Fourier series of Loggamma function

We are now equipped to compute the fourier series of the Loggamma function. This fourier expansion is known as Kummer´s fourier series for $\log(\Gamma(x))$ . It was discovered in 1847 by Kummer, here is a link to the original paper. In actuality it was already known to the Sweden mathematician Karl Malmesten who derived it in 1842 before Kummer.

Since it´s derivation involves the evaluation of three hard and beautiful integrals, and the computation of each one of them is lengthy, I decided to break down the proof in separate posts.

Our goal in these posts is to be able to write the Loggama function as a fourier expansions, i.e.

$\boxed{\log(\Gamma(x))=\frac{a_{0}}{2}+ \sum_{k=1}^{\infty}a_{k}\cos(2 \pi k x) + \sum_{k=1}^{\infty}b_{k}\sin(2 \pi k x)}$

where

$a_{0}= 2\int_{0}^{1}\log(\Gamma(x))dx$

$a_{k}= 2\int_{0}^{1}\log(\Gamma(x))\cos(2 \pi k x)dx$

and

$b_{k}= 2\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx$

In the end of the day, our objective is to evaluate the three integrals above. Today we will focus in the first and second integrals, and the third one which is the harder, will be computed in the part 2. So, let´s get started!

1. Evaluation of $a_{0}$

Consider

$I=\int_{0}^{1}\log(\Gamma(x))dx$ (1.1)

and observe the following fact:

$\int_{0}^{a}f(x)dx= \int_{0}^{a}f(a-x)dx$

which can be easily proved making the substitution $x\longmapsto a-x$ , therefore, we can rewrite (1) as

$I(x)=\int_{0}^{1}\log(\Gamma(1-x))dx$ (1.2)

Adding (1.1) and (1.2) we get

$2I=\int_{0}^{1}\log(\Gamma(x))dx+\int_{0}^{1}\log(\Gamma(1-x))dx$

$=\int_{0}^{1}\log(\Gamma(x) \cdot \Gamma(1-x))dx$

Recalling the reflection formula of the gamma function proved in this post

$\Gamma(x) \cdot \Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$

we get

$2I=\int_{0}^{1}\log(\frac{\pi}{\sin(\pi x)})dx$

$=\int_{0}^{1}\log(\pi)dx-\int_{0}^{1}\log(\sin(\pi x))dx$

$=\log(\pi)-\int_{0}^{1}\log(\sin(\pi x))dx$

To evaluate this integral, recall the expansion of $\log(\sin( \pi x))$ derived in this post (the last equation of the post)

$-\log(\sin(\pi x))= \log(2) + \sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k}$

Plugging this expansion in the last integral we get

$2I=\log(\pi)+\int_{0}^{1}\log(2)dx+\int_{0}^{1}\sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k}dx$

$=\log(\pi)+\log(2)+\sum_{k=1}^{\infty}\frac{1}{k}\int_{0}^{1}\cos(2 \pi k x)dx$

$=\log(\pi)+\log(2)+\sum_{k=1}^{\infty}\frac{1}{k} \underbrace{\sin(2 \pi k x) \Big|_{0}^{1}}_{=0} dx$

$I=\frac{\log(2 \pi)}{2}$ (1.3)

Plugging (3) in the expression for $a_{0}$

$\boxed{a_{0}= 2\int_{0}^{1}\log(\Gamma(x))dx=\log(2)}$

2. Evaluation of $a_{k}$

Our first integral is done. Let´s now jump into the second one. I haven´t seen the evaluation of this one the way I´m going to evaluate it here. Consider

$I =\int_{0}^{1} \log \Gamma(x) \cos (2 \pi k x) d x$ (2.1)

integrating by parts where

$u =\log \Gamma(x) \, \Rightarrow \,d u=\psi(x) d x$

$d v =\cos (2 \pi k x) d x \Rightarrow v=\frac{\sin (2 \pi k x)}{2 \pi k}$

$I= \underbrace{\left.\frac{\sin (2 \pi k x)}{2 \pi k} \cdot \log \Gamma(x)\right|_{0} ^{1}}_{=0}-\frac{1}{2 \pi k} \int_{0}^{1} \psi(x) \sin (2 \pi k x) d x$

