Is this sequence bounded ? (An open problem between my schoolmates !) 
Let $f$ be a smooth function (say $\mathcal{C}^{\infty}$) in its two real variables ($t$ and $T$). I consider the following sequence defined by
  $$A_n:=\lim_{T \to \infty} \int_{0}^{1} e^{-n t} f(t,T)dt \quad \quad (n\geq 1)$$
  where we suppose that this limit exists and is finite for all integer $n$.
  I would like to conclude that the sequence $(A_n)$ is bounded. Is that true ?

This is not an obvious problem : assuming certain conditions on $f$ then the exercise becomes classic. However, very little can be said about $(A_n)$ in the general case. In fact I deeply think that the answer is no and that we can find a $f$ such that $A_n$ tends to infinity. 
If one takes $f(t,T)=2\sin(tT)/\pi t$, then $(A_n)$ is the constant sequence equal to $1$. This shows that $f$ does not necessarily tend to $0$.
Thanks for your help !
 A: The sequence $\{A_n\}$ need not to be bounded. To see this, one could for example as $f(t,T)$ choose something that approximates a derivative of a delta distribution as $T\to+\infty$. I wish to give credits to my colleague Tomas Persson who came up with that idea.
I will give such an approximating example. My example is non-smooth, but that is just to make the calculations more transparent.
Let 
$$
g(t,T)=
\begin{cases}
\frac{T}{2} & |t|\leq\frac{1}{T}\\
0 & |t|>\frac{1}{T}.
\end{cases}
$$
This is an approximation of the delta distribution as $T\to+\infty$. We then let $f$ be the following difference quotient:
$$
f(t,T)=\frac{g(t-1/T,T)-g(t-2/T,T)}{1/T}
$$
It is then a simple matter to calculate the integral
$$
\int_0^1 e^{-nt}f(t,T)\,dt=\frac{T^2}{2n}\Bigl(1+e^{-3n/T}-e^{-2n/T}-e^{-n/T}\Bigr)
$$
Hence,
$$
A_n=\lim_{T\to+\infty}\int_0^1 e^{-nt}f(t,T)\,dt = n,
$$
which of course is unbounded.
Update
Let me, for completeness, add a smooth function $f$ that also gives $A_n=n$:
$$
f(t,T)=(T^2-T^3t)e^{-Tt}.
$$
The argument is the same, it approximates a derivative of the delta distribution.
A: $0 \lt \mathrm e^{-nt} \le 1 \quad \forall x \in [0,1] \\ \implies  A_n \lt \lim_{T \to \infty} \int_0^1 f(t,T) \,\mathrm d t$
so it's bounded above if the limit exists. 
A similar result gives a lower bound if the limit is negative.
