1

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?

Alvin L-B
  • 417

2 Answers2

4

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

Jack D'Aurizio
  • 353,855
  • Very clever! I am able to fully understand everything except the first equality. How is that integral evaluated to $4^{-m}{2m\choose m}$? (I can see that the asymptotic formula follows from Stirling's approximation, but I am not clear on how the integral is evaluated in the first place.) – Alvin L-B Jan 09 '20 at 22:46
  • 2
    You use that $\cos x = 1/2 e^{ix} + 1/2e^{-ix}$. When you raise this to the $2m$ power and integrate from $0$ to $2\pi$, all terms vanish except for the middle term, and the integral of that term will be given in terms of the appropriate binomial coefficient. The same principle will apply in the case at hand with appropriate modifications. – Zarrax Jan 09 '20 at 23:17
0

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.

Zarrax
  • 44,950