A very useful integral representation of the Digamma function

In this post I will derive an Integral representation for the digamma function which will turn out to be very useful in the computation of the coefficients of the Fourier series of the $\log(\Gamma(x))$ function that I plan to derive soon in a future post.

In these notes I try as much as possible to make each one of them as self content as possible, therefore, although it will be a little long, I will start this derivation almost from scratch, i.e., I will start proving first a product representation for the gamma function.

Ok, enough talking, lets do some math!

1. Gauss product for the Gamma function

The Gauss product for the Gamma function is given by the following formula

$\Gamma(z)= \lim_{n \rightarrow \infty} \,\frac{n!\, n^z}{z (z+1)\cdots (z+n)}$ (1.1)

To prove this formula, recall the well known fact that

$e^{-t}= \lim_{n \rightarrow \infty}\Big( 1-\frac{t}{n} \Big)^n$ (1.2)

and the standard integral representation of the Gamma function that we are assuming here as the definition of the Gamma function

$\Gamma(z) = \int_{0}^{\infty}e^{-t} t^{z-1}dt$ (1.3)

substituting (1.2) in (1.3) and taking the limit $\lim_{n \rightarrow \infty}$ , we can rewrite $\Gamma(x)$ as

$\Gamma(z) = \lim_{n \rightarrow \infty} \, \int_{0}^{n} \Big( 1-\frac{t}{n} \Big)^n t^{z-1}dt$

Now lets consider the following integral

$I_{n}=\int_{0}^{n} \Big( 1-\frac{t}{n} \Big)^n t^{z-1}dt$

and integrate it by parts letting $u= \Big( 1-\frac{t}{n} \Big)^n$ and $v = \frac{t^{z}}{z}$ to get

$I_{n}= \Big( 1-\frac{t}{n} \Big)^n \cdot \frac{t^{z}}{z} \bigg|_{0}^{n} \, + \, \frac{n}{nz} \int_{0}^{n}\Big( 1-\frac{t}{n} \Big)^{n-1} t^{z}dt$

$I_{n}=\frac{1}{z} \int_{0}^{n}\Big( 1-\frac{t}{n} \Big)^{n-1} t^{z}dt$

integrating by parts once again with $u= \Big( 1-\frac{t}{n} \Big)^{n-1}$ and $v = \frac{t^{z+1}}{(z+1)}$ we get

$I_{n}= \frac{1}{z}\bigg(\Big( 1-\frac{t}{n} \Big)^{n-1} \cdot \frac{t^{z+1}}{(z+1)} \bigg|_{0}^{n} \, + \, \frac{(n-1)}{n(z+1)} \int_{0}^{n}\Big( 1-\frac{t}{n} \Big)^{n-2} t^{z+1}dt \bigg)$

$I_{n}=\frac{(n-1)}{nz(z+1)} \int_{0}^{n}\Big( 1-\frac{t}{n} \Big)^{n-2} t^{z+1}dt$

We can already see a pattern emerging in this process. If we keep integrating by parts n-times we get

$I_{n} = \frac{(n-1)(n-2)\cdots(n-(n-1))}{n^{n-1}z(z+1)\cdots(z+n-1)} \, \int_{0}^{n}t^{z+n-1}dt$

$I_{n} = \frac{(n-1)(n-2)\cdots(n-(n-1))}{n^{n-1}z(z+1)\cdots(z+n-1)} \, \Big[\frac{t^{z+n}}{(z+n)}} \Big|_{0}^{n}\,\,\Big]$

$I_{n} = \frac{n(n-1)(n-2)\cdots(n-(n-1))\,n^{z}}{z(z+1)\cdots(z+n-1)(z+n)}$

and finally

$I_{n}=\frac{n!\, n^z}{z (z+1)\cdots (z+n)}$

Taking the $\lim_{n \rightarrow \infty}$ in the last expression we prove (1.1)

