Can I imply absolute integrability from integration with exponential function?

Question

I have a piecewise continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ and $\varepsilon>0$ such that $$ \int_0^\infty e^{\varepsilon t}f(t) dt < \infty $$ holds.

Can I somehow imply absolute integrability of $f$, or that $f$ goes to $0$ (or that for all $\varepsilon_2>0$ the set $A:=\{x\in\mathbb{R}:|f(x)|>\varepsilon_2\}$ has zero measure)? Or is it possible, that the function $f$ oscillates infinitely? (my guess would be, that it should be absolute integrable, but I do not have an idea how to show it)

I got the existence of the integral, because I have, that the Laplace transform of $f$ is defined and analytical on $\mathbb{C}_{Re\geq -\varepsilon}$ for some $\varepsilon > 0$.

Answer 1

I have left the trials and errors in a second part but the answer is that $f$ is not necessarily (absolutely) integrable, nor does it have to converge to $0$ at $+\infty$:

Based on the function you gave in comment: let us consider the function $f(t):=\sin(e^{2\epsilon t})$ which is not integrable as it does not even converge to $0$. Nevertheless, one checks by integration by parts that the integral with the exponential converges $$\begin{aligned} \int_0^{+\infty} e^{\epsilon t} f(t) dt &= \int_0^{+\infty} e^{-\epsilon t} \frac{d}{dt} \left(-\frac{\cos (e^{2\epsilon t})}{2\epsilon}\right) dt \\ &=\left[ -e^{-\epsilon t} \frac{\cos (e^{2\epsilon t})}{2\epsilon}\right]^{+\infty}_0 - \int_0^{+\infty} e^{-\epsilon t} \frac{\cos (e^{2\epsilon t})}{2} dt \end{aligned}$$ Both term are finite, the second being absolutely convergent (dominate $\cos$ by $1$).

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• The usual construction of a continuous function with finite integral but which does not converge to $0$ at infinity also works: sum bump functions with constant height, but sufficiently fast decreasing width, moving to the right. Let us take $$b:\left\lbrace \begin{aligned} \ [-1,1] &\longrightarrow \mathbb{R},\\ x &\longmapsto \begin{cases} 1+x & \text{if} x\leq 0\\ 1-x & \text{otherwise}\end{cases} \end{aligned}\right.$$ as the "bump" function (although, the convention is usually that $b\in \mathcal{C}^{\infty}_c$, but it's just for simplicity, it does not change anything). Define $b_n(x)= b\big(n^2e^{n\epsilon}(x-n)\big)$ which is a "bump" centered at $n+1$, with height $1$. When multiplied by $e^{\epsilon x}$ and integrated: $$ \int_{\mathbb{R}} e^{\epsilon x} b_n(x) dx \leq \int_{\mathbb{R}} e^{(n+1)\epsilon } b(y) \frac{dy}{n^2e^{n\epsilon}} = \frac{e^{\epsilon}}{n^2} $$ which is the general term of a convergent series.

• The following is actually excluded by the assumption $f$ piecewise continuous but if $f$ is not supposed to be continuous at $0$ (or at a "junction" between the "pieces" where $f$ is continuous) then $\int_0^{M} \frac{sin(1/x)}{x}\, dx< \infty $ (Wolfram Alpha computation), but I suspect the integrand not to be integrable (still to be checked).

• An attempt that was doomed to fail: For $t$ large enough, $f$ will be continuous (normally a function is piecewise continuous if there are finitely many pieces, and it is continuous up to the boundary of each) and by definition $\displaystyle \int_M^{+\infty}e^{\epsilon t} f(t) dt := \lim_{R\to\infty} \int_M^{R}e^{\epsilon t} f(t) dt$. Let us now use the following "trick": consider the r.h.s. integral as a function of $R$, it has a limit at $+\infty$ so for any increasing sequence $(R_n)_{n\in \mathbb{N}},\ \left(\int_M^{R_n}e^{\epsilon t} f(t) dt\right)_{n\in \mathbb{N}}$ is Cauchy, or directly, without introducing a sequence $(R_n)_{n\in \mathbb{N}}$ $$\forall \varepsilon > 0,\ \exists R_0 > 0,\ \forall\ R_1, R_2 > R_0, \ \left\lvert \int_M^{R_2} e^{\epsilon t} f(t) dt - \int_M^{R_1}e^{\epsilon t} f(t) dt\right\rvert = \left\lvert \int_{R_1}^{R_2}e^{\epsilon t} f(t) dt \right\rvert < \varepsilon \tag{A}\label{A}$$ (assuming without lost of generality that $R_1 < R_2$). Let us now consider the open subset $f^{-1}\big(]-\infty,0[\big) \cap\ ]R_0,+\infty[ $: as an open subset of $\mathbb{R}$ it is at most a countable union of disjoint intervals, so that $]R_0,+\infty[$ is cut into a countable number of intervals, closed ones on which $f\geq 0$ and open ones on which $f<0$. (subtility on the boundary $]R_0,...$, closed or open in the subset topology.) On each of these pieces $[R_n,R_{n+1}]$ (closed or open) where $f$ has a constant sign, one has $$ e^{\epsilon R_n} \left\lvert \int_{R_n}^{R_{n+1}} f(t) dt \right\rvert \leq \left\lvert \int_{R_n}^{R_{n+1}}e^{\epsilon t} f(t) dt \right\rvert < \varepsilon \quad \Longrightarrow\quad \left\lvert \int_{R_n}^{R_{n+1}} f(t) dt \right\rvert < \varepsilon e^{-\epsilon R_n} \tag{B}\label{B}$$ We know that $\displaystyle \int_{R_0}^{+\infty}e^{\epsilon t} f(t) dt = \pm \sum_{n\in \mathbb{N}} (-1)^n \left\lvert \int_{R_n}^{R_{n+1}}e^{\epsilon t} f(t) dt \right\rvert$ converges and we want to show that $\displaystyle \int_{R_0}^{+\infty} \lvert f(t) \rvert dt = \sum_{n\in \mathbb{N}} \left\lvert\int_{R_n}^{R_{n+1}} f(t) dt \right\rvert$ converges. After a failed attempt with "summation by parts" a.k.a. Abel transformation (e.g. Principles of real Analysis (1976), W. Rudin, Thm 3.42 p.70), I think we should examine $\sum_{n\in \mathbb{N}} e^{-\epsilon R_n}$.

• In your question, you probably want to intersect what you defined as $A$ by something like $[R,+\infty[$ and look at the behavior when $R\to\infty$;

• One should check if there are theorems about Laplace transform analogous to the Paley-Wiener one for Fourier transform as you talked about analyticity.)