When Does The Integral Converge?

To get started, since $0 \leq \cos^2(x) \leq 1$ , we have

$\large \int_0^{\infty}\left(\cos^2x\right)^{ \left( \pi\left \lceil{\frac{x}{\pi}}\right \rceil \right)^{n}}dx<\int_0^{\infty}\left(\cos^2x\right)^{x^n}dx<\int_0^{\infty}\left(\cos^2x\right)^{\left( \pi\left \lfloor\frac{x}{\pi}\right \rfloor \right)^{n}}dx$

Now we can break the integrals into sums

$\large \sum_{j=1}^{\infty}\int_0^{\pi}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx<\int_0^{\infty}\left(\cos^2x\right)^{x^n}dx<\sum_{j=0}^{\infty}\int_0^{\pi}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx$

Since $\cos(x)$ is periodic, we can shift the integrals

$\large \sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx<\int_0^{\infty}\left(\cos^2x\right)^{x^n}dx<\sum_{j=0}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx$

Which means the integral converges if and only if $\displaystyle \sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx$ converges

Now, to make an upper bound, we can use the very good approximation $e^{-x^2} \geq \cos^2(x)$ . To make a lower bound, we can use a rough, but still useful $e^{-2x^2} \leq \cos^2(x)$ . Both hold in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ . We get

$\large \sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{-2\left(j\pi\right)^nx^2}dx<\sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx<\sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{-\left(j\pi\right)^nx^2}dx$

Now, its not very hard to show that $\large \displaystyle \frac{1}{2}\int_{-\infty}^{\infty}e^{-2\left(j\pi\right)^nx^2}dx<\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{-2\left(j\pi\right)^nx^2}dx$ , so I'll skip this step to make this solution less cumbersome. If you want to get a feel for it, I'd suggest looking at the graph of the integrand. Also, $\large \displaystyle \int_{-\infty}^{\infty}e^{-\left(j\pi\right)^nx^2}dx>\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{-\left(j\pi\right)^nx^2}dx$ . This gives us

$\large \frac{1}{2}\sum_{j=1}^{\infty}\int_{-\infty}^{\infty}e^{-2\left(j\pi\right)^nx^2}dx<\sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx<\sum_{j=1}^{\infty}\int_{-\infty}^{\infty}e^{-\left(j\pi\right)^nx^2}dx$

Solving the gaussian integrals, we get

$\large \frac{1}{2}\sum_{j=1}^{\infty}\sqrt{\frac{\pi}{2\left(j\pi\right)^n}}<\sum_{j=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^2x\right)^{\left(j\pi\right)^n}dx<\sum_{j=1}^{\infty}\sqrt{\frac{\pi}{\left(j\pi\right)^n}}$

Which shows the term in the middle converges if and only if $\displaystyle \large \sqrt{\pi^{1-n}}\sum_{j=1}^{\infty}\frac{1}{j^{\frac{n}{2}}}$ also converges, which converges if and only if $\zeta ( \frac{n}{2} )$ also converges.

But $\zeta(x)$ converges for every $x > 1$ , so our initial integral converges for every $n$ greater than $\boxed{2}$ .

Bonus: It turns out $\sqrt{\pi^{1-n}}\zeta(\frac{n}{2})$ is actually a pretty good approximation to $\displaystyle \large \int_{\pi}^{\infty}\left(\cos^2x\right)^{x^n}dx$ . The first term, $\displaystyle \large \int_{0}^{\pi}\left(\cos^2x\right)^{x^n}dx$ goes to $1$ as you can see here , but it approaches slowly.

I do not know whether this is a solution or not .

But I think that it is an analogy.

Let us think of the series $\sum_{r=1}^{\infty} (cos(r))^{2r^{n}}$ .

Now by applying Cauchy's root test we see that the series converges for $n \geq 1$ .

But in the series we see that $(cos(r))^{2r^{n}}$ is never equal to $1$ for any $r$ and $n \geq 1$ .

But in the case of the integral, the function would have values of 1 when say $x=k.\pi$ .

So to eliminate that possibility we take $n \geq 2$ . Hence we arrive at the answer that the smallest possible $k$ is $2$

I will give a very intuitive (and not rigorous) explanation.

Clearly, if $n\leq 0$ the integral diverges. So let $n>0$ .

Because $|\cos(x)|<1$ except for all $x=k\pi, \ k\in\mathbb{N}$ where it is one, and $x^n\underset{x\to +\infty}{\longrightarrow}+\infty, \quad (\cos^2(x))^{x^n}$ goes to $0$ super fast except on the points $k\pi$ where is remains $1$ .

I will not, but you can prove that if you remove all the intervals $[k\pi-\varepsilon;k\pi+\varepsilon]$ for $\varepsilon>0$ the integral converges for all $n>0$ . (It is because we have now this inequality: $|\cos(x)|\leq a<1$ for some $a$ .)

Therefore the heart of the matter is around the points $x=k\pi$ where there are some little peaks of height $1$ . I will try to estimate there width.

Around $k\pi, \ (|x|\ll 1), \ |\cos(k\pi +x)| \approx 1-\frac{x^2}{2}$ .

So,

$\begin{aligned} (\cos^2(k\pi+x))^{(k\pi+x)^n} & \approx (1-\frac{x^2}{2})^{2(k\pi)^n}\\ &=e^{2(k\pi)^n\ln(1-\frac{x^2}{2})}\\ &\approx e^{-(k\pi)^n x^2} \end{aligned}$

Roughly the peak exists where $(k\pi)^n x^2\leq 1$ (either the decreasing exponential is almost $0$ ) ie for $|x|\leq \frac{1}{(k\pi)^{n/2}}$ . That gives us the width.

Resume; peak at $k\pi$ :

• height $1$
• width $\approx \frac{1}{k^{n/2}}$
• $\Rightarrow \ Area\approx \frac{1}{k^{n/2}}$

So intuitively, our integral behaves like $\displaystyle\sum_{k=1}^{+\infty}\frac{1}{k^{n/2}}$ which converges if and only if $\boxed{n>2}$ .