$I=-\frac{1}{2 \pi k} \int_{0}^{1} \psi(x) \sin (2 \pi k x) d x$ (2.2)

Recall now the integral representation of the Digamma Function proved previously in another post

$\psi(x)=\int_{0}^{\infty} \frac{e^{-t}}{t}-\frac{e^{-x t}}{1-e^{-t}} d t$ (2.3)

Plugging (2.3) in (2.2)

$I=-\frac{1}{2 \pi k} \int_{0}^{1} \color{red} \bigg\{{\int_{0}^{\infty} \frac{e^{-t}}{t}-\frac{e^{-x t}}{1-e^{-t}} d t} \,\, \bigg\}\color{black}\sin (2 \pi k x) d x$

Switching the order of integration

$I=-\frac{1}{2 \pi k} \int_{0}^{\infty} \!\! {\int_{0}^{1} \sin (2 \pi k x) \bigg[\frac{e^{-t}}{t}-\frac{e^{-x t}}{1-e^{-t}} \bigg] \,\,d x \, d t$ (2.4)

Let´s now look into the inner integral, call it

$J= {\int_{0}^{1} \sin (2 \pi k x) \bigg[\frac{e^{-t}}{t}-\frac{e^{-x t}}{1-e^{-t}} \bigg] \,\,d x$

$=\frac{e^{-t}}{t}\underbrace{\int_{0}^{1} \sin (2 \pi k x)dx}_{=0}-\frac{1}{1-e^{-t}}\int_{0}^{1} \sin (2 \pi k x) e^{-x t}dx$

$=-\frac{1}{1-e^{-t}}\int_{0}^{1} \sin (2 \pi k x) e^{-x t}dx$

This integral was already evaluated in this post and it is equal to

$J=-\frac{1}{(1-e^{-t})}\cdot\frac{ 2 \pi k (1-e^{-t})}{t^2 + (2 \pi k)^2}$

$J=-\frac{2 \pi k}{t^2 + (2 \pi k)^2}$ (2.5)

Plugging (2.5) in (2.4)

$I=\frac{1}{2 \pi k}\int_{0}^{\infty}\frac{2 \pi k}{t^2 + (2 \pi k)^2}dt$

$=\frac{1}{(2 \pi k)^2}\int_{0}^{\infty}\frac{dt}{1+\frac{t^2} {(2 \pi k)^2}}$

Changing variable $u=\frac{t}{2 \pi k} \Rightarrow dt= 2 \pi k du$

$I =\frac{1}{(2 \pi k)^2}\int_{0}^{\infty}\frac{2 \pi k}{1+u^2}du$

$=\frac{1}{2 \pi k}\int_{0}^{\infty}\frac{du}{1+u^2}$

$=\frac{1}{2 \pi k} \Big[ \arctan(u) \Big|_{0}^{\infty}\Big]$

$=\frac{1}{2 \pi k} \cdot \frac{\pi}{2}$

$\boxed{\int_{0}^{1} \log \Gamma(x) \cos (2 \pi k x) d x=\frac{1}{4k}}$ (2.6)

Now, plugging (2.6) into the formula of $a_{k}$

$\boxed{a_{k}= 2\int_{0}^{1}\log(\Gamma(x))\cos(2 \pi k x)dx=\frac{1}{2k}}$

And we are done!!

Next post, Part 2 we will compute

$b_{k}= 2\int_{0}^{1}\log(\Gamma(x))\sin(2 \pi k x)dx$

Reference

Blagouchine, Iaroslav V. (2014). "Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results". Ramanujan J. 35 (1): 21–110.

Ricardo Albahari

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