$\int_0^1 \min\{X, ||\alpha||^{-1}\} \, d\alpha \ll \log(X)$? I'm currently reading a book and the author used the following fact without mentioning why $$ \int_0^1 \min\{X, ||\alpha||^{-1}\} \, d\alpha \ll \log(X).$$
In here, $X > 0$ is a large real number and $||\alpha||$ represents $\min\limits_{n \in \mathbb{Z}} |\alpha - n|$. Also, $\ll$ means that there exists a $C > 0$ so that $$\int_0^1 \min\{X, ||\alpha||^{-1}\} \, d\alpha \leq C\log(X).$$
Can someone explain this to me?
 A: The integration is over $\alpha\in (0,1)$ then
$$||\alpha|| = 
\begin{cases} 
\alpha \qquad  \quad \alpha\in \left(0,\frac{1}{2} \right)\\
1-\alpha \qquad \alpha\in \left(\frac{1}{2},1 \right)\\
\end{cases}
$$
Then,
$$\int_0^1 \min\{X, ||\alpha||^{-1}\} \, d\alpha = 
\int_0^{\frac{1}{2}} \min\{X, \frac{1}{\alpha}\} \, d\alpha
+\int_{\frac{1}{2}}^1 \min\{X, \frac{1}{1-\alpha}\} \, d\alpha  \tag{1}
$$
The calculation of the two integrals in $(1)$ is not difficult. With the help of Mathematica, we have
$$\int_0^1 \min\{X, ||\alpha||^{-1}\} \, d\alpha = 
\begin{cases} 
X \qquad  \qquad  \quad X \le 2 \\
2+2\ln \frac{X}{2} \qquad X > 2 \\
\end{cases}  \tag{2}
$$
But from $(2)$, I don't think there exists $C$ satisfying your condition
$$\int_0^1 \min\{X, ||\alpha||^{-1}\} \, d\alpha \leq C\log(X) \tag{3}$$
Indeed, take $X =1 + \epsilon$ for $\epsilon$ very small. Then the LHS of $(3)$ equal to $1-\epsilon$ and tends to $1$ whilst the RHS of $(3)$ equal to $C \log (1-\epsilon) $ and tends to $0$ when $\epsilon \to 0$.
Note: if we know $X$ is a large real number, then it suffices to choose $C = 3$.
