12.4. Let $I_{n}=\int_{1}^{n} \frac{[x]}{x^{2}+1} d x, n \in \mathbb{N}, n \geq 2$. Calculate: $\lim _{n \rightarrow \infty} \frac{I_{n}}{\ln n}$.
## Solution:

According to the Stolz-Cesàro lemma, we obtain

$$
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{I_{n}}{\ln n}=\lim _{n \rightarrow \infty} \frac{I_{n+1}-I_{n}}{\ln (n+1)-\ln n} \\
I_{n+1}-I_{n}=\int_{n}^{n+1} \frac{[x]}{x^{2}+1} d x=n \int_{n}^{n+1} \frac{d x}{x^{2}+1}= \\
=n(\operatorname{arctg}(n+1)-\operatorname{arctg} n)= \\
=n \cdot \operatorname{arctg} \frac{(n+1)-n}{1+(n+1) n}=n \cdot \operatorname{arctg} \frac{1}{n^{2}+n+1}
\end{gathered}
$$

Then

$$
\lim _{n \rightarrow \infty} \frac{I_{n+1}-I_{n}}{\ln \left(1+\frac{1}{n}\right)}=\lim _{n \rightarrow \infty} \frac{n \cdot \operatorname{arctg} \frac{1}{n^{2}+n+1}}{\ln \left(1+\frac{1}{n}\right)}=\lim _{n \rightarrow \infty} \frac{n \cdot \frac{1}{n^{2}+n+1}}{\frac{1}{n}}=1
$$