Convergence of $\int_1^\infty(\cos^2(\pi x))^x\ dx$

Question

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?

Answer 1

$$\int_{0}^{1}\cos^2(\pi x)^m\,dx = \frac{1}{4^m}\binom{2m}{m}\sim \frac{1}{\sqrt{\pi m}}$$ 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 $\alpha>2$.

Answer 2

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.