Integral representation of Digamma function

Question

A similar question was already asked about 2 years ago: Integral representation of the Digamma function

Someone asked for the derivation of the integral representation of the Digamma-function and it was answered, but I don't see how you get from here:

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

To here:

$$ \psi (x)=\int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt $$

I really thought a lot about it, but I just don't see how it is done... :(

Thank you so much for your much-appreciated help :)

Answer 1

Here is another derivation of the second formula:

We can use a Frullani Integral to evaluate the red part. The green part is a simple integral of an exponential. $$ \begin{align} &\int_0^\infty\left(\frac1t-\frac1{1-e^{-t}}\right)e^{-t}\,\mathrm{d}t\\ &=\lim_{n\to\infty}\sum_{k=1}^n\int_0^\infty\left(\color{#C00}{\frac1t}-\color{#090}{\frac1{1-e^{-t}}}\right)\left(e^{-kt}-e^{-(k+1)t}\right)\,\mathrm{d}t\\ &=\lim_{n\to\infty}\sum_{k=1}^n\left(\color{#C00}{\log\left(\frac{k+1}k\right)}-\color{#090}{\frac1k}\right)\\[6pt] &=\lim_{n\to\infty}\left(\color{#C00}{\log(n+1)}-\color{#090}{H_n}\right)\\[9pt] &=-\gamma\tag1 \end{align} $$ where $\gamma$ is the Euler-Mascheroni constant.

This integral evaluates to the extension of the Harmonic Numbers to all of $\mathbb{C}$ (except the negative integers). $$ \begin{align} \int_0^\infty\frac{e^{-t}-e^{-zt}}{1-e^{-t}}\,\mathrm{d}t &=\sum_{k=1}^\infty\int_0^\infty\left(1-e^{-(z-1)t}\right)e^{-kt}\,\mathrm{d}t\\ &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z-1}\right)\\[6pt] &=H_{z-1}\tag2 \end{align} $$ If $\mathrm{Re}(z)\gt0$, each of the integrals in the sum converges and the sum of those integrals converges.

Putting these together, we get that for $\mathrm{Re}(z)\gt0$, $$ \begin{align} &\int_0^\infty\left(\frac{e^{-t}}t-\frac{e^{-zt}}{1-e^{-t}}\right)\mathrm{d}t\\ &=\int_0^\infty\left(\left(\frac1t-\frac1{1-e^{-t}}\right)e^{-t}+\frac{e^{-t}-e^{-zt}}{1-e^{-t}}\right)\mathrm{d}t\\[6pt] &=H_{z-1}-\gamma\tag3 \end{align} $$ and this is the digamma function, as shown in this answer.

Answer 2

I'll give a more detailed version of Jack D'Aurizio's answer on the other question.Start with the Weierstrass product for the $\Gamma$ function $$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1}.$$

By definition, \begin{align*} \psi(z+1) &= \frac{d}{dz} \log \Gamma(z+1) \\ &= \frac{d}{dz} \log\left( e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z} {n}\right)^{-1}e^{z/n}\right)\\ &= \frac{d}{dz}\left( -\gamma z + \sum_{n\geq 1}\left(\frac{z}{n} + \log \left(1+\frac{z} {n}\right)^{-1}\right) \right)\\ &= -\gamma + \sum_{n\geq 1}\left(\frac{1}{n} - \frac{1}{z+n}\right) \end{align*}

Now note that \begin{align*}\gamma &= \lim_{k\rightarrow \infty} - \log(k) + \sum_{n=1}^k 1/n \\ &= \lim_{k \rightarrow \infty} -\sum_{n=1}^{k-1} \big( \log(n+1) - \log(n)\big) + \sum_{n=1}^{k} \frac{1}{n} \\ &= \sum_{n=1}^{\infty} \big( \log(n) - \log(n+1) + \frac{1}{n} \big) \end{align*} since the inner sum is a telescoping series.

And so

\begin{align*} \psi(z+1) &= \sum_{n\geq 1}\left[\log(n+1)-\log(n)+\frac{1}{n+z}.\right] \end{align*}

Now he cites the identity

$$ \log(n+1)-\log(n) = \int_{0}^{+\infty}\frac{e^{-nt}-e^{-(n+1)t}}{t}\,dt$$

(by Frullani), and

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

by a Laplace transform (or just by hand).

And so we get

\begin{align*} \psi(z+1) &= \sum_{n\geq 1}\int_{0}^{+\infty}\frac{e^{-nt}-e^{-(n+1)t}}{t} + e^{-(n+z)t}\,dt\\ &=\int_{0}^{+\infty} \sum_{n\geq 1}\frac{e^{-nt}-e^{-(n+1)t}}{t} + e^{-(n+z)t}\,dt\\ &=\int_{0}^{+\infty} \left( \frac{1- e^{-t}}{t} + e^{-zt} \right)\sum_{n\geq 1} e^{-nt} \,dt \\ &=\int_{0}^{+\infty} \left( \frac{1- e^{-t}}{t} + e^{-zt} \right) \frac{e^{-t}}{1-e^{-t}} \,dt \\ &=\int_{0}^{+\infty} \left(\frac{e^{-t}}{t} + \frac{e^{-(z+1)t}}{1-e^{-t}} \right) \,dt \\ \end{align*}

and we are done (modulo whatever typos/mistakes I've made.)