If $\lim_{t\to +\infty} \int_{0}^{\pi} f(x)e^{xt} \, dx=0$ then $f=0$? 
Question, Let $f \in L^2(0, \pi)$ such that $\lim_{t\to +\infty} \int_{0}^{\pi} f(x)e^{xt} \, dx=0$. Is it necessary that $f$ is identically zero almost everywhere?

If true, how to use the fact $\lim_{t\to +\infty} e^{xt} = +\infty, \hspace{0.5em} \forall x>0$ ?
If $f$ is positive, we can use that $\forall x>0, \hspace{0.5em} e^{tx}\geq t$ for large $t$ and we deduce
$$ \int_{0}^{\pi} f(x)e^{xt} \, dx
\geq t\int_{0}^{\pi} f(x) \, dx
\to +\infty$$
if $\int_{0}^{\pi} f(x) \, dx  \neq 0$. But we know $\int_{0}^{\pi} f(x) \, dx = 0 \iff f = 0 \hspace{0.5em} \text{a.e.} $
 A: Remark
Who knows the theorems about the rate that a Laplace transform goes to $0$?  Is that relevant?
In this problem, write $g(x) = f(\pi-x)$ for $x \in [0,\pi]$ and $g(x) = 0$ elsewhere. Then change variable $y=\pi - x$:
$$
\int_0^\pi f(x)e^{tx}\;dx = e^{t\pi}\int_0^\infty g(y) e^{-ty}\;dy
=e^{t\pi}\mathcal L\{g\}(t)
$$
where $\mathcal L\{g\}$ is the Laplace transform of $g$.
Now there are theorems (I don't remember them offhand) relating (a) how fast $\mathcal L\{g\}(t)$ goes to $0$ as $t \to +\infty$ to (b) regularity properties of $g$.  In our case, we have $\mathcal L\{g\}(t)$ goes to zero exponentially fast: faster than $e^{-t\pi}$.  What does that say about $g$ [and therefore about $f$] ?
A: $\newcommand{\d}{\,\mathrm{d}}$EDIT 2: There is one final issue with this post that needs to be resolved. My estimate $J(t)\ge Ce^{-T\eta}e^{t\eta}$ is very intuitive, and matches all experiments, but I do not yet know how to rigorously demonstrate it.
A partial response. EDIT: Now complete, thanks to a comment from the OP.
Suppose $f\neq0\in L^2(0,\pi)$, that is, $\|f\|>0$. By Holder's inequality, for any $0<\eta<\pi$ and any $t>0$ we know: $$\left|\int_0^\eta f(x)e^{xt}\d x\right|\le\|f\|\cdot\frac{1}{2t}\sqrt{e^{2t\eta}-1}$$Fix $\eta$ such that $f$ is not identically zero Let $\varepsilon>0$ be arbitary. We know there is $T>0$, such that $t\ge T$ implies: $$0\le\left|\underset{I(t)}{\underbrace{\left|\int_0^\eta f(x)e^{xt}\d x\right|}}-\underset{J(t)}{\underbrace{\left|\int_\eta^\pi f(x)e^{xt}\d x\right|}}\right|\le\left|\int_0^\pi f(x)e^{xt}\d x\right|<\varepsilon$$Let $C:=\int_\eta^\pi f(x)e^{xT}\d x$. For $t>T$ I know $J(t)\ge Ce^{-T\eta}e^{t\eta}=:Ke^{t\eta}$. Suppose $\eta$ can be chosen so that $J(t)\neq0$ for all large $t$. Then we know: $$\frac{I(t)}{J(t)}\to1$$As $J(t)\to\infty$. But: $$0\le\frac{I(t)}{J(t)}\le\frac{\|f\|}{2K}\frac{\sqrt{e^{2t\eta}-1}}{t\cdot e^{t\eta}}\to0$$Which is a contradiction. So it must be that $J(t)$ is eventually zero for any choice of $\eta$ (if it is nonzero just once, at $t_0$, then the bound $J(t)\ge Ke^{-t\eta}$ shows it is nonzero for all $t>t_0$, leading to a contradiction). That is, $\int_\eta^\pi f(x)e^{xt}\d x=0$ for all large $t$.
EDIT: Thanks to the OP, we have a way to finish off my argument.
Fix $0<\eta<\pi$: $\hat f:\Bbb C\to\Bbb C$ be the function $t\mapsto\int_\eta^\pi f(x)e^{xt}\d x$. This function is complex-analytic. Then $\hat f(t)\equiv0$ for all large $t\in\Bbb R$ means, by the identity principle, $\hat f\equiv0$ for all $t\in\Bbb C$. That is true for any $\eta$, so we may take it as true for $\eta=0$. In particular this implies all of $f$'s Fourier coefficients are zero. By the Parseval identity, $\|f\|_2=0$, so $f=0$ almost everywhere, as desired.
Old continuation/conjecture (if any reader can see how to finish off that line of argument, I'd really appreciate a comment!):

