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

Question

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.

Answer 1

Random Variable has kindly pointed out that the integral has been evaluated by Cody on this site. Here is a slightly different method of evaluating this integral.

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.

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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.

Answer 2

Not a direct answer for the $b_n$'s (don't have enough reputation to make this a comment) but a clarification on the calculation of the $a_n$'s for those, like me, who are interested in working through the details and are greatly put off by hand-waving of the ilk of "clearly", "obviously", "easy to see" etc.

Suppose we define $I_n := \int_{-\pi}^{\pi} \sin(2nx) \frac{\cos(x)}{\sin(x)} dx$; then \begin{aligned} \require{cancel} I_{n+1} - I_n &= \int_{-\pi}^{\pi} [\sin(2(n+1)x) - \sin(2nx)] \frac{\cos(x)}{\sin(x)} dx \\ &= \int_{-\pi}^{\pi} [\sin(2nx + 2x) - \sin(2nx)] \frac{\cos(x)}{\sin(x)} dx \\ &= \int_{-\pi}^{\pi} [\sin(2nx)\cos(2x) + \cos(2nx)\sin(2x) - \sin(2nx)] \frac{\cos(x)}{\sin(x)} dx &&\text{(addition formula for $\sin$)} \\ &= \int_{-\pi}^{\pi} [\cos(2nx)\cdot \sin(2x) + \sin(2nx)\cdot(\cos(2x) - 1)] \frac{\cos(x)}{\sin(x)} dx \\ &= \int_{-\pi}^{\pi} [\cos(2nx)\cdot 2\sin(x)\cos(x) + \sin(2nx)\cdot -2\sin^2(x)] \frac{\cos(x)}{\sin(x)} dx &&\text{(double angle formulae for $\sin$ and $\cos$)} \\ &= \int_{-\pi}^{\pi} [\cos(2nx)\cos(x) - \sin(2nx)\sin(x)] \cdot 2\cancel{\sin(x)}\frac{\cos(x)}{\cancel{\sin(x)}} dx \\ &= 2\int_{-\pi}^{\pi} \cos((2n+1)x)\cos(x) dx &&\text{(addition formula for $\cos$)} \\ &= \cancel{2}\int_{-\pi}^{\pi} \frac{1}{\cancel{2}}[\cos((2n+2)x) + \cos((2n+0)x)] dx &&\text{(product-to-sum formula for $\cos$)} \\ &= \int_{-\pi}^{\pi} \cos(2(n+1)x) dx + \int_{-\pi}^{\pi} \cos(2nx) dx \end{aligned} Each of the final 2 integrals is of a $2\pi$-periodic function over an integer number of its periods, and are therefore both 0, yielding $I_{n+1} - I_n = 0$, as desired $\blacksquare$

Answer 3

Let $S(x)=\sin(2n\pi x)$. For each $n\in\Bbb N$, $$\int_n^{n+1}\ln\Gamma(x)S(x)dx\\ =\int_{n-1}^n\ln\Gamma(x+1)S(x+1)dx\\ =\int_{n-1}^n\ln xS(x)dx+\int_{n-1}^n\ln\Gamma(x)S(x)dx $$ By induction $$\int_n^{n+1}\ln\Gamma(x)S(x)dx= \int_0^n\ln xS(x)dx+\int_0^1\ln\Gamma(x)S(x)dx $$ Let $n\to\infty$, then LHS$\to0$ and we have $$\int_0^1\ln\Gamma(x)S(x)dx\\ =-\int_0^\infty\ln xS(x)dx\\ =-\Im\int_9^\infty\ln xe^{i2n\pi x}dx\\ =-\Im\frac{-\gamma-\ln(-2\pi i n)}{-2n\pi i}\\ =\frac{\gamma+\ln(2n\pi)}{2n\pi}$$