Fourier series of $\log(\Gamma(x))$

Question

To calculate Fourier coefficients with cosine is quite easy and I wonder if there is any simple (using only basic propertes of $\Gamma$, without any integral representation for $\log(\Gamma(x))$ and so on) way to evaluate the coefficients with sine, namely $$b_n=2\int\limits_{0}^{1}\sin(2\pi nx)\log(\Gamma(x))dx.$$

Answer 1

An alternative approach is to use infinite products for $\Gamma$ (either one), or rather the formulae for $\psi(x)=\Gamma'(x)/\Gamma(x)$ obtained by taking the logarithmic derivative. Integrating by parts, we get $$b_n=-\frac{1}{n\pi}\int_0^1(1-\cos 2n\pi x)\psi(x)\,dx$$ and, using Euler's infinite product, we find $$\psi(x)=-\frac1x+\sum_{k=1}^\infty\left[\log\left(1+\frac1k\right)-\frac{1}{k+x}\right],$$ so that \begin{align*} n\pi b_n&=\int_0^1\frac{1-\cos 2n\pi x}{x}\,dx-s_n, \\s_n&=\sum_{k=1}^\infty\left[\log\left(1+\frac1k\right)-\int_0^1\frac{1-\cos 2n\pi t}{k+t}\,dt\right] \\&=\sum_{k=1}^\infty\int_0^1\frac{\cos 2n\pi t}{k+t}\,dt\underset{k+t=x}{\phantom{[}\quad=\quad\phantom{]}}\int_1^\infty\frac{\cos 2n\pi x}{x}\,dx. \end{align*}

Since $\int_0^\infty\frac{\cos ax-\cos bx}{x}\,dx=\log b-\log a$ for $a,b>0$, we obtain $$n\pi b_n=\log(2n\pi)+B,\quad B=\int_0^1\frac{1-\cos x}{x}\,dx-\int_1^\infty\frac{\cos x}{x}\,dx.$$

Finally the (known) fact that $B=\gamma$ can be taken from somewhere else: $$\bbox[5pt,border:1pt solid]{b_n=\frac{\gamma+\log(2n\pi)}{n\pi}.}$$

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Also, the connection between $b_n$ and $b_1$ can be obtained from the multiplication theorem: $$\log\Gamma(nx)=\left(nx-\frac12\right)\log n-\frac{n-1}{2}\log(2\pi)+\sum_{k=0}^{n-1}\log\Gamma\left(x+\frac{k}{n}\right).$$ We get \begin{align*} b_1&=2\int_0^1\sin 2\pi t\log\Gamma(t)\,dt\underset{t=nx}{\phantom{[}=\phantom{]}}2n\int_0^{1/n}\sin 2\pi nx\log\Gamma(nx)\,dx \\&=2n\int_0^{1/n}\sin 2\pi nx\left(nx\log n\color{LightGray}{-(0)}+\sum_{k=0}^{n-1}\log\Gamma\left(x+\frac{k}{n}\right)\right)dx \\&=2\log n\underbrace{\int_0^1t\sin 2\pi t\,dt}_{t=nx}+2n\sum_{k=0}^{n-1}\underbrace{\int_{k/n}^{(k+1)/n}\sin 2\pi nt\log\Gamma(t)\,dt}_{t=x+k/n} \\&=-\frac{\log n}{\pi}+2n\int_0^1\sin 2\pi nt\log\Gamma(t)\,dt=nb_n-\frac{\log n}{\pi}\\\implies b_n&=\frac{b_1}{n}+\frac{\log n}{n\pi}. \end{align*}