# Showing that $\gamma = -\int_0^{\infty} e^{-t} \log t \,dt$, where $\gamma$ is the Euler-Mascheroni constant.

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.

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*}

• Thank you. That's a nice trick at the beginning there. – Antonio Vargas Feb 23 '12 at 5:32
• 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 :) – Thinking Nov 7 '18 at 9:49
• @Thinking, That is precisely why I am considering $f_n$ instead of $(1-\frac{x}{n})^n$. We have $$\int_{0}^{n}f_n(x)\,dx=\int_{0}^{\infty} f_n(x)\,dx.$$ – Sangchul Lee Nov 7 '18 at 16:01
• @SangchulLee Aaaah nice, I’ve completely missed that. I understand better Antonio’s comment then. Indeed that’s a nice trick. – Thinking Nov 7 '18 at 16:20

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

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

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

• 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 – Mark Viola May 27 '16 at 3:14
• @MarkViola Thanks. I didn't go back to this answer for awhile. I just read your comment. $\textsf{Fixed}$. – Felix Marin Mar 14 '18 at 21:55

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

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

• 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? – Dzoooks Jul 4 '19 at 2:46