Throughout, let $f$ be a Lebesgue measurable function (or continuous if you wish, but this is probably no easier). (Questions with distributions etc. are possible also but I want to keep things simple here).
FINAL CLARIFICATION/REWRITE!!
Thanks to all who have commented so far. I will need some more time to digest it properly. The original forms of the questions are at the end; here I have rewritten the questions, hopefully more clearly; sorry for my poor explanation before!
Definition $\mathcal{M}_a$, the space of absolutely null moment functions, is the set of all measurable $f$ satisfying
$$
\int_0^\infty t^n |f(t)| dt < \infty,
\quad \int_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots.
$$
$\mathcal{M}$, the space of null moment functions, is the set of all measurable $f$ satisfying
$$
\int_0^T |f(t)| dt < \infty \quad \forall \, T>0,
\qquad
\lim_{R \to \infty} \int_0^R t^n f(t) dt = 0, \quad \forall \, n=0,1,2,\ldots
$$
Thus, trivially $0 \in \mathcal{M}_a \subseteq \mathcal{M}$, but $\mathcal{M}_a$ contains many other non-trivial functions. It seems certain that $\mathcal{M}_a \ne \mathcal{M}$ (I would be amazed if the spaces were equal), although constructing an explicit example seems tricky.
Definition Given a function $\psi \geq 0$, let $G(\psi)$ be the set of all $f$ such that $|f| \leq \psi$.
(Of course we're identifying functions equal a.e., so really we should consider equivalence classes etc. just as for $L^p$ spaces).
Rephrased Question 1 Find general simple necessary and/or sufficient conditions on $\psi$ with the property
$$
G(\psi) \cap \mathcal{M}_a = \{ 0 \}.
$$
Rephrased Question 2 The same as Question 1, but with $G(\psi) \cap \mathcal{M}$ instead.
Or, if $G(\psi)$ is not the appropriate space for these problems, consider $\int_T^{2T} |f| dt \leq g(T) $ or $\int_n^{n+1} |f| dt \leq A_n$ or something similar instead. Finding the correct kind of restrictions on $f$ is part of the problem.
Thus, $G(\psi) \cap \mathcal{M}_a = \{ 0 \}$ for $\psi(t) = \exp(-\delta t)$, by the discussion below; and also for any compactly supported $\psi \in L^1$.
But $G(\psi) \cap \mathcal{M}_a \ne \{ 0 \}$ for $\psi(t) = \exp(-t^{1/4}) |\sin(t^{1/4})|$.
Original QUESTION 1
If
$$
\int_0^\infty t^n |f(t)| dt < \infty,
\quad \int_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots,
$$
when must $f \equiv 0$ almost everywhere?
EDIT: CLARIFICATION: this is really about classes of functions, expressed in terms of a growth/decay rate function $\phi$, which give unique solutions to the moment problem. I am NOT asking how to solve the moment problem itself!
If possible, find necessary and sufficient conditions on functions $\phi \searrow 0$ with the property that
$$
\int_R^\infty |f(t)| dt \leq \phi(R) \qquad \Rightarrow \qquad f \equiv 0.
$$
So, $\phi(R) = \int_R^\infty \exp(-t^{1/4}) |\sin(t^{1/4})| dt$ is not enough (by Example 2 below); nor is $\phi(R) = \int_R^\infty |f(t)| dt$ with f given by "coudy" in his/her answer below.
But $\phi = \chi_{[0,b]}$ is enough (by Example 1 below); moreover $\phi(R) = \exp(-\delta R)$ would be enough for any fixed $\delta > 0$, by my discussion of Example 1 below, because the relevant Laplace transform $F = \mathcal{L}f$ is analytic on the half-plane $\{ \mathrm{Re}(z) > -\delta \}$. So we want to know about the gap between $\exp(-\delta R)$ and functions like that given by "coudy" below.
FURTHER CLARIFICATION: by analogy, maybe an example from PDEs will explain better (I'm not saying this is related to my problem; I'm saying that this is the same kind of result as what I want):
Definition A null temperature function is a continuous function $u = u(x,t) : \mathbb{R} \times [0, \infty) \to \mathbb{R}$ such that the heat equation is satisfied, i.e. $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ in $\{ t>0 \}$, and $u(x, 0) = 0$ for all $x$.
Theorem There exists a null temperature function satisfying $|u(x,t)| \leq \exp(A/t)$ with $A>0$, such that $u(x,t) \not\equiv 0$ for all $t>0$.
Theorem Let $u$ be a null temperature function satisfying $|u(x,t)| \leq A \exp(B t^{-\delta})$, for some $A,B>0$ and $\delta<1$. Then $u \equiv 0$.
So here the "critical" growth rate for null temperature functions is roughly $\exp(A/t)$. I am looking for a similar thing with "null moment functions".
Note that this is a totally different problem to: given $v$, find some $u$ satisfying the heat equation such that $u(x,0)=v(x)$.
Original QUESTION 2
If instead we have only $\int_0^R |f(t)| dt < \infty$ for each $R>0$, and
$$
\lim_{R \to \infty} \int_0^R t^n f(t) dt = 0, \qquad n=0,1,2,\ldots
$$
when must $f \equiv 0$ almost everywhere? (I have very little idea about this).
I think these questions are clearly very natural, interesting, and important, but Googling etc. didn't work well (I tried "vanishing moments" and other phrases, but there's just too much stuff out there). Standard known examples/methods follow.
Example 1: if $f$ is compactly supported on $[a,b]$, say, then $f \equiv 0$ a.e. because polynomials are dense in $C[a,b]$.
Example 2: by taking imaginary parts of $\int_0^\infty t^{4n+3} \exp(-(1+i)t) dt \in \mathbb{R}$, we get
$$
\int_0^\infty t^{4n+3} e^{-t} \sin t \, dt = 0,
$$
and so by substituting $t = x^{1/4}$,
$$
\int_0^\infty x^n \exp(-x^{1/4}) \sin(x^{1/4}) dx = 0, \qquad n=0,1,2,\ldots
$$
Alternative method for Example 1: consider the Laplace transform $F(z) = \int_0^\infty e^{-zt} f(t) dt$. In Example 1, $F$ is an entire function such that $F^{(n)}(0) = 0$ for all $n$, so $F \equiv 0$ and thus $f \equiv 0$ a.e. as required.
So, any condition on $f$ forcing $F$ to be analytic on some disc with centre $0$ is enough; but can we do better?
In Example 2, $f \in L^1(0,\infty)$ and so $F$ is bounded and analytic on $\{ \mathrm{Re}(z) > 0 \}$, and continuous on the boundary, with $\lim_{z \to 0} F^{(n)}(z) = 0$ for all $n$. But this is still not enough to force $F \equiv 0$.
Best Answer
The answer to the first question is no. The following example is standard in probability theory, see e.g. Billingsley "probability and measure", example 30.2.
$$f(x) = {1\over \sqrt{2\pi}\ x}\ e^{-{(\ln x)^2\over 2}}\ \sin(2\pi \ln x) \ {\bf 1}_{[0,\infty[}(x)$$
You can check that all the moments are zero using the change of variable $\ln x = s+k$.
This example is used to show that a probability measure is not always determined by its moments.