Prove $\int_{0}^{1}\!f(x)\cos(k\pi x)dx>0$ I have a positive continuous function $f(x)$ that is strictly monotonically decreasing on the interval $(0,1)$. That is, $f(x)>0$ and $f'(x)<0$ for $x\in(0,1)$. I want to show that the integral $$I = \int_{0}^{1}\! f(x)\cos(k\pi x)dx$$
is always positive, for all $k = 0,1,2,...$
Edit:
As pointed out in the comments, it's not possible to show this for general $f(x)$, but I have numerical eveidence to suggest that this is true for $$f(x) = {\frac {\left( 3\,{\epsilon}^{2}-\cos \left( \pi\,x
  \right) +1 \right) }{ \left( 2\,{\epsilon}^{2}
  -\cos \left( \pi\,(x) \right) +1 \right) ^{3/2}}}$$
where $0<\epsilon<1$.
 A: $I$ can be negative. For instance, with $k=2$, take the function $g(x)=1$ if $x\le\frac34$, and $g(x)=0$ if $x>\frac34$. Then
$$\int_0^1 g(x)\cos(2\pi x)dx=\int_0^\frac34\cos(2\pi x)dx=-\frac{1}{2\pi}$$
Of course, $g$ doesn't satisfy your conditions; but you can easily construct a continuous, differentiable, strictly decreasing $f$ that does satisfy them, and is sufficiently close to $g$ that the integral is still negative.
You can use this idea to construct a counter-example for any integer $k\ge 2$ $-$ just replace $\frac34$ with $\frac{3}{2k}$.
A: For $k=1$:
$$I=\int_{0}^{1} f(x) \cos \pi x ~ dx ~~~(1)$$
Use: $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$, then
$$I=\int_{0}^{1} -f(1-x) \cos \pi x ~dx~~~~~(2)$$
Add (1) and (2), to gt
$$2I=\int_{0}^{1}[f(x)-f(1-x)] \cos \pi x ~dx~~~~(3)$$
$x <1-x, x \in (0,1/2)$, as $f(x)$ is decreasing function, we get
$$f(x)>f(1-x), x \in (0,1/2), ~f(x)< f)1-x), x \in (1/2,1)~~~~~~(4)$$
Then $$2I=\int_{0}^{1/2} [f(x)-f(1-x)] \cos \pi x~dx+\int_{1/2}^{1} [f(x)-f(1-x)] \cos \pi x ~dx~~~~(5)$$
$\cos \pi x <0, ~in~ (0,1/2)$ and $\cos \pi x >0 ~ in~ (1/2,1)$.
So it follows that $$2I>0$$
