0
$\begingroup$

enter image description here

I am having trouble trying to understand a step in this theorem. Basically how do they justify $\int_{1}^{1+x} \frac{(1-t)^n}{t} dt\leq \int_{1}^{1+x} (t-1)^n dt$. I tried solving the left side integral but it is pretty ugly. I'm sure there is something obvious i'm not seeing. Clarifying this step will be great thanks.

$\endgroup$
1
$\begingroup$

If $f(x) \leq g(x)$ on $[a,b]$ for all $x$, then $\int_a^b f(t) dt \leq \int_a^b g(t) dt$. In this case, $\frac{(1 - t)^n}{t} \leq (t - 1)^n$ on $[1, 1 + x]$ since $t \geq 1$ on this interval.

$\endgroup$
  • $\begingroup$ Thank you so much I forgot about that theorem. $\endgroup$ – user60887 May 2 '14 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.