Show $\lim\limits_{n \to \infty} \int_0^n \frac{dx}{x^n + x^2 + 1} = \frac{\pi}{4}$

Question

Show that

$$\lim_{n \to \infty} \int_0^n \frac{dx}{x^n + x^2 + 1} = \frac{\pi}{4}$$

Answer 1

Hint. One may write, for $n>1$, $$ \begin{align} 0\le\int_0^n \frac{dx}{x^n + x^2 + 1} &=\int_0^1 \frac{dx}{x^n + x^2 + 1} +\int_1^n \frac{dx}{x^n + x^2 + 1} \end{align} $$ we have $$ 0\le\int_1^n \frac{dx}{x^n + x^2 + 1} \le\int_1^n \frac{dx}{x^n} \le\frac1{n-1} $$ and, by the Dominated Convergence Theorem, one has $$ \lim_{n \to \infty}\int_0^1 \frac{dx}{x^n + x^2 + 1} =\int_0^1 \frac{dx}{x^2 + 1} \:dx=\frac \pi4. $$

Answer 2

Just to mention that one can do without dominated convergence (in case one does not know about it).

First, divide the integral $$ \int_0^n\frac{1}{x^n+x^2+1}\,dx= \int_0^1\frac{1}{x^n+x^2+1}\,dx+ \int_1^n\frac{1}{x^n+x^2+1}\,dx $$ and estimate the second one, just as Olivier did.

For the integral between $0$ and $1$, we just add and subtract with what we want, $$ \begin{aligned} \int_0^1\frac{1}{x^n+x^2+1}\,dx &=\int_0^1\frac{1}{x^2+1}\,dx +\int_0^1\frac{1}{x^n+x^2+1}-\frac{1}{x^2+1}\,dx\\ &=\frac\pi4-\int_0^1\frac{x^n}{(x^n+x^2+1)(x^2+1)}\,dx \end{aligned} $$ Since, on $[0,1]$, $$ \Bigl|\frac{x^n}{(x^n+x^2+1)(x^2+1)}\Bigr|\leq x^n $$ we find by the triangle inequality that $$ \biggl|\int_0^1\frac{x^n}{(x^n+x^2+1)(x^2+1)}\,dx\biggr|\leq\int_0^1 x^n\,dx=\frac{1}{n+1}. $$ The statement in your question follows by putting the pieces together.