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Old 2009-07-20, 12:58   #3
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by cheesehead View Post
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).
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