Upperbound for integral of a function times a cosine

Question

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.

Answer 1

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 $$ c_1\int_0^ag(x)\cos(x)dx \leq \int_0^af(x)g(x)\cos(x)dx \leq c_2\int_0^ag(x)\cos(x)dx. $$