Upperbound for integral of a function times a cosine (specific case) Let $f$ be a function such that $f(x)=x^4\times\sum(x)$ where $\sum$ is a series satisfying $0<c_1\leqslant \sum(x)\leqslant c_2$ for all $x\in\left[0,a\right]$, where $a$ is large. Let $c$ be a constant. Is it possible to upper bound the following integral?
$$\int_0^a f(x)e^{-(x/c)^2}\cos x\,\mathrm{d}x$$
We can compute $\int_0^\infty x^4e^{-(x/c)^2}\cos x\,\mathrm{d}x$, but i don't know if/how we can upperbound the above integral in terms of $\int_0^a x^4 e^{-(x/c)^2}\cos x \, \mathrm{d}x$, then i would evaluate the remaining part of the integral.
Thanks in advance.
 A: This is related to my earlier answer to the general case.
Since $f$ is postivie, we have the simple estimate
$\DeclareMathOperator{\erf}{erf}$
\begin{eqnarray}
\left|\int_0^a f(x)e^{-(x/c)^2}\cos x\,\mathrm{d}x\right|
&\leq&
\int_0^a f(x)e^{-(x/c)^2}\,\mathrm{d}x
\\&\leq&
\int_0^a c_2x^4e^{-(x/c)^2}\,\mathrm{d}x
\\&\leq&
c_2a^4\int_0^a e^{-(x/c)^2}\,\mathrm{d}x
\\&=&
c_2a^4\frac12c\sqrt{\pi}\erf(a/c).
\end{eqnarray}
This is of course coarse in the sense that it doesn't make any use of the oscillation of the cosine.
For a better estimate, you need some kind of a continuity estimate (like Lipschitz) for $\Sigma$.
Let us study the function $g(x)=x^4e^{-(x/c)^2}$.
It reaches is maximum $4c^4e^{-2}$ at $x=\sqrt{2}c$.
It seems to be quite sharply peaked at its maximum, so the integral mainly comes from those values of $x$.
You can estimate $g$ for $x\gg c$ to see that the tail is very small (there is such an $A$ that $x^4\leq e^{Ax}$ for $x\geq c$, for example).
With the estimate for $\Sigma$ you could then say how little $f(x)e^{-(x/c)^2}$ changes over a period of the cosine.
Then you will get some cancellation due to the oscillation of the cosine, and you can bound your integral in terms of an estimate for the tails of $g$ and the continuity of $\Sigma$.
Working out the details is a bit laborious and requires more knowledge of $\Sigma$, so I will not try it here.
I'm not sure if this idea gives a significantly better estimate than the simple calculation, though.
