On a certain integral related to Euler-Masheroni constant When trying to derive Mertens' theorem using Perron's formula, I encountered the following integral:
$$
\int_0^1{1-e^{-t}\over t}\mathrm dt-\int_1^\infty{e^{-t}\over t}\mathrm dt
$$
which, according to Titchmarsh's The Theory of Riemann Zeta Function, evaluates to $\gamma$ (a.k.a. Euler-Mascheroni constant).
To begin with, I converted this integral into the following limit:
$$
\lim_{N\to\infty}\left[\int_0^1{1-(1-t/N)^N\over t}\mathrm dt-\int_1^N{(1-t/N)^N\over t}\mathrm dt\right]
$$
so that I can eventually simplify the expression into
$$
\lim_{N\to\infty}\left[\sum_{m=1}^N\frac1m-\log N\right]
$$
which is identical to the definition of $\gamma$.
I wonder what tools can be used to justify the limit step.
 A: METHODOLOGY $1$:
To show that
$$\lim_{N\to\infty}\int_1^N \frac{\left(1-\frac tN\right)^N}{t}\,dt=\int_1^\infty \frac{e^{-t}}{t}\,dt$$
we will show that
$$\lim_{N\to\infty}\int_1^N \frac{e^{-t}}t \left(e^t\left(1-\frac tN\right)^N-1\right)\,dt=0\tag 1$$
Proceeding, we use the inequalities
$$\left(1-\frac tN\right)^N\le e^{-t}\tag 2$$
and
$$\begin{align}
e^t\left(1-\frac tN\right)^N &\ge \left(1+\frac tN\right)^N\left(1-\frac tN\right)^N\\\\
&=\left(1-\frac{t^2}{N^2}\right)^N\\\\
&\ge 1-\frac{t^2}{N}\tag3
\end{align}$$
Using $(2)$ and $(3)$, we have the estimates for the integral in $(1)$
$$-\frac1N \int_1^N te^{-t}\,dt \le \int_1^N \frac{e^{-t}}t \left(e^t\left(1-\frac tN\right)^N-1\right)\,dt\le 0\tag4$$
whence application of the squeeze theorem to $(4)$ yields $(1)$.

To show that
$$\lim_{N\to\infty}\int_0^1 \frac{1-\left(1-\frac tN\right)^N}{t}\,dt=\int_0^1 \frac{1-e^{-t}}{t}\,dt$$
we exploit the uniform convergence of $\left(1-\frac tN\right)^N$ to $e^{-t}$ for $t\in [0,1]$.  Then, we may interchange the limit and integral to obtain the result.

Putting it together, we have
$$\lim_{N\to \infty}\left(\int_1^N \frac{\left(1-\frac tN\right)^N}{t}\,dt-\int_0^1 \frac{1-\left(1-\frac tN\right)^N}{t}\,dt\right)=\int_0^1 \frac{1-e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-t}}{t}\,dt$$
as was to be shown.


METHODOLOGY $2$:
There is another way to proceed.  We can write the integral of interest as
$$\begin{align}
I&=\int_0^1 \frac{1-e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-t}}{t}\,dt\\\\
&=\lim_{\varepsilon\to 0^+}\int_\varepsilon^1  \frac{1-e^{-t}}{t}\,dt-\lim_{L\to \infty}\int_1^L \frac{e^{-t}}{t}\,dt\tag1
\end{align}$$
Integrating by parts both of the integrals on the right-hand side of $(1)$ we find
$$\begin{align}
I&=\lim_{\varepsilon\to 0^+}\left(-(1-e^{-\varepsilon})\log(\varepsilon)-\int_\varepsilon^1  e^{-t}\log(t)\,dt\right)\\\\
&-\lim_{L\to \infty}\left(e^{-L}\log(L)-\int_1^L e^{-t}\log(t)\,dt\right)\\\\
&=-\int_0^\infty e^{-t}\log(t)\,dt\\\\
\end{align}$$
Finally, using the development in the NOTE section of THIS ANSWER, we see that $-\int_0^\infty e^{-t}\log(t)\,dt=\gamma$ as was to be shown!
