Convergence of $\int_1^\infty(\cos^2(\pi x))^x\ dx$ I am looking for ways to figure out whether the integral
$$\int_1^\infty(\cos^2(\pi x))^x\ dx$$
converges.
If not, are there other similar integrands, for example
$$\int_1^\infty(\cos^2(\pi x))^{x^x}\ dx$$
for which the integral does converge?
 A: $$\int_{0}^{1}\cos^2(\pi x)^m\,dx = \frac{1}{4^m}\binom{2m}{m}\sim \frac{1}{\sqrt{\pi m}}$$
so the first integral is divergent. The trick is to break the integration range into intervals with unit length, then exploit the fact that 
$$ \int_{m}^{m+1}\cos^2(\pi x)^x \,dx \sim \int_{m}^{m+1}\cos^2(\pi x)^m \,dx \sim \frac{1}{\sqrt{m}}.$$
Since $\sum_{m\geq 1}\frac{1}{\sqrt{m}}$ is divergent, so it is $\int_{1}^{+\infty}\cos^2(\pi x)^x\,dx.$ The same argument also shows that
$$ \int_{1}^{+\infty}\cos^2(\pi x)^{x^\alpha}\,dx $$
is convergent for any $\alpha>2$.
A: I realize I might be a little late, but anyhow, for a more elementary argument, if we shift by $n$ we get that
$$\int_{n}^{n+1}(\cos^2(\pi x))^x\ dx = \int_0^1 (\cos^2(\pi x))^{x + n}\ dx$$
$$> \int_0^1 (\cos^2(\pi x))^{n + 1}\ dx$$
$$\geq \int_0^{1 \over \sqrt{n}} (\cos^2(\pi x))^{n + 1}\ dx$$
Using that $\cos x > 1 - {x^2/2}$, one then has 
$$\int_0^{1 \over \sqrt{n}} (\cos^2(\pi x))^{n + 1}\ dx > \int_0^{1 \over \sqrt{n}}\bigg(1 - {x^2 \over 2}\bigg)^{2n + 2}\ dx$$
By Bernoulli's inequality, $(1 + r)^k > 1 + rk$ whenever $1 + r > 0$, so we are led to
$$\int_0^{1 \over \sqrt{n}}\bigg(1 - {x^2 \over 2}\bigg)^{2n + 2}\ dx > \int_0^{1 \over \sqrt{n}}1 - (n+1) x^2\ dx$$
$$= {1 \over \sqrt{n}} - {n+ 1 \over 3 n^{3 \over 2}}$$
This is greater than ${1 \over 2\sqrt{n}}$ if $n$ is large enough. Hence for large enough $n$ we have
$$\int_{n}^{n+1}(\cos^2(\pi x))^x\ dx > {1 \over 2\sqrt{n}}$$ Since $\sum_{n=1}^{\infty} {1 \over 2\sqrt{n}} = \infty$ the integral diverges.
