Prove $\lim_{n\to\infty} \int_0^1 \frac{\cos(x^n)}{1+x^n}\,dx = 1$

Question

Please help me prove the following equation:

$$\lim_{n\to \infty} \int_{0}^{1} \frac {\cos(x^n)}{1+x^n}\,dx=1.$$

I think I need to prove that the integrand uniformly converges to 1 (besides the endpoint) and then use a theorem but I don't know how.

Any ideas?

Answer 1

Try doing this without any theorems. As a hint, break it up into a two main pieces: $$\int_0^{1 - \epsilon}\frac{\cos(x^n)}{1+x^n}\,dx + \int_{1-\epsilon}^{1}\frac{\cos(x^n)}{1+x^n} \,dx.$$

You should give it a shot from here; for a complete solution, hover over the spoiler below.

The latter is less than $\epsilon$ since the integrand is bounded above by $1$, so we just have to deal with the former. For each $\delta > 0$, we may take $n$ sufficiently large so that $x^n < \delta$ for all $x \in [0,1-\epsilon]$. Since the function $y \mapsto \frac{\cos(y)}{1 + y}$ is continuous on $[0,1]$, there exists a $\delta$ so that $|y| < \delta$ implies $$\left|\frac{\cos(y)}{1 + y} - 1\right| < \epsilon\,.$$Thus, for $n$ sufficiently large, $$\left|\frac{\cos(x^n)}{1 + x^n} - 1\right| < \epsilon$$ for all $x \in [0,1-\epsilon]$. Plugging in this bound completes the proof.

Answer 2

For $x\geq0$, we have $$\left|\frac{\cos(x^n)}{1+x^n}-1\right|=\frac{1-\cos(x^n)+x^n}{1+x^n} \leq\frac{\frac{1}{2}x^{2n}+x^n}{1}=\frac{1}{2}x^{2n}+x^n.$$ So $$\left|\int_{0}^{1} \frac{\cos(x^n)}{1+x^n}\,dx-1\right| =\int_{0}^{1} \left(1-\frac{\cos(x^n)}{1+x^n}\right)\,dx\\ \leq\int_{0}^{1}\left(\frac{1}{2}x^{2n}+x^n\right) =\frac{1}{2(2n+1)}+\frac{1}{n+1}\to0.$$ Hence $$\lim_{n\to\infty}\int_{0}^{1} \frac{\cos(x^n)}{1+x^n}\,dx=1$$