I would like to prove the following:
$$\lim_{x\to\infty}\frac{1}{x}\int_0^x\left(\frac{x}{t}-1\right)^{1/2}\;dt=\infty$$
It might not be true.
I suppose I could try to give a lower bound via Riemann sums, and hope they tend to something order larger than x. But it seems like there should be an easier way.
Thoughts? Thanks!
It's not true. Let $x>0$; and do the change of variable $u=t/x$. The integral becomes $$ \frac{1}{x}\int_0^x \left(\frac{x}{t}-1\right)^{1/2} dt = \int_0^1 \left(\frac{1}{u}-1\right)^{1/2} du = \frac{\pi}{2} $$ independent of $x$. (The value matters little; but it's finite, and independent of $x$.)
To see that the integral is finite (i.e., $f(u)=\left(\frac{1}{u}-1\right)^{1/2}$ is integrable on $(0,1]$), note that $f$ is continuous on $(0,1]$ (therefore, the only question is integrability at $0$), and moreover $$ f(u) \operatorname*{\sim}_{u\to 0} \frac{1}{\sqrt{u}} $$ so by comparison $f$ is integrable at $0$, since $1/\sqrt{u}$ is.