Integral $\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha$

Question

Hi I am trying to prove$$ I:=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x =\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0. $$ I am not sure whether to use an integral representation or to somehow use the Euler reflection formula $$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z} $$ since a previous post used that to solve this kind of integral. Other than this method, we can use the integral representation $$ \Gamma(t)=\int_0^\infty x^{t-1} e^{-x}\, dx. $$

Also note $\Gamma(n)=(n-1)!$.

Answer 1

Another way to show $$\int_{0}^{1} \log \Gamma(x+ \alpha) \, dx = \int_{0}^{1} \log \Gamma(x) \, dx + \alpha \log \alpha - \alpha $$

is to rewrite the integral as

$$ \begin{align} \int_{0}^{1} \log \Gamma (x+\alpha) \, dx &= \int_{\alpha}^{\alpha+1} \log \Gamma(u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{1}^{\alpha+1} \log \Gamma (u) \, du - \int_{0}^{\alpha} \log \Gamma (u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{0}^{\alpha} \log \Gamma (w+1) \, dw - \int_{0}^{\alpha} \log \Gamma (u) \, du \end{align}$$

and then combine the 2nd and 3rd integrals and use the functional equation $\frac{\Gamma(x+1)}{\Gamma (x)} = x.$

Answer 2

This one is deceptively simple. Differentiate with respect to $\alpha$ and note that your integrand becomes $\dfrac{\Gamma'(x+\alpha)}{\Gamma(x+\alpha)} $. You can view this also as $(\log\Gamma(x+\alpha))'$ (where the derivative is taken with respect to $x$ now). At this point you have

$$\begin{align}\int_0^1(\log\Gamma(x+\alpha))'dx &= \log\Gamma(x+\alpha)\bigg|_0^1 \\ &= \log\Gamma(1+\alpha)-\log\Gamma(0+\alpha) \\ &= \log(\alpha\Gamma(\alpha))-\log\Gamma(\alpha) \\ &= \log\alpha+\log\Gamma(\alpha)-\log\Gamma(\alpha) \\ &=\log\alpha \end{align}$$

So $I'(\alpha) = \log(\alpha)$ which gives that $I(\alpha) = \alpha\log\alpha-\alpha+C$. To determine the constant of integration, take $\alpha = 0$. This gives

$$I(0) = C = \int_0^1\log\Gamma(x)dx.$$

From here, refer to achille's answer on a different question to evaluate this integral.

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(#1\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\totald{}{\alpha}\int_{0}^{1} \ln\pars{\Gamma\pars{x + \alpha}}\,\dd x =\int_{0}^{1}\partiald{\ln\pars{\Gamma\pars{x + \alpha}}}{\alpha}\,\dd x =\int_{0}^{1}\partiald{\ln\pars{\Gamma\pars{x + \alpha}}}{x}\,\dd x \\[3mm]&=\ln\pars{\Gamma\pars{1 + \alpha} \over \Gamma\pars{\alpha}} =\ln\pars{\alpha} \end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x =\alpha\ln\pars{\alpha} - \alpha + \int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x \end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x =\int_{0}^{1}\ln\pars{\Gamma\pars{1 - x}}\,\dd x =\int_{0}^{1}\ln\pars{\pi \over \Gamma\pars{x}\sin\pars{\pi x}}\,\dd x \\[3mm]&=\ln\pars{\pi} - \int_{0}^{1}\ln\pars{\sin\pars{\pi x}}\,\dd x -\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x \end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x =\half\,\ln\pars{\pi} -{1 \over 2\pi}\ \underbrace{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x}_{\ds{-\pi\ln\pars{2}}} =\half\,\ln\pars{2\pi} \end{align} The above $\ds{\ul{\ln\pars{\sin\pars{\cdots}}\!\mbox{-integral}}}$ is a well known result and it appears frequently in M.SE.

$$\color{#66f}{\large% \int_{0}^{1}\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x ={\ln\pars{2\pi} \over 2} + \alpha\ln\pars{\alpha} - \alpha} $$

Answer 4

Here's a general method you could use to calculate $I(\alpha)$ if you already know $I(0),I(1)$.

After you've differentiated w.r.t. $\alpha$ under the integral, you could always use that $$(\log\Gamma(x+\alpha))'=-\gamma+\sum_{k \ge1}\frac{1}{k}-\frac{1}{(k+x+\alpha-1)}$$ and differentiate again to give $$(\log\Gamma(x+\alpha))''=\sum_{k \ge1}\frac{1}{(k+x+\alpha-1)^2}.$$ Thus by Tornelli we swap integral and summation order, giving $$I''(\alpha)=\sum_{k \ge 1}\int_0^1\frac{dx}{(k+x+\alpha-1)^2}=\sum_{k \ge 1}\frac{1}{(k+\alpha-1)}-\frac{1}{(k+\alpha)}=\frac{1}{\alpha}$$ $$I'(\alpha)=\log(\alpha)+k$$ $$I(\alpha)=\alpha\log(\alpha)+k\alpha+c$$ $$I(\alpha)=\alpha\log(\alpha)+(I(1)-I(0))\alpha+I(0).$$