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 :)

  • $\begingroup$ In the answer you linked to, the second integral is not derived from the first. Instead it's derived directly from the product representation of $\Gamma$. $\endgroup$ – Jair Taylor Feb 14 at 20:12
  • $\begingroup$ So it is derived from the following, right? wikimedia.org/api/rest_v1/media/math/render/svg/… But I still don't see how it is done, I'm so sorry. $\endgroup$ – ansebene Feb 14 at 20:29
  • $\begingroup$ No, Jack D'Aurizio derives it from $$\Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}.$$ If no one answers sooner, I might fill in the few missing details later when I have time. $\endgroup$ – Jair Taylor Feb 14 at 20:34
  • $\begingroup$ @ansebene Is it clear to you the main point with $\Gamma$ is to show $\int_0^\infty t^{z} e^{-t}dt = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}$. This is non-trivial and needs some complex analysis/Fourier analysis. Then the series and integral representation of $\Gamma'/\Gamma$ follows directly from the product $\endgroup$ – reuns Feb 15 at 4:08

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.)

  • $\begingroup$ Thank you so much :) Now it is perfectly clear :) $\endgroup$ – ansebene Feb 15 at 9:08

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