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I am trying to show that, given a function $f:\mathbb{R}\rightarrow\mathbb{C}$ that is in the Schwartz space of functions with rapid decay at infinity, the following function is holomorphic for all $s\in\mathbb{C}$:

$$F(s) = \int_1^\infty \sum_{n=1}^\infty f(ny) y^{s-1}dy$$

So far I have tried changing variables $y \rightarrow \frac{y}{n}$ to obtain:

$$\int_1^\infty \sum_{n=1}^\infty f(ny) y^{s-1}dy = \int_1^\infty \sum_{n=1}^\infty \frac{1}{n^s}f(y) y^{s-1}dy = \zeta(s) \int_1^\infty f(y) y^{s-1}dy$$

This is where I am lost. I know the $\zeta(s)$ is holomorphic except at $s=1$, but I do not see how the integral goes to $0$ at $s=1$ to cancel out the pole, if that is indeed how I am suppose to solve the problem. For example, $f(x)=e^{-x^2}$ seems like a counterexample.

Edit: Following the comment below, I will try to show that $\sum_{n=1}^\infty f(ny)$ is in the Schwartz space. Consider the following, for any $N,M \in \mathbb{Z}^+$:

$$\lim_{y\rightarrow \infty}y^N\frac{d^M}{dy^M}\sum_{n=1}^\infty f(ny) = \sum_{n=1}^\infty \lim_{y\rightarrow \infty}y^N\frac{d^M}{dy^M} f(ny) = \sum_{n=1}^\infty 0 = 0$$

So, $g(y)=\sum_{n=1}^\infty f(ny)$ is in the Schwartz space.

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    I see there is an error in the bounds of integration. The lower bounds should be $\frac{1}{n}$, so perhaps I am not going down the correct path. – 2017-02-23
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    Show $\sum_{n=1}^\infty f(ny)$ is in the Schwarz space. – 2017-02-23
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    If $\sum_{n=1}^\infty f(ny)$ is in the Schwartz space, what does this tell me about the integral in question? – 2017-02-23
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    $F(y) = \sum_{n=1}^\infty f(ny)$ is Schwarz, then $F(y)y^n \to 0$ for all $n$ as $y \to \infty$. Therefