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

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

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

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

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*}

• You are very good. Thank you Chen Wang for your clever solutions – Jeff Faraci May 6 '14 at 4:01
• 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. – abnry May 6 '14 at 4:11
• @nayrb if its not broken, don't fix it :) – Jeff Faraci May 6 '14 at 4:56
• @Integrals, but it is potentially broken. At the very least it is lacking detail. – abnry May 6 '14 at 5:14
• 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. – Jean-Claude Arbaut May 6 '14 at 6:07


\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}

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