$\boxed{\Gamma(z)= \lim_{n \rightarrow \infty} \,\frac{n!\, n^z}{z (z+1)\cdots (z+n)}}$

2. The Digamma function

The Digamma function is defined as the logarithmic derivative of the gamma function:

$\psi(z)= \frac{d}{dz}\log(\Gamma(z))=\frac{\Gamma^{\prime}(z)}{\Gamma(z)}$ (2.1)

Taking log in (1.1) we get

$\log(\Gamma(z))= \lim_{n \rightarrow \infty} \, \Big[ \log(n!) + z \log(n)-\log(z)- \cdots - \log(z+n)\Big]$ (2.2)

differentiating (2.2) with respect to

$\psi(z)= \frac{d}{dz}\log(\Gamma(z))= \lim_{n \rightarrow \infty} \, \Big[\log(n) - \frac{1}{z}-\frac{1}{z+1}- \cdots - \frac{1}{z+n} \Big]$ (2.3)

Now note the following facts

$\frac{1}{z+n}=\int_{0}^{\infty}e^{-(z+n)t}dt$ (2.4)

and

$\log(n)=\int_{0}^{\infty}\frac{e^{-t}-e^{-nt}}{t}dt$ (2.5)

(2.4) is straightforward. (2.5) is known as Frulani integral, to prove it consider

$\int_{1}^{n}e^{-xt}dx=- \frac{e^{-xt}}{t}\bigg|_{1}^{n}=\frac{e^{-t}-e^{-nt}}{t}$ (2.6)

Substituting (2.6) in (2.5)

$\int_{0}^{\infty} \!\!\!\int_{1}^{n}e^{-xt}dx \,dt=\int_{1}^{n} \!\!\!\int_{0}^{\infty}e^{-xt}dt \,dx$

$\int_{1}^{n} \bigg( - \frac{e^{-xt}}{x}\bigg|_{0}^{\infty}\bigg)dx=\int_{1}^{n}\frac{1}{x}dx=\log(n)$

Proving (2.5). We can now rewrite (2.3) in terms of (2.4) and (2.5)

$\psi(z)= \lim_{n \rightarrow \infty} \, \Big[\int_{0}^{\infty}\frac{e^{-t}-e^{-nt}}{t}dt - \int_{0}^{\infty}e^{-zt}dt \cdots - \int_{0}^{\infty}e^{-(z+n)t}dt \Big]$

$\psi(z)= \lim_{n \rightarrow \infty} \, \Big[\int_{0}^{\infty}\bigg(\frac{e^{-t}-e^{-nt}}{t} - e^{-zt} \cdots - e^{-(z+n)t} \bigg)\,dt \Big]$

$\psi(z)= \lim_{n \rightarrow \infty} \, \Big[\int_{0}^{\infty}\bigg(\frac{e^{-t}-e^{-nt}}{t} - \sum_{k=0}^{n}e^{-(z+k)t} \bigg) \,dt \Big]$

The sum inside the integral is a geometric sum, the solution is given by

$\sum_{k=0}^{n} e^{-(z+k) t}=\frac{e^{-z t}-e^{-(z+n+1) t}}{1-e^{-t}}$

I´ll provide the proof as an appendix in the end of the post, for the time being take it for granted if you are not familiar with it. We can now rewrite the last expression as

$\psi(z)=\lim _{n \rightarrow \infty}\left[\int_{0}^{\infty} \frac{e^{-t}-e^{-n t}}{t}-\frac{e^{-z t}-e^{-(z+n+1) t}}{1-e^{-t}} d t\right]$

$\psi(z)=\lim _{n \rightarrow \infty}\left[\int_{0}^{\infty} \frac{e^{-t}}{t}-\frac{e^{-z t}}{1-e^{-t}} d t-\int_{0}^{\infty} \frac{e^{-n t}}{t}-\frac{e^{-(z+n+1) t}}{1-e^{-t}} d t\right]$