Thread: Integral Variation View Single Post
2009-07-20, 12:58   #3
R.D. Silverman

Nov 2003

3·13·191 Posts

Quote:
 Originally Posted by cheesehead The variable ${t}$ is inside the integral. The variable ${x}$ is outside the integral. Let ${f(x)} = \int_2^{x^{1/3}}\frac{dt}{\log^3 t}$. The claim is that ${f(x)}$ varies as 1.5x/log x.
[mathematical gibberish deleted]

If "varies" means "is asymptotic to", then the claim is wrong.

int 1/log^3 t dt = -t/(2 log t) - t/(2 log^2 t) + li(t)/2. := g(t)

Now evaluate g(x^1/3) - g(2).