Calculus – Convergence of Improper Integral and Limit of xf(x)

calculusimproper-integrals

I need to prove (as I wrote in the title):
for monotonic $f$: if the improper integral $\int_0^\infty f(x)dx$ converges, then $\lim_{x\to\infty}xf(x)=0$

hints please? tried to think of Cauchy's…..

Best Answer

For large $x$ $f$ will either be nonnegative or nonpositive. So assume WLOG $f(x)\ge 0$. Then it clearly must be monotone decreasing. Note that since $\int_0^\infty f(x)\,dx<\infty$ and $f\ge 0$ we have $\lim\limits_{t\to\infty}\int_{\frac{t}{2}}^{t} f(x)\,dx=0$. Then since $f$ is monotone decreasing we have $0\le \displaystyle \frac{tf(t)}{2}\le \int_{\frac{t}{2}}^{t} f(x)\,dx\to 0$ at $t\to\infty$, giving the result.

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