21
$\begingroup$

I'm trying to show that

$$\lim_{n \to \infty} \left[\sum_{k=1}^{n} \frac{1}{k} - \log n\right] = -\int_0^{\infty} e^{-t} \log t \,dt.$$

In other words, I'm trying to show that the above definitions of the Euler-Mascheroni constant $\gamma$ are equivalent.

In another post here (which I can't seem to find now) someone noted that

$$\int_0^{\infty} e^{-t} \log t \,dt = \left.\frac{d}{dx} \int_0^{\infty} t^x e^{-t} \,dt \right|_{x=0} = \Gamma'(1) = \psi(1),$$

where $\psi$ is the digamma function. This may be a good place to start on the right-hand side.

For the left-hand side I was tempted to represent the terms with integrals. It is not hard to show that

$$\sum_{k=1}^{n} \frac{1}{k} = \int_0^1 \frac{1-x^n}{1-x} \,dx,$$

but I'm not sure this gets us anywhere.

Any help would be greatly appreciated.

$\endgroup$
28
$\begingroup$

It is easy to prove that the function

$$ f_n(x) = \begin{cases} \left( 1 - \frac{x}{n}\right)^n & 0 \leq x \leq n \\ 0 & x > n \end{cases}$$

satisfies $0 \leq f_n(x) \uparrow e^{-x}$. Thus by dominated convergence theorem,

$$ \int_{0}^{\infty} e^{-x} \log x \; dx = \lim_{n\to\infty} \int_{0}^{n} \left( 1 - \frac{x}{n}\right)^n \log x \; dx. $$

Now by the substitution $x = nu$, we have

$$\begin{align*} \int_{0}^{n} \left( 1 - \frac{x}{n}\right)^n \log x \; dx &= n\int_{0}^{1} \left( 1 - u\right)^n (\log n + \log u) \; du \\ &= \frac{n}{n+1}\log n + n\int_{0}^{1} \left( 1 - u\right)^n \log u \; du \\ &= \frac{n}{n+1}\log n + n\int_{0}^{1} v^n \log (1-v) \; dv \\ &= \frac{n}{n+1}\log n - n\int_{0}^{1} v^n \left( \sum_{k=1}^{\infty} \frac{v^k}{k} \right) \; dv \\ &= \frac{n}{n+1}\log n - n \sum_{k=1}^{\infty} \frac{1}{k(n+k+1)} \\ &= \frac{n}{n+1}\log n - \frac{n}{n+1} \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{n+k+1}\right) \\ &= \frac{n}{n+1} \left( \log n - \sum_{k=1}^{n+1} \frac{1}{k} \right). \end{align*}$$

Therefore taking $n \to \infty$ yields $-\gamma$. If you are not comfortable with the interchange of integral and summation, you may perform integration by parts as follows:

$$ \begin{align*} \int_{0}^{1} v^n \log (1-v) \; dv &= \left. \frac{v^{n+1} - 1}{n+1} \log (1-v) \right|_{0}^{1} - \int_{0}^{1} \frac{v^{n+1} - 1}{n+1} \cdot \frac{1}{v - 1} \; dv \\ &= - \frac{1}{n+1} \int_{0}^{1} \frac{1 - v^{n+1}}{1 - v} \; dv \end{align*}$$

$\endgroup$
  • 2
    $\begingroup$ Thank you. That's a nice trick at the beginning there. $\endgroup$ – Antonio Vargas Feb 23 '12 at 5:32
  • 1
    $\begingroup$ The dominated convergence theorem says we can interchange limit and integral on a fixed interval $I$ (ie. $\lim_{n \to \infty} \int_I f_n = \int_I \lim_{n \to \infty} f_n$) yet here $[0, \infty)$ is transformed into : $\lim_{n \to \infty} [0,n)$, so I might be mistaken but there must extra justifications to say that we can take the limit on the bounds of the interval and on the functions $f_n$, since : $\int_0^n \lim_{n \to \infty} f_n$ doesn’t mean anything ? Thank you :) $\endgroup$ – Thinking Nov 7 '18 at 9:49
  • 1
    $\begingroup$ @Thinking, That is precisely why I am considering $f_n$ instead of $(1-\frac{x}{n})^n$. We have $$\int_{0}^{n}f_n(x)\,dx=\int_{0}^{\infty} f_n(x)\,dx.$$ $\endgroup$ – Sangchul Lee Nov 7 '18 at 16:01
  • $\begingroup$ @SangchulLee Aaaah nice, I’ve completely missed that. I understand better Antonio’s comment then. Indeed that’s a nice trick. $\endgroup$ – Thinking Nov 7 '18 at 16:20
8
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

