itegrable, converge, improper integral

Question

$f(x)$ is a itegrable function on $(0,+\infty)$ and $\int_{0}^{+\infty}f(x)dx$ coverge. Prove that $$\lim_{t\to 0^+}{\int_{0}^{+\infty }e^{-tx}f(x)dx}=\int_{0}^{+\infty }f(x)dx$$

I got that problem when I learned about improper-integals with a parameter. $\int_{x\to 0^+}^{+\infty}e^{-tx}f(x)dx$ is uniformly converge but $f(x)$ is not continuous (if we had, it would be easier).

Surb · Answer 1 · 2016-09-20T13:29:44.033

1

BIG Hint

When $t>0$, $$|e^{-tx}f(x)|\leq |f(x)|.$$

Second way (without DCT)

$$|f(x)e^{-tx}-f(x)|=|f(x)||e^{-tx}-1|.$$ Using MVT, $$|e^{-tx}-1|\leq |t|\underset{t\to 0}{\longrightarrow }0,$$ and thus, the convergence is uniform (in $x$). Therefore, you can permute limit and integral.

edited Sep 20 '16 at 13:29

answered Sep 20 '16 at 07:21

Surb

55,662

So the left integral uniformly converge but $f(x)$ is just integrable – Noun Sep 20 '16 at 07:53
No ! Use DCT ! – Surb Sep 20 '16 at 08:44
Thank you :) but I haven't learned about that. Can you prove more easily? – Noun Sep 20 '16 at 10:07
@Noun: I edited my answer. – Surb Sep 20 '16 at 13:29
Can you add a link which is about permuting limit and integral, please? – Noun Sep 21 '16 at 10:42
https://en.wikipedia.org/wiki/Uniform_convergence – Surb Sep 21 '16 at 14:30
You mean $lim_{n\to+\infty}\int_{0}^{+\infty }f(x)(e^{-t_nx})dx=\int_{0}^{+\infty}f(x)dx$ with $t_n$ is a sequence comes to $0$, don't you? I can prove overall $b>0$ but for $+\infty $. – Noun Sep 22 '16 at 09:57

itegrable, converge, improper integral

1 Answers1