TITLE: Showing that $\gamma = -\int_0^{\infty} e^{-t} \log t \,dt$, where $\gamma$ is the Euler-Mascheroni constant.
QUESTION [23 upvotes]: 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.

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
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$\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 $\ds{\int_{0}^{\infty}\expo{-t}\,\dd t = 1}$:
$$\bbox[15px,border:1px dotted navy]{\displaystyle
\lim_{n \to \infty}\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}=
     -\int_{0}^{\infty}\expo{-t}\ln\pars{t}\,\dd t}
$$