How to prove $ I_t=\int_{0}^{\infty}g(u)\,\cos\bigl(\frac{\pi\,u}{2t }\bigr)\,du$ is positive?

Question

We have the following integral $$I_t=\int_{0}^{\infty}g(u)\,\cos\left(\frac{\pi\,u}{2t }\right)\,du\,\,\,\text{with}\,\,\,\,t>0$$

where $g(u)$ is continuous on $(0,\infty)$, $g(0)=1$ and $g(u)$ is decreasing on $[t,\infty)$. This implies that $g(u)$ attains it maximum on $(0,t)$, the first branch where $\displaystyle \cos\left(\frac{\pi\,u}{2t }\right)$ is positive.

I wonder how can use this to prove that $I_t$ is positive ($I_t>0$).

Answer 1

Let $f(u)=g(tu)$. By a change of variable, $$\begin{eqnarray*} \frac{1}{t}\,I_t=\int_{0}^{+\infty}f(x)\cos\left(\frac{\pi}{2}x\right)\,dx &=& \sum_{n\geq 0}\int_{2n}^{2n+2}f(x)\cos\left(\frac{\pi}{2}x\right)\,dx\\&=&\sum_{n\geq 0}(-1)^n\int_{0}^{2}f(x+2n)\cos\left(\frac{\pi}{2}x\right)\,dx\tag{1}\end{eqnarray*}$$ so, if we set $A_n=\int_{0}^{2}f(x+2n)\cos\left(\frac{\pi}{2}x\right)\,dx$ we have: $$ \frac{1}{t}\,I_t = \sum_{n\geq 0}(-1)^n A_n \tag{2}$$ as well as: $$ \forall n\geq 1,\qquad 0<A_{n+1}<A_{n},\tag{3} $$ since $\cos\left(\frac{\pi}{2}x\right)$ is positive on $(0,1)$ and $f(x)$ is decreasing on $[1,+\infty)$.

If you manage to prove that $A_0>0$, $(2)$ and $(3)$ prove your claim.

Answer 2

But if

$$\frac{1}{t}I_t=A_0-A_1+A_2-A_3+A_4+\ldots$$

and $A_2<A_1$, $A_4<A_3$...

although I'll prove that $A_0>0$, this not assurance that $I_t>0$ because maybe that $-A_1+A_2-A_3+A_4+\ldots$ (which is a negative number) will be bigger than $A_0$.

I think there is something that I don't understand...