Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $f(x)$ is integrable on any $[0,a]$, such that $\lim_{x\to+\infty} f(x)=1$

Prove that: $$\lim_{t\to 0+}\int_0^{\infty}t\text{e}^{-tx}f(x)\text{d}x=1$$

share|cite|improve this question
What does that title mean? Might you replace it with something less obscure, more informative? – Gerry Myerson Feb 3 '13 at 9:46
@GerryMyerson : It is a question in Rudin's special function chapter...I dont see anything relevant to that chapter :( eventhough I know a proof – Ryan Feb 3 '13 at 9:57
Ryan, that doesn't answer my questions. Also, Rudin wrote several books. – Gerry Myerson Feb 3 '13 at 10:03
@GerryMyerson : His book on Analysis – Ryan Feb 3 '13 at 10:08
@Ryan: Rudin wrote several books about analysis... – Hans Lundmark Feb 3 '13 at 10:40

1 Answer 1

up vote 3 down vote accepted

The function $g:z\mapsto\sup\limits_{x\geqslant z}|f(x)-1|$ is finite everywhere and, for every $z$, $$ \left|\int_0^{+\infty}t\mathrm e^{-tx}f(x)\mathrm dx-1\right|\leqslant(1-\mathrm e^{-tz})\cdot g(0)+\mathrm e^{-tz}\cdot g(z)\leqslant t\cdot z\cdot g(0)+ g(z) $$ Thus, for every $z$, $$ \limsup\limits_{t\to0}\ \left|\int_0^{+\infty}t\mathrm e^{-tx}f(x)\mathrm dx-1\right|\leqslant g(z). $$ Since $g$ has limit $0$ at $+\infty$, the proof is complete.

share|cite|improve this answer
So, was that special function, or was it Weierstrass? – Gerry Myerson Feb 3 '13 at 11:08
@GerryMyerson Why the name of Weierstrass appears on this page eludes me. Sorry. – Did Feb 3 '13 at 13:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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