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

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

• I get the feeling that you want to try differentiating under the integral here. – Cameron Williams May 7 '14 at 4:24
• Look at the first half of my answer to a completely unrelated question. It shows how to evaluate your integral. – achille hui May 7 '14 at 4:31
• Just thought I'd note the interesting identity that $$\int_0^1 \log \Gamma(x+\mathrm{e})\, \mathrm{d}x = \int_0^1 \log \Gamma(x)\, \mathrm{d}x$$ – Bennett Gardiner May 7 '14 at 13:31
• Ps, check my document :p This integral is given as a problem and both solutions below are displayed :p Altso this integral is nice $$\int_0^1 \int_0^1 \log B(x+1,y+1)\,\mathrm{d}x \, \mathrm{d}y$$ Where you of course instead could look at $B(x+\alpha,y+\beta)$. If you want a messier result – N3buchadnezzar Jun 17 '14 at 18:47
• @N3buchadnezzar what document are you talking about? Thanks! – Jeff Faraci Jun 17 '14 at 19:09

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.

• Can the details be fleshed out? I'm not up on my gamma function identities and the definition does not appear immediately useful here. – abnry May 7 '14 at 4:57
• @nayrb I filled in more details. Does this help? – Cameron Williams May 7 '14 at 17:12
• @CameronWilliams, oh yes! It's clear now. It works because for any $f(x+a)$. – abnry May 7 '14 at 17:17
• @CameronWilliams +1, and I checked this as the answer also because you were the first to post:) Thanks. – Jeff Faraci May 7 '14 at 17:19

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 $\frac{\Gamma(x+1)}{\Gamma (x)} = x.$

• I always appreciate your solutions Thanks. I apologize for not checking this as the correct answer. I still give +1 though. It is because the user Cameron had posted the solution before you. Sorry again my friend. – Jeff Faraci May 7 '14 at 17:19

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

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

• Felix, why is your text in a different font? I've noticed this about all of your posts (as well as a couple of others). (I'm not trying to be argumentative or anything, just curious.) – Cameron Williams Jul 13 '14 at 3:27
• This result is known as $\large\tt\mbox{Raabe Integral of the Gamma Function}$. It was originally derived by Joseph Ludwig Raabe (1801-1859) – Felix Marin Jul 13 '14 at 3:29
• @CameronWilliams I don't really know why. I'm using Safari Browser in a MacBook Pro and I don't know if it has any influence. I've seen than my posts look too clear like a book. I'll ask to my friend guru about it. – Felix Marin Jul 13 '14 at 3:34
• I just realized why. You use have the custom command \pars and you put \, before and after the argument. I was thinking there was a lot of white space in your answers. That's why! Nothing was wrong, you just customized it. – Cameron Williams Jul 13 '14 at 3:40
• That first line is really cutely insightful :) – user3002473 Jul 27 '15 at 4:45

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 $$I''(\alpha)=\sum_{k \ge 1}\int_0^1\frac{dx}{(k+x+\alpha-1)^2}=\sum_{k \ge 1}\frac{1}{(k+\alpha-1)}-\frac{1}{(k+\alpha)}=\frac{1}{\alpha}$$ $$I'(\alpha)=\log(\alpha)+k$$ $$I(\alpha)=\alpha\log(\alpha)+k\alpha+c$$ $$I(\alpha)=\alpha\log(\alpha)+(I(1)-I(0))\alpha+I(0).$$