19
$\begingroup$

$$ I:=\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}. $$ 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.

$\endgroup$
  • $\begingroup$ Where $n>0$ and an integer, presumably. For $n=0$, $$ \int_0^1 \ln \Gamma(z) \ \mathrm{d}z = \frac{1}{2}\log 2\pi $$ $\endgroup$ – Bennett Gardiner May 6 '14 at 4:14
15
$\begingroup$

$$\log\Gamma(x)=(\frac12-x)(\gamma+\log 2)+(1-x)\log\pi-\frac12\log\sin\pi x+\frac1\pi\sum^{\infty}_{k=1}\frac{\log k\sin (2\pi kx)}{k}$$

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

$\endgroup$
  • $\begingroup$ You are very good. Thank you Chen Wang for your clever solutions $\endgroup$ – Jeff Faraci May 6 '14 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 '14 at 4:11
  • $\begingroup$ @nayrb if its not broken, don't fix it :) $\endgroup$ – Jeff Faraci May 6 '14 at 4:56
  • $\begingroup$ @Integrals, but it is potentially broken. At the very least it is lacking detail. $\endgroup$ – abnry May 6 '14 at 5:14
  • 1
    $\begingroup$ 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. $\endgroup$ – Jean-Claude Arbaut May 6 '14 at 6:07
23
+50
$\begingroup$

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

\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}

$\endgroup$
  • 5
    $\begingroup$ Very beautiful solution! $\endgroup$ – user111187 May 6 '14 at 7:13
  • 5
    $\begingroup$ Nice solution. Much more elegant than mine. $\endgroup$ – Chen Wang May 6 '14 at 11:14
  • 2
    $\begingroup$ Very nice thanks Felix!+1 $\endgroup$ – Jeff Faraci May 6 '14 at 11:38
  • 5
    $\begingroup$ WOW ! Just wow ! $\endgroup$ – Lucian May 6 '14 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.