Euler Mascheroni Constant is Zero $$
H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n \frac{dy}{y}
$$
Let $x = \frac{y - 1}{n - 1}$, or $y = (n-1)x + 1$. Then, $dy = (n - 1) dx$.
$$
H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_0^1 \frac{n - 1}{(n - 1) x + 1} dx = \int_0^1 (\frac{1 - x^n}{1 - x} - \frac{n - 1}{(n - 1) x + 1}) dx
$$
$$
\gamma = \lim_{n \rightarrow \infty} \int_0^1 (\frac{1 - x^n}{1 - x} - \frac{n - 1}{(n - 1) x + 1}) dx = \int_0^1 (\lim_{n \rightarrow \infty} \frac{1 - x^n}{1 - x} - \lim_{n \rightarrow \infty} \frac{n - 1}{(n - 1) x + 1}) dx = \int_0^1 (\frac{1}{1 - x} -  \frac{1}{x}) dx
$$
(The first limit relies on the fact that $0 < x < 1$, so that $\lim_{n \rightarrow \infty} x^n = 0$. The second is solved via. L'Hospital's rule.)
Let $t = 1 - x$. Then, $dt = -dx$.
$$
\gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x} = - \int_1^0 \frac{dt}{t} - \int_0^1 \frac{dx}{x} = \int_0^1 \frac{dt}{t} - \int_0^1 \frac{dx}{x} = 0
$$
Evidently something's wrong with this, but I genuinely can't find it. The only part I find even fishy is how the limit is split in two. Any ideas?
 A: You obviously took a wrong turn along the way to conclude
$$\gamma = \int_0^1 \frac{dx}{1 - x} - \int_0^1 \frac{dx}{x},$$
since each integral is improper and divergent.  The limit/integral switch cannot be justified by uniform convergence, monotone convergence, etc.
As an example where the switch is valid, change variables in the first integral with $x = (1-y/n)$ to get
$$H_n - \ln n = \int_0^1 \frac{1 - x^n}{1 - x} dx - \int_1^n  \frac{dy}{y} \\= \int_0^n \left[1 - \left(1- \frac{y}{n}\right)^n\right]\frac{dy}{y}-\int_1^n  \frac{dy}{y}\\=\int_0^1 \left[1 - \left(1- \frac{y}{n}\right)^n\right]\frac{dy}{y}-\int_1^n  \left(1- \frac{y}{n}\right)^n\frac{dy}{y}.$$
Now you can justify a switch of limit and integral using LDCT to obtain
$$\gamma = \lim_{n \to \infty}(H_n - \ln n)= \int_0^1 \left(1 - e^{-y}\right)\frac{dy}{y}-\int_1^\infty  e^{-y}\frac{dy}{y},$$
where the improper integrals converge.  
Further manipulation leads to the well-known result
$$\gamma = \int_0^1 \left( \frac{1}{1-x}+ \frac{1}{\ln x}\right) \, dx.$$
Here the improper integral converges, as the singular behavior of $1/(1-x)$ is offset by that of $1/\ln x$ near $x=1$.
