Upperbound for integral of a function times a cosine Let $f$ be a function such that $0<c_1\leqslant f(x)\leqslant c_2$ for all $x$, and $g$ be a positive function. Assuming that we know the integral $\int_0^\infty g(x)\cos{x}\,\mathrm{d}x$, is it possible to upper bound the integral
$$\int_0^af(x)g(x)\cos{x}\,\mathrm{d}x$$
where $a$ is large in terms of $\int_0^a g(x)\cos{x}\,\mathrm{d}x$? Then I would evaluate the remaining part of the integral.
 A: I assume $a>0$ and $0<c_1<c_2$.
If $a>\pi/2$, I think there is no easy way to bound your integral without additional assumptions.
Let $M\in\mathbb R$ be a number, which we shall adjust later.
If $a\geq\pi$, let $g=M\chi_{[0,\pi]}$, and otherwise let $g=M\chi_{[\pi-a,a]}$.
Then $\int_0^ag(x)\cos(x)dx=\int_0^\infty g(x)\cos(x)dx=0$.
Now let $f=c_1+(c_2-c_1)\chi_{[0,\pi/2]}$, so that
$$
\int_0^af(x)g(x)\cos(x)dx=(c_2-c_1)M\int_{\max\{0,\pi-a\}}^{\pi/2}\cos(x)dx.
$$
The last cosine integral is strictly positive.
Thus the integral you want to estimate can be made arbitrarily large by increasing $M$, although the control quantity remains zero.
You can also make the same phenomenon appear with smooth functions $f$ and $g$, but I chose piecewise constant functions to make the argument more transparent.
Similarly, I think there will be problems even for $a\leq\pi/2$ if $g$ is allowed to change sign.
But if $g$ is positive and $a\leq\pi/2$, there are bounds.
Since all functions are now positive, we have
$$
c_1\int_0^ag(x)\cos(x)dx
\leq
\int_0^af(x)g(x)\cos(x)dx
\leq
c_2\int_0^ag(x)\cos(x)dx.
$$