$\ds{\lim_{n \to \infty}\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}= -\int_{0}^{\infty}\expo{-t}\ln\pars{t}\,\dd t:\ {\large ?}}$

\begin{align} \sum_{k = 1}^{n}{1 \over k} &= \sum_{k = 1}^{n}\int_{0}^{1}t^{k - 1}\,\dd t = \int_{0}^{1}\sum_{k = 1}^{n}t^{k - 1}\,\dd t =\int_{0}^{1}{1 - t^{n - 1} \over 1 - t}\,\dd t =\int_{\infty}^{1}{1 - t^{1 - n} \over 1 - 1/t}\,\pars{-\,{\dd t \over t^{2}}} \\[3mm]&=\int_{1}^{\infty}{t^{-1} - t^{-n} \over t - 1}\,\dd t =\int_{0}^{\infty}{\pars{1 + t}^{-1} - \pars{1 + t}^{-n} \over t}\,\dd t \\[3mm]&=-\int_{0}^{\infty}\ln\pars{t} \bracks{-\pars{1 + t}^{-2} + n\pars{1 + t}^{-n - 1}}\,\dd t \\[3mm]&=\int_{0}^{\infty}{\ln\pars{t} \over \pars{1 + t^{2}}}\,\dd t -\int_{0}^{\infty}\ln\pars{t \over n}\pars{1 + {t \over n}}^{-n - 1}\,\dd t \end{align}


The first integral vanishes out: Just split $\ds{\pars{0,\infty}}$ in $\ds{\pars{0,1}}$ and $\ds{\pars{1,\infty}}$ and we'll see that the 'pieces' cancels each other: \begin{align} \sum_{k = 1}^{n}{1 \over k} - \ln\pars{n} & = \ln\pars{n}\bracks{\overbrace{\int_{0}^{\infty}\pars{1 + {t \over n}}^{-n - 1}\,\dd t}^{\ds{=\ 1\,,\ \forall\ n\ >\ 0}}\ -\ 1}\ -\ \int_{0}^{\infty}\ln\pars{t}\pars{1 + {t \over n}}^{-n - 1}\,\dd t \end{align}

Note that $\ds{\lim_{n \to \infty}\pars{1 + {t \over n}}^{-n - 1} = \expo{-t}}$ and $\ds{\int_{0}^{\infty}\expo{-t}\,\dd t = 1}$: $$\bbox[15px,border:1px dotted navy]{\displaystyle \lim_{n \to \infty}\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}= -\int_{0}^{\infty}\expo{-t}\ln\pars{t}\,\dd t} $$

$\endgroup$
  • $\begingroup$ Felix, this is an excellent development ... +1. If I may make one observation; the bracketed term in the highlighted box is actually zero since the integral is $1$ for all $n$. Therefore, there is no need to pass to the limit to annihilate that term. It is simply zero for all $n$. -Mark $\endgroup$ – Mark Viola May 27 '16 at 3:14
  • $\begingroup$ @MarkViola Thanks. I didn't go back to this answer for awhile. I just read your comment. $\textsf{Fixed}$. $\endgroup$ – Felix Marin Mar 14 '18 at 21:55
2
$\begingroup$

Since

$$\frac{{\Gamma '\left( x \right)}}{{\Gamma \left( x \right)}} = - \gamma - \frac{1}{x} + \sum\limits_{v = 1}^\infty {\frac{x}{{v\left( {x + v} \right)}}} $$

We evaluate the expression at $x=1$ to get

$$\frac{{\Gamma '\left( 1 \right)}}{{\Gamma \left( 1 \right)}} = \Gamma '\left( 1 \right) = - \gamma - 1 + \sum\limits_{v = 1}^\infty {\frac{1}{{v\left( {1 + v} \right)}}} $$

But since $$\sum\limits_{v = 1}^\infty {\frac{1}{{v\left( {1 + v} \right)}}}=1 $$

we get

$$\Gamma '\left( 1 \right) = - \gamma $$

This would be an instant consequence of the proof that the digamma function is defined by

$$\psi \left( x \right) = \frac{{\Gamma '\left( x \right)}}{{\Gamma \left( x \right)}} = - \gamma - \frac{1}{x} + \sum\limits_{v = 1}^\infty {\frac{x}{{v\left( {x + v} \right)}}} $$

$\endgroup$
  • $\begingroup$ But what is your proof that $\frac{\Gamma'(x)}{\Gamma(x)}= - \gamma + \dots$ that does not rely on showing what's in the question prompt? $\endgroup$ – Dzoooks Jul 4 at 2:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.