I claim this Improper Integral converges

[Rudeboy37]

Homework Statement

Does the integral from 1 to infinity of ([(Cos[Pi x])^(2x)]/x)dx converge?

Homework Equations

N/A

The Attempt at a Solution

I claim it converges (based on how small the values of the function get when x is not an integer), but I'm not really sure how to rigorously justify it. It doesn't look fun (or possible) to integrate, so I was trying to do it by a comparison of some sort, but that didn't pan out particularly well (I couldn't get functions like ((cosx)^2k)/x to not diverge or justify that they do converge). Any help or ideas? Thanks

 

[hunt_mat]

Note:

<br /> \int_{1}^{\infty}\frac{\cos^{2x}\pi x}{x}dx\leqslant\left|\int_{1}^{\infty}\frac{\cos^{2x}\pi x}{x}dx\right|\leqslant\int_{1}^{\infty}\left|\frac{\cos^{2x}\pi x}{x}\right| dx<br />

as |\cos\pi x|\leqslant 1, we have:

<br /> \int_{1}^{\infty}\frac{\cos^{2x}\pi x}{x}dx\leqslant\int_{1}^{\infty}\frac{1}{x}dx =\left[\log x\right]_{1}^{\infty}<br />

So it may not converge.

 

[Dick]

You can show it converges. Estimate the integral over periods of the cos(pi*x). For example, what's the behavior of the integral of cos(pi*x)^(2*n) for x from 0 to 1 for large n?