Wikipedia:Reference desk/Archives/Mathematics/2011 November 30

= November 30 =

Integral limit
If f is continuous on [0,1], how to show that
 * $$\lim_{x \to 0^{+}} x \int_x^1 \frac{f(t)}{t} dt = 0?$$

I've been asked to compute the integral, which is easily done by choosing f = 0; but how could you show that it's true for all such f? Please assume only a knowledge of the fundamental theorem of calculus, if possible; I'd like to avoid log. — Anonymous Dissident  Talk 12:29, 30 November 2011 (UTC)


 * You can use
 * $$\left|\int_x^1 \frac{f(t)}{t} dt \right|\le \max|f|\int_x^1 \frac{1}{t} dt$$
 * So it's enough to calculate
 * $$\lim_{x\to 0^+} x\int_x^1\frac{1}{t}\,dt.$$
 * You can either do this by L'Hopital's rule, or, if that's not allowed, write it as
 * $$x\int_x^1\frac{1}{t}\,dt = \sqrt{x}\left(\sqrt{x}\int_x^1\frac{1}{t}\,dt\right).$$
 * By taking derivatives, show that the factor in parenthesis is increasing and bounded below, and therefore has a limit as $$x\to 0^+$$. Sławomir Biały  (talk) 12:46, 30 November 2011 (UTC)


 * (e.c.) Just use the fact that a continuous function on a closed bounded interval is bounded: for some $$C$$ there holds $$|f(t)|\le C$$ for all $$t\in [0,1]$$. So you can bound the absolute value of that expression replacing $$f(t)$$ by the constant $$C$$. In this case integrating then get $$-Cx \log x$$, that notoriously converges to $$0$$ as $$\scriptstyle x\to 0^+$$; you conclude thanks to the sandwich theorem.  --pm a  12:53, 30 November 2011 (UTC)