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$$ I:=\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}. $$ Thank you. The Gamma function is given by $\Gamma(n)=(n-1)!$ and its integral representation is $$ \Gamma(x)=\int_0^\infty t^{x-1} e^{-t}\, dt. $$ If we write the gamma function as an integral we end up with a more complicated double integral. And I am not too equipped with tools for dealing with gamma functions inside integrals.

We can possibly try $$ \Re\bigg[\int_0^1 \log \Gamma(x)e^{2\pi i n x}\, dx\bigg]=\frac{1}{4n}. $$ but I still do not where to go from here. Thanks.

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  • $\begingroup$ Where $n>0$ and an integer, presumably. For $n=0$, $$ \int_0^1 \ln \Gamma(z) \ \mathrm{d}z = \frac{1}{2}\log 2\pi $$ $\endgroup$ May 6, 2014 at 4:14

2 Answers 2

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{1}\ln\pars{\Gamma\pars{x}}\cos\pars{2\pi n x}\,\dd x ={1 \over 4n}:\ {\large ?}}$

\begin{align} I&=\int_{0}^{1}\ln\pars{\pi \over \Gamma\pars{1 - x}\sin\pars{\pi x}} \cos\pars{2\pi n x}\,\dd x \\[5mm]&=\ln\pars{\pi}\ \overbrace{\int_{0}^{1}\cos\pars{2\pi nx}\,\dd x} ^{\ds{=\ \color{#c00000}{0}}}\ -\ \overbrace{% \int_{0}^{1}\ln\pars{\Gamma\pars{x}}\cos\pars{2\pi n\bracks{1 - x}}\,\dd x} ^{\ds{=\ \color{#c00000}{I}}} \\[5mm]&-{1 \over \pi}\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x \end{align}

\begin{align} I&=-\,{1 \over 2\pi}\ \overbrace{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x} ^{\ds{-\,{\pi \over 2n}}} =\color{#00f}{\large{1 \over 4n}} \end{align}

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    $\begingroup$ Very beautiful solution! $\endgroup$
    – user111187
    May 6, 2014 at 7:13
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    $\begingroup$ Nice solution. Much more elegant than mine. $\endgroup$
    – Chen Wang
    May 6, 2014 at 11:14
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    $\begingroup$ Very nice thanks Felix!+1 $\endgroup$ May 6, 2014 at 11:38
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    $\begingroup$ WOW ! Just wow ! $\endgroup$
    – Lucian
    May 6, 2014 at 13:02
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$$\log\Gamma(x)=(\frac12-x)(\gamma+\log 2)+(1-x)\log\pi-\frac12\log\sin\pi x+\frac1\pi\sum^{\infty}_{k=1}\frac{\log k\sin (2\pi kx)}{k}$$

Exploiting the orthogonality of $\{\sin(2n \pi x),\cos(2n\pi x)\mid n\in\mathbb{Z}^+\}$ on $[0,1]$, we have

$$\begin{align*} I&=\int^1_0\log\Gamma(x)\cos(2n\pi x)dx\\ &=-\frac12\int^1_0\log(\sin(\pi x))\cos(2n\pi x)dx\\ &=-\frac1{4n\pi}\int^1_0\log(\sin(\pi x))d(\sin(2n\pi x))\\ &=\frac1{4n\pi}\int^1_0\sin(2n\pi x)d(\log(\sin(\pi x)))\\ &=\frac1{4n}\int^1_0\sin(2n\pi x)\cot(\pi x)dx\\ &=\frac1{4n}\int^1_0\frac{\sin(2n\pi x)}{\sin (\pi x)}\cos(\pi x)dx\\ &=\frac1{2n}\int^1_0\left(\sum_{k=1}^{n}\cos((2k-1)\pi x)\right)\cos(\pi x)dx\\ &=\frac1{2n}\int^1_0\cos^2(\pi x)dx\\ &=\frac1{4n}. \end{align*}$$

Edit: $$\begin{align*} \int^1_0x\cos(2\pi n x)dx&=\frac12\left(\int^1_0x\cos(2\pi n x)dx+\int^1_0(1-x)\cos(2\pi n (1-x))dx\right)\\ &=\frac12\left(\int^1_0x\cos(2\pi n x)dx+\int^1_0(1-x)\cos(2\pi n x)dx\right)\\ &=\frac12\int^1_0\cos(2\pi n x)dx\\ &=0 \text{ for }n\in\mathbb{Z}^+. \end{align*}$$

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  • $\begingroup$ You are very good. Thank you Chen Wang for your clever solutions $\endgroup$ May 6, 2014 at 4:01
  • $\begingroup$ You appeal to the orthogonality of the Fourier basis functions to disregard terms. It is not obvious to me that you can disregard the $x$ terms. I haven't thought carefully about this but it seems to me since the Fourier series of $x$ is nontrivial it should in fact matter. $\endgroup$
    – abnry
    May 6, 2014 at 4:11
  • $\begingroup$ @nayrb if its not broken, don't fix it :) $\endgroup$ May 6, 2014 at 4:56
  • $\begingroup$ @Integrals, but it is potentially broken. At the very least it is lacking detail. $\endgroup$
    – abnry
    May 6, 2014 at 5:14
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    $\begingroup$ Actually, $\int_0^1 (ax+b) \cos(2n\pi x) \mathrm{d}x=0$ for $n>0$. So except for the case $n=0$, it looks good. For the $x$ alone, remember it's an odd function. $\endgroup$ May 6, 2014 at 6:07

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