I don't know if this conjecture is well know but let me try it :
Let $f(x)$ be a continuous , differentiable function on $[0;+\infty[ $ with $f(x)\geq 0 \quad\forall x\geq 0$ and such that :$$\int_{0}^{\infty}\frac{\sin(f(x))}{f(x)}dx=\frac{\pi}{2}$$ then the equation $f(x)=x$ have at least one solution on $[0;+\infty[ $.
I have tried to find counter-example like : $f(x)=x^n+y$ and it seems to work
$f(x)=x+\alpha e^{\beta}$ works too
I have not tried counter-example like : $f(x)=x+ \sin^2(x)$ where we have an infinity of solution .
If it works maybe we can use power series of $h(x)=\frac{\sin(x)}{x}$ and make the substitution $f(x)=x+g(x)$ but I don't know what to do next .
Another idea is to use Lipschitzian maps but it becomes really difficult for me.
My question :
First I want to know : Is there a counter-example ?
Any helps is greatly appreciated .
Thanks a lot .
Update:
Since :$$\int_{0}^{\infty}\frac{\sin(f(x))}{f(x)}-\frac{\sin(x)}{x}dx=0$$
We deduce that there is an $y\in [0;+\infty[$ such that :
$$\frac{\sin(f(y))}{f(y)}-\frac{\sin(y)}{y}=0$$
Or: $$\frac{\sin(f(y))}{f(y)}=\frac{\sin(y)}{y}$$
Wich have as solution :$f(y)=y$
It is right ?