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.
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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 $\ds{\int_{0}^{\infty}\expo{-t}\,\dd t = 1}$:
$$\bbox[15px,border:1px dotted navy]{\displaystyle
\lim_{n \to \infty}\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}=
     -\int_{0}^{\infty}\expo{-t}\ln\pars{t}\,\dd t}
$$
A: 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:
$$ \begin{align*}
\int_{0}^{1} v^n \log (1-v) \; dv
&= \left. \frac{v^{n+1} - 1}{n+1} \log (1-v) \right|_{0}^{1} - \int_{0}^{1} \frac{v^{n+1} - 1}{n+1} \cdot \frac{1}{v - 1} \; dv \\
&= - \frac{1}{n+1} \int_{0}^{1} \frac{1 - v^{n+1}}{1 - v} \; dv
\end{align*}$$
A: 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
$$\psi \left( x \right) = \frac{{\Gamma '\left( x \right)}}{{\Gamma \left( x \right)}} =  - \gamma  - \frac{1}{x} + \sum\limits_{v = 1}^\infty  {\frac{x}{{v\left( {x + v} \right)}}} $$
