# 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.

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

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*}

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

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

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

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

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