# 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.

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

• Thanks for answering. I put general functions $f$ and $g$ because i thought maybe there is some result general and trivial that i don't see. I will edit my post with my specific functions. Sep 9 '14 at 16:40
• @coco, in case someone is interested in the general case you asked, I think it's good to let the old question remain. (Changing a question will also invalidate most answers, which can make the answerers feel that they worked for no reason.) You can make an addition if you want, but you can also ask a follow-up question. Sep 9 '14 at 16:45