Abel summability of integrals

Solution 1:

No further assumption is required if we understand the defining integral in Laplace transform as an improper one. Indeed, the following proposition holds.

Proposition. Assume $f$ is locally integrable on $(0, \infty)$ and the improper integral

$$\int_{0}^{\infty} f(x) \; dx := \lim_{p\to\infty} \int_{0}^{p} f(x) \; dx$$

exists. Then its Laplace transform

$$ A_{s} = \mathcal{L}f(s) := \int_{0}^{\infty} f(x)\,e^{-sx} \; dx = \lim_{p\to\infty} \int_{0}^{p} f(x)\,e^{-sx} \; dx $$

exists for $s \geq 0$ and

$$\lim_{s\downarrow 0}A_{s} = A_{0} = \int_{0}^{\infty} f(x) \; dx.$$

For the proof, let $h(x) = \int_{0}^{x} f(t) \; dt$. Then $h(x)$ is absolutely integrable, and is bounded since the limit $h(\infty) := \lim_{p\to\infty} h(p)$ exists. By integration by parts, we have

$$ \begin{align*} \int_{0}^{p} f(x)\,e^{-sx}\;dx &= \left[h(x)\,e^{-sx}\right]_{0}^{p} + s \int_{0}^{p} h(x) \, e^{-sx} \; dx \\ &= h(p)\,e^{-sp} + s \int_{0}^{p} h(x) \, e^{-sx} \; dx. \end{align*} $$

Here, note that

$$ |h(x) \, e^{-sx}| \leq \| h \|_{L^{\infty}} e^{-sx}. $$

Thus by Lebesgue's dominated convergence theorem, we have

$$\lim_{p\to\infty} \int_{0}^{p} f(x)\,e^{-sx}\;dx = s \int_{0}^{\infty} h(x) \, e^{-sx} \; dx. $$

This proves the existence of $A_{s}$. To prove that $A_{s} \to A_{0}$ as $s \downarrow 0$, we observe that

$$ F(0) = h(\infty) = s \int_{0}^{\infty} h(\infty) \, e^{-sx} \; dx. $$

Thus we have

$$ \begin{align*} |A_{s} - A_{0}| &\leq s \int_{0}^{\infty} |h(x) - h(\infty)| \, e^{-sx} \; dx \\ &= \int_{0}^{\infty} |h(u/s) - h(\infty)| \, e^{-u} \; du. \qquad(u = sx) \end{align*}$$

Since this integrand is bounded by the dominating function $2\|h\|_{L^{\infty}} \,e^{-u}$, we can apply Lebesgue's dominated convergence theorem again, yielding

$$\lim_{s\downarrow 0} |A_{s} - A_{0}| \leq \int_{0}^{\infty} \lim_{s\downarrow 0} |h(u/s) - h(\infty)| \, e^{-u} \; du = 0.$$

Therefore $A_{s} \to A_{0}$ as desired.

Of course, in the reference book the Laplace transform is defined as a Lebesgue integral sense. Even in this case, only a mild auxiliary condition, such as

$$f(x) = O(e^{\epsilon x}) \quad \text{for all} \ \epsilon > 0$$

or

$$\int_{0}^{x} |f(t)| \; dt = O(e^{\epsilon x}) \quad \text{for all} \ \epsilon > 0$$

guarantees the proposition, since we only need to show that the improper integral defining $A_{s}$ in fact converges in a Lebesgue sense. Actually, as you read that page, you will find that the book also requires some conditions on $f$ so that $A_{\epsilon}$ is defined for all $\epsilon > 0$.