I am trying to solve Rudin 8.11:

Suppose $f$ is Riemann-integrable on $[0,A]$ for all $A<\infty$, and $f(x) \rightarrow 1$ as $x \rightarrow \infty$. Prove that $$\lim_{t \rightarrow 0} \;\int_{0}^{\infty}t e^{-tx} f(x) dx =1.$$

This is easy if one assumes $f$ is differentiable (it is a special case of the Final Value Theorem for Laplace transforms), but I'm apparently not even allowed to assume continuity here. Anyone have any ideas?

  • 1
    $\begingroup$ The formula is not true for $f$ constant equal to $1$. In this case, the integral is $\frac{1}{t}$ for every $t>0$. You might want to correct that. Multiply the lhs by $t$. $\endgroup$ – Julien Apr 10 '13 at 5:27
  • $\begingroup$ @copper.hat It is true with an extra $t$ on the left. $\endgroup$ – Julien Apr 10 '13 at 5:28
  • $\begingroup$ ah yes that was a typo $\endgroup$ – mikefallopian Apr 10 '13 at 5:35

Assuming you left out the $t$, split the interval $[0,\infty)$ into $[0,A]$ and $[A,\infty)$ and choose $A$ so that $|f(x)-1|<\epsilon$ for all $x>A$. Write the integral as the sum of integrals over either interval. On the first interval, you can use dominated convergence theorem and the fact that $f$ is integrable on $[0,A]$. On the infinite interval, use the fact that $f$ is close to 1.

  • 2
    $\begingroup$ +1. On the first interval, the boundedness of $f$ is enough. $\endgroup$ – Did Apr 10 '13 at 5:31
  • $\begingroup$ this is great! thanks $\endgroup$ – mikefallopian Apr 10 '13 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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