frac{2a+1}{2} H_n +\ln \bigg( \frac{ e^{a+1} a!}{(a+n+1)^{a+\frac{1}{2}} \sqrt{2\pi} } \bigg) \bigg)$

$\displaystyle = \lim_{n\to \infty} (\frac{2a+1}{2} H_n -\frac{2a+1}{2} \ln (n+a+1) ) +\ln \bigg( \frac{ e^{a+1} a!}{\sqrt{2\pi} } \bigg)$

Recall the definition of the Euler-Mascheroni Constant:

$\displaystyle = \frac{2a+1}{2} \gamma + a+1 + \ln a! - \frac{1}{2} \ln (2\pi)$

Now substitute this back in our expression of $I$ :

$\displaystyle I= a\ln a +\ln a! +1+\frac{2a+1}{2}\gamma -\sum_{k=1}^{\infty} \int_{a}^{a+1} \frac{x^2 dx }{k(x+k)}$

$\displaystyle = a\ln a +\ln a! +1+\frac{2a+1}{2}\gamma -\bigg(\frac{2a+1}{2} \gamma + a+1 + \ln a! - \frac{1}{2} \ln (2\pi) \bigg)$

$\displaystyle \boxed{I = \frac{1}{2} \ln (2\pi) +a\ln a - a }$

Let $a=2$ , we get our desired answer to this problem.

The integral is a Raabe's integral named after Joseph Ludwig Raabe and it is given by:

$\begin{aligned} \int_a^{a+1} {\ln{(\Gamma (x))} dx} & = \frac {1}{2} \ln {2\pi} + a\ln{a} - a \\ \Rightarrow \int_2^{3} {\ln{(\Gamma (x))} dx} & = \frac {1}{2} \ln {2\pi} + 2\ln{2} - 2 \\ & = \frac {1}{2} \ln {2} + \frac {1}{2} \ln {\pi} + 2\ln{2} - 2 \\ & = \frac {5}{2} \ln {2} + \frac {1}{2} \ln {\pi} - 2 \\ \Rightarrow a + b + c & = 5 + 1 + 2 = \boxed{8} \end{aligned}$