# Integral that arises from the derivation of Kummer's Fourier expansion of $\ln{\Gamma(x)}$

I am trying to prove that for $0<x<1$, $$\color{blue}{\ln{\Gamma(x)}=\frac{1}{2}\ln(2\pi)+\sum^\infty_{n=1}\left\{\frac{1}{2n}\cos(2\pi nx)+\frac{\gamma+\ln(2\pi n)}{n\pi}\sin(2\pi nx)\right\}}$$ It is quite straightforward to compute $a_0$ and $a_n$. $a_0$ is \begin{align} a_0 =2\int^1_0\ln{\Gamma(x)}\ {\rm d}x =\int^1_0\ln\left(\frac{\pi}{\sin(\pi x)}\right)\ {\rm d}x =\ln(2\pi)\\ \end{align} As for $a_n$, \begin{align} a_n =&2\int^1_0\ln{\Gamma(x)}\cos(2\pi nx)\ {\rm d}x =\int^1_0\ln\left(\frac{\color{grey}{\pi}}{\sin(\pi x)}\right)\cos(2\pi nx)\ {\rm d}x\\ =&-\frac{1}{\pi}\int^\pi_0\ln(\sin{x})\cos(2nx)\ {\rm d}x =\frac{1}{4n\pi}\int^\pi_{-\pi}\frac{\sin(2nx)\cos(x)}{\sin{x}}\ {\rm d}x \end{align} If we name the remaining integral $\mathcal{I}_n$, it is easy to see that $\mathcal{I}_{n+1}-\mathcal{I}_n=0$. Hence $\mathcal{I}_n=\mathcal{I}_1=2\pi$, and $$a_n=\frac{2\pi}{4n\pi}=\frac{1}{2n}$$ However, I have trouble calculating $b_n$ and proving that $$\color{red}{2\int^1_0\ln{\Gamma(x)}\sin(2n\pi x)\ {\rm d}x=\frac{\gamma+\ln(2n\pi)}{\pi n}}$$ The only idea that I can think of is to use the series representation of $\ln{\Gamma(x)}$ \begin{align} \ln{\Gamma(x)} =&-\gamma x-\ln{x}+\sum^\infty_{k=1}\left\{\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right\}\\ =&-\gamma x-\ln{x}+\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m}x^m \end{align} and multiply throughout by $\sin(2n\pi x)$ then integrating term by term. However, arriving to the final result using this method definitely seems arduous. Therefore, I seek your assistance in evaluating the integral in red. Help will be greatly appreciated. Thank you.

• It looks like if you do integration by parts you can turn this into something that looks like the digamma function, and you can also take the imaginary part of the exponential, maybe that'll help. – Kainui Nov 8 '14 at 9:18

Begin with the infinite product representation of the gamma function. \begin{align} \Gamma(x)=\frac{e^{-\gamma x}}{x}\prod^\infty_{k=1}e^\frac{x}{k}\left(1+\frac{x}{k}\right) \end{align} Take the logarithm and multiply throughout by $\sin(2n\pi x)$. \begin{align} \ln{\Gamma(x)}\sin(2n\pi x)=-(\gamma x+\ln{x})\sin(2n\pi x)+\sum^\infty_{k=1}\left\{\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right\}\sin(2n\pi x) \end{align} Integrating the non-sum terms from $0$ to $1$, \begin{align} \int^1_0(-\gamma x-\ln{x})\sin(2n\pi x)\ {\rm d}x =&\frac{\gamma}{2n\pi}+\frac{\ln{x}\cos(2n\pi x)}{2n\pi}\Bigg{|}^1_0-\int^1_0\frac{\cos(2n\pi x)}{2n\pi x}{\rm d}x\\ =&\frac{\gamma}{2n\pi}+\left[\frac{\ln{x}\cos(2n\pi x)-{\rm Ci}(2n\pi x)}{2n\pi}\right]^1_0\\ =&\frac{\gamma-{\rm Ci}(2n\pi)}{2n\pi}-\lim_{\epsilon\to 0}\frac{\ln{\epsilon}-{\rm Ci}(2n\pi \epsilon)}{2n\pi}\\ =&\frac{\gamma-{\rm Ci}(2n\pi)}{2n\pi}-\lim_{\epsilon\to 0}\frac{\ln{\epsilon}-\ln(2n\pi)-\ln{\epsilon}-\gamma+\mathcal{O}(\epsilon)}{2n\pi}\\ =&\frac{2\gamma+\ln(2n\pi)-{\rm Ci}(2n\pi)}{2n\pi} \end{align} Integrate the remaining terms from $0$ to $1$. \begin{align} &\int^1_0\sum^\infty_{k=1}\left\{\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right\}\sin(2n\pi x)\ {\rm d}x\\ =&-\frac{1}{2n\pi}\sum^\infty_{k=1}\left\{\frac{1}{k}+{\rm Ci}(2n\pi k+2n\pi)-{\rm Ci}(2n\pi k)-\ln\left(1+\frac{1}{k}\right)\right\}\\ =&-\frac{1}{2n\pi}\left[\sum^\infty_{k=1}\left\{\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right\}+\lim_{N\to\infty}\left(\sum^{N+1}_{k=2}{\rm Ci}(2n\pi k)-\sum^{N}_{k=1}{\rm Ci}(2n\pi k)\right)\right]\\ =&-\frac{1}{2n\pi}\left[\gamma+\lim_{N\to\infty}\left({\rm Ci}(2n\pi+2n\pi N)-{\rm Ci}(2n\pi)\right)\right] =\frac{{\rm Ci}(2n\pi)-\gamma}{2n\pi} \end{align} Adding them together, \begin{align} \color{red}{\int^1_0\ln{\Gamma(x)}\sin(2n\pi x)\ {\rm d}x}=\frac{2\gamma+\ln(2n\pi)-{\rm Ci}(2n\pi)+{\rm Ci}(2n\pi)-\gamma}{2n\pi}\color{red}{=\frac{\gamma+\ln(2n\pi)}{2n\pi}} \end{align} I am looking forward to seeing cleaner and more interesting approaches to this integral.
As an aside, I derived that $$\int^1_0\psi_0(x+a)\sin(2n\pi x)={\rm Si}(2an\pi)-\frac{\pi}{2}$$ This isn't really related to the problem though.