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I've been studying the digamma function lately and I often see it defined as the following:

$$ \psi(x) = \frac{\Gamma'(x)}{\Gamma(x)} $$ or

$$ \psi(x) = \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}} dt $$

I've seen this shown but rarely justified how these expressions are the same. Perhaps I don't have the tools to recognize it, but it is not immediately obvious to me. I'd like a proof laid out. Showing why these are equivalent.

I understand the identity $\psi(x) = -\gamma+\sum_{n=0}^{\infty}\frac{1}{n+1}-\frac{1}{n+x} $ comes from differentiating $\ln(\Gamma(x))$ and using the Weierstrass product formula for $\Gamma$ as explained in this post here (for those like me learning from the bottom up I found this PDF which explains the product formula; also $\gamma$ is the Euler-Mascheroni constant). I also accept $\Gamma'(1)=-\gamma$ (some good proofs in this post).

Now I tried first to split the integral into two:

$$ \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}} dt = \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-t}}{1-e^{-t}} dt +\int_0^\infty \frac{e^{-t}}{1-e^{-t}} - \frac{e^{-xt}}{1-e^{-t}} dt $$

For $0<t<\infty$ then $0<e^{-t}<1$ thus it follows that

$$ \int_0^\infty \frac{e^{-t}}{1-e^{-t}} - \frac{e^{-xt}}{1-e^{-t}} dt=\sum_{n=0}^{\infty}\int_0^\infty e^{-(n+1)t}-e^{-(n+x)t}dt = $$

$$ \sum_{n=0}^{\infty} \frac{1}{n+1} - \frac{1}{n+x} $$

Now clearly last integral should be $-\gamma$ and we're done. However I can't satisfactorily do this. I suspect it's because I'm not being rigorous enough somewhere or I've been thinking about it too much and can't see an obvious mistake. To start, I split the integral in two and use integration by parts $u=e^{-t}$ and $dv=\frac{1}{t}dt$.

$$ \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-t}}{1-e^{-t}} dt = \int_0^\infty \frac{e^{-t}}{t} dt - \int_0^\infty \frac{e^{-t}}{1-e^{-t}} dt = $$

$$ \ln(t)e^{-t}|_{0}^\infty+\int_0^\infty \ln(t)e^{-t}dt-\sum_{n=0}^{\infty} \frac{1}{n+1}=\lim_{a\to0}-\ln(a)-\sum_{n=1}^{\infty} \frac{1}{n}+\Gamma'(1) $$

Here I had two thoughts a) the limit goes to $\infty$ and the harmonic series does the same so we can say they cancel each other out. This is too much hand waving for my liking, but b) was that $\lim_{a\to0}-\ln(a)=\lim_{a\to0}\ln(\frac{1}{a})=\lim_{n\to\infty}\ln(n)$ and hence we have

$$ \lim_{a\to0}\ln(a) - \sum_{n=1}^{\infty} \frac{1}{n} = \lim_{n\to\infty}\ln(n) - \sum_{k=1}^n \frac{1}{k} \equiv -\gamma $$

$$ \therefore \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-t}}{1-e^{-t}} dt = -2\gamma $$

This second approach seems less dodgy to me but is wrong.

Could someone point out my error in this method, or give an alternative proof using Calc II techniques? Please justify any identities for $\Gamma$, $\psi$, or $\gamma$ not used here or provide a link to a source which does.

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Let $F(x)$ be represented by the integral

$$F(x)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\right)\,dt\tag 1$$


Differentiating under the integral in $(1)$, we find that

$$\begin{align} F'(x)&=\int_0^\infty \frac{te^{-xt}}{1-e^{-t}}\,dt\\\\ &=\sum_{n=0}^\infty \int_0^\infty te^{-(n+x)t}\,dt\\\\ &=\sum_{n=0}^\infty \frac{1}{(n+x)^2}\tag 2 \end{align}$$


Next, integrating both sides of $(2)$ reveals

$$\begin{align} F(x)-F(1)&=\int_1^x F'(t)\,dt\\\\ &=\sum_{n=0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+x}\right)\\\\ F(x)&=F(1)+\sum_{n=0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+x}\right)\tag3 \end{align}$$

where $F(1)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}\right)\,dt$.