Let $A\sqcup B$ be the nonnegative/negative Hahn decomposition for $f$. We then know: $$\int_A|f(x)|e^{xt}\d x=\int_B|f(x)|e^{xt}\d x$$For all large $t$, and neither side is zero.
Since these sets are disjoint, I conjecture this is impossible. For some majority of $x$ values in, say, $B$, will be larger than those in $A$, and increasing $t$ will compound this difference so that it is impossible for the integrals to remain equal. If $A,B$ were intervals then this would be very easy to show, for instance. While I know we can approximate $A,B$ closely in measure by unions of intervals, I can't see a way to make that work at the moment. I instinctively feel that the above equality can hold for at most one value of $t$, if $|f|$ is not almost everywhere zero.

A: I give a very detailed answer to my question, a little modified, to be more general. I had suffered to understand all the details, that's why I am giving this answer to whoever needs it

Question, Let $f \in L^1(\alpha, \beta)$ , $\beta>0$ and $F(x)= \int_{\alpha}^{\beta} f(t)e^{xt} \, dt$ such that $\displaystyle \lim_{x\to +\infty}F(x)=0$. Is it necessary that $f$ is identically zero almost everywhere?


First show that F is analytic on $\mathbb C$
Let $z=a+ib$ it suffices to prove that $\frac{\partial F} {\partial b} = i\frac{\partial F} {\partial a}$. We can use to see it, differentiation under the integral sign because $\forall t\in(\alpha,\beta), \quad |tf(t)e^{t(a+ib)}|\leq M|f(t)|\in L^1(\alpha,\beta)$


If we show that F est bounded on $\mathbb C$, then by Liouville's theorem F must be constant on $\mathbb C$ and this constant is given by $\lim_{x\to +\infty} F(x)=0$. Now $F(z)=0,\, \forall z\in\mathbb C$. Extend f by 0 outside $[\alpha,\beta]$,  we have  $\hat f(x)=F(-ix)=0,\, \forall x\in \mathbb R $. By injectivity of the Fourier transform on $L^1$, we conclude that $f$ is identically zero almost everywhere.


Now, i show that that F est bounded on $\mathbb C$ by using Phragmén-Lindelöf Principle  ( I have adapted the proof drawing inspiration from  the link given by @dezdichado and answer givin by @Pavel Gubkin  )


First i show that F is an exponential type, put $z=a+ib$
$|F(z)|\leq \int_{\alpha}^{\beta} |f(t)|e^{at}\ \, dt$ . Note that
$\forall t\in[\alpha,\beta], \quad  |t|\leq c:=max(|\alpha|,|\beta|)$ and $e^{ta}\leq e^{c|a|} \leq e^{c|z|}$  thus $|F(z)|\leq ||f||_{L^1} e^{c|z|}, \forall z\in \mathbb C$


To conclude with Phragmén-Lindelöf Principle , it suffices to show that F is bounded on both axes.
On the real axis: it's clear that F est bounded for x<0 and also for  $x\geq 0$ , because   F  is continuous and $F(x)\to 0$ if $x\to +\infty$ . On the imaginary axis  $|F(ix)|\leq \int_{\alpha}^{\beta} |f(t)e^{itx}|\ \, dt=\int_{\alpha}^{\beta} |f(t)|\ \, dt=||f||_{L^1}$

A: It seems that $f$ is necessarily zero.
Assume the converse and let $\operatorname{ess\,supp}f = [a,b]\subset[-\pi, \pi]$. Then the Fourier transform $F(z) = \int_a^b f(x) e^{izx}\,dx$ of $f$ is a entire function from the Paley-Wiener space of entire functions. Moreover, exponential type of $F$ equals $\max(|a|, |b|)$.
We have $F(it) = \int_a^b f(x)e^{-tx}\,dx$. Since $f\in L^2$ we have $\displaystyle\lim_{t\to+\infty}F(it) = 0$. On the other hand, because of our assumption we also have $\displaystyle\lim_{t\to-\infty}F(it) = 0$. Exponential type of $F$ can be calculated by
$$
\text{type}(F) = \limsup_{y\to+\infty}\frac{\log\max(|F(iy)|, |F(-iy)|)}{y},
$$
which is non-positive for our $F$. This is a contradiction.
