$\lim_{n \to \infty}\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$ Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$
For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge to $\infty$ as $n\rightarrow \infty$.
1 - How to find a bound for $f(n)$?
2- $f(n)\rightarrow\infty$, as $n\rightarrow\infty$?
Thank you for your time.

$$ \frac{x + n}{x + n^{-1}} = 1 + \frac{n-1}{nx + 1}$$
Using the expansion of $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$
We see that the resulting expansion of the integrand is $$\frac{n - 1}{nx + 1} - \frac{(n - 1)^2}{2(n x+ 1)^2} + O\left (\left(\frac{n-1}{nx + 1} \right)\right)^3$$
Squaring this again, we get $(\frac{n - 1}{nx + 1})^2 + O((\frac{n-1}{n x + 1})^3)$
The integral of the left term is just $$\left.-\frac{(n - 1)^2}{n(1 + n x)}\right|^{\infty}_0 = \frac{(n-1)^2}{n}$$