Then, we can evaluate $F(1)$ by simple application of integration by parts. Proceeding, we find that

$$\begin{align} F(1)&=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}\right)\,dt\\\\ &=\lim_{\epsilon\to 0^+}\left(\int_\epsilon^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}\right)\,dt\right)\\\\ &=\lim_{\epsilon\to 0^+}\left(-\log(\epsilon)+\int_\epsilon^\infty \log(t)e^{-t}\,dt+\log(1-e^{-\epsilon})\,dt\right)\\\\ &=\int_0^\infty \log(t)e^{-t}\,dt\tag4\\\\ &=-\gamma \tag5 \end{align}$$

where in going from $(4)$ to $(5)$, I appealed to the note at the end of THIS ANSWER.


Finally, substituting $(5)$ in $(3)$ yields

$$F(x)=-\gamma +\sum_{n=0}^\infty\left( \frac{1}{n+1}-\frac{1}{n+x}\right)=\psi(x)$$

as was to be shown!

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Non-detailed proof: (copied from my old answer here. Thinking of adding more detail and thus copying this if its not clear enough)

Using the definition $\Gamma'(z)$ and by replacing the $\text{log}(t)$ term with the integral $$ \text{log}(t)= \int_0^{\infty}\frac{e^{-x}-e^{-xt}}{x}, dx $$ we can show that $$ \Gamma'(z)=\int_0^\infty \frac{dx}{x}\left[ e^{-x}\Gamma(z)-\int_0^\infty e^{-t(x+1)}t^{z-1} dt \right]. $$ Using the substitution $u=t(x+1)$ we get $$ \psi(z)=\int_0^\infty\left[e^{-x}-\frac{1}{(x+1)^{z}} \right]\frac{dx}{x}. $$ Using this we can find $$ \psi(z)=\lim_{\epsilon \to 0} \left[ \int_\epsilon^\infty \frac{e^{-x}}{x}dx-\int_\epsilon^\infty \frac{1}{(x+1)^{z}x} dx \right]. $$ Substituting $1+x = e^u$ for the last integral we get $$ \psi(z)=\lim_{\epsilon \to 0} \left[ \int_{log(1+\epsilon)}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du -\int_{log(1+\epsilon)}^\epsilon \frac{e^{-u}}{u} du \right]. $$ Here the last integral goes to $0$ as $\epsilon \to 0$ so we get $$ \psi(z)= \int_{0}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du. $$

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I just was thinking about this several months later and it came to me what I wasn't seeing before: $$ \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-xt}}{1-e^{-t}} dt = \ln(t)e^{-t}|_0^\infty + \int_0^\infty\ln(t)e^{-t}dt + \int_0^\infty \frac{e^{-t}-e^{-xt}}{1-e^{-t}}dt - \int_0^\infty\frac{e^{-t}}{1-e^{-t}}dt $$ $$ =-\lim_{n\to0}\ln(n)+\Gamma'(1)+\int_0^1\frac{1-u^{x-1}}{1-u}du-\int_0^1\frac{1}{w}dw = -\lim_{n\to0}\ln(n)-\gamma + \int_0^1\frac{1-u^{x-1}}{1-u}du+\lim_{n\to0}\ln(n) $$ $$ =-\gamma + \sum_{n=0}^{\infty}\int_0^1u^n-u^{n+x-1} du =-\gamma + \sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{n+x} = \psi(x) $$

The last identity as explained in my question can be show using differentiation of $\ln(\Gamma(x))$ and using the Weierstrass formula of $\Gamma$ (see links in question for detailed explanation).

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  • $\begingroup$ This analysis is flawed. $\endgroup$ – Mark Viola Nov 12 '18 at 17:37

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