# Prove that $\lim_{t \rightarrow 0} t \int_{0}^{\infty} e^{-tx} f(x) dx =1$

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?

• 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 $t$. Apr 10, 2013 at 5:27
• @copper.hat It is true with an extra $t$ on the left. Apr 10, 2013 at 5:28
• ah yes that was a typo Apr 10, 2013 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.
• +1. On the first interval, the boundedness of $f$ is enough.
Observe $$\int_0^{\infty}te^{-tx}dx=[-e^{-tx}]_0^{\infty}=1$$
so that $$t\int_0^{\infty}e^{-tx}f(x)dx-1=t\int_0^{\infty}e^{-tx}f(x)dx-\int_0^{\infty}te^{-tx}dx=\int_0^{\infty}te^{-tx}[f(x)-1]dx$$
By $$\lim_{x \to \infty}f(x)=1$$, there exists $$A>0$$ such that $$x>A \implies |f(x)-1|<\frac{\epsilon}{2}$$. We then have $$|t\int_0^{\infty}e^{-tx}f(x)dx-1|=|\int_0^A te^{-tx}[f(x)-1]dx+\int_A^{\infty} te^{-tx}[f(x)-1]dx|$$ $$\le t\int_0^A |f(x)-1|dx+\frac{\epsilon}{2}\int_A^{\infty} te^{-tx}dx=t\int_0^A |f(x)-1|dx+\frac{\epsilon}{2}e^{-tA}$$ Since $$\lim_{t \to 0} (t\int_0^A |f(x)-1|dx+\frac{\epsilon}{2}e^{-tA})=\frac{\epsilon}{2} < \epsilon$$ Then, there exsits $$\delta>0$$ such that $$0