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November 30

Integral limit

If f is continuous on [0,1], how to show that

$\underset{x\to 0^{+}}{\lim }x\underset{x}{\overset{1}{\int }}\frac{f(t)}{t}dt=0?$

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

$\left|\underset{x}{\overset{1}{\int }}\frac{f(t)}{t}dt\right|\le \max |f|\underset{x}{\overset{1}{\int }}\frac{1}{t}dt$

So it's enough to calculate

$\underset{x\to 0^{+}}{\lim }x\underset{x}{\overset{1}{\int }}\frac{1}{t}dt.$

You can either do this by L'Hopital's rule, or, if that's not allowed, write it as

$x\underset{x}{\overset{1}{\int }}\frac{1}{t}dt=\sqrt{x}\left(\sqrt{x}\underset{x}{\overset{1}{\int }}\frac{1}{t}dt\right).$

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 $x\to 0^{+}$; you conclude thanks to the sandwich theorem. --pma 12:53, 30 November 2011 (UTC)