Can $A(s)=\int_{0}^{\infty}f(x,s)\space dx $ have any zeros if $f(x,s)≠0$ for all complex $s$? $f(x,s)$ is a holomorphic function. Intuitively I would say $A(s)$ can't have any zeroes because the "area" between the function and the x-axis isn't zero since it never crosses zero, its bounds are not equal and $f(x,s)≠0$. The issue is that I can't really prove this and therefore I can't know for sure if I'm missing something. In short, what I'm wondering is the following: Does there exist an $s$ such that $A(s)=0$, and if not, could anyone provide a proof?
Thank you.
This is not really well-formulated as one can easily do the following: note that $\int_{-\infty}^{\infty}ye^{-y^2}dy=0$ and the integral is absolutely convergent, while the function inside has only one finite zero at $y=0$.
Change variables $y=\ln x, y=\ln x -1$ and get absolute convergent integrals of functions $h,g$ on $[0,\infty)$, where $h(0)=h(1)=0, g(0)=g(e)=0, \int_{0}^{\infty}h(x)dx=\int_{0}^{\infty}g(x)dx=0$ and $h,g$ have no other zeroes. Then shift variables for $h$ say, so for example we now take $h_1(x)=h(x-1), x \ge 1$,while we take say $h_1(x)=(x-1)(x-\frac{1}{3}), 0 \le x \le 1$ with integral zero on $[0,1]$ too (and we can smooth it more at $1$ if we want)
Now $h_1(x)+ig(x)$ has no zero when $x \ge 0$ and then if one takes $f(x,s)=e^s(h_1(x)+ig(x))$, $f(x,s)$ is holomorphic in $s$ and nonzero anywhere and $A(s)=0$.