How to show this integral is $O(\epsilon)$ Let $p>1$. How to prove that
$$ \int_{0}^{1}\left|\log{\frac{x+\epsilon}{x}}\right|^{p}dx\leq C \epsilon $$
for some constant $C>0$.
By the dominated convergence theorem we know that
$$\lim_{\epsilon\to 0^{+}} \int_{0}^{1}\left|\log{\frac{x+\epsilon}{x}}\right|^{p}dx=0. $$
This suggests using l'Hopital to show that
$$\lim_{\epsilon\to 0^{+}} \frac{\int_{0}^{1}(\log{\frac{x+\epsilon}{x}})^{p}dx} {\epsilon}=C $$
for some constant $C$.
My question then is about the limit
$$\lim_{\epsilon\to 0^{+}}\int_{0}^{1}\left(\log{\frac{x+\epsilon}{x}}\right)^{p-1}\frac{1}{x+\epsilon}dx.$$
Taking into account the factor $1/(x+\epsilon)$, is the dominated convergence theorem applicable here?
 A: The substitution $x=\epsilon/(e^t-1)$ gives $$\frac1\epsilon\int_0^1\left(\log\frac{x+\epsilon}x\right)^pdx=\int_{\log(1+\epsilon)}^\infty\frac{t^p e^t\,dt}{(e^t-1)^2}.$$ As $\epsilon\to0^+$, this tends to $\int_0^\infty$, which converges since $p>1$. In fact we get $$\lim_{\epsilon\to0^+}\frac1\epsilon\int_0^1\left(\log\frac{x+\epsilon}x\right)^pdx=\Gamma(1+p)\zeta(p)$$ in terms of the gamma function and the Riemann zeta function.
A: You can rewrite the integral as
$$
\int_0^1 dx\, \log\left(1+\frac{\varepsilon}{x}\right)^p = 
\underbrace{\int_0^\varepsilon dx\, \log\left(1+\frac{\varepsilon}{x}\right)^p}_{A}
+ \underbrace{\int_\varepsilon^1 dx\, \log\left(1+\frac{\varepsilon}{x}\right)^p}_{B}
$$
The second term is easy: using $\log(1+u)\leq u$,
$$
B \leq \varepsilon^p \int_\varepsilon^1 \frac{dx}{x^p} = \varepsilon^p \cdot \frac{1}{p-1}\left(\frac{1}{\varepsilon^{p-1}}-1\right) \leq  \frac{\varepsilon}{p-1}
$$
The first is a bit more annoying: letting $u := x/\varepsilon$,
$$
A = \varepsilon \int_0^1 du\, \log\left(1+\frac{1}{u}\right)^p
$$
but now, the integral $C_p := \int_0^1 du\, \log\left(1+\frac{1}{u}\right)^p$ converges (and is just a constant, depending only on p).
So $$\boxed{A+B \leq \left(\frac{1}{p-1}+C_p\right) \varepsilon}$$ showing the result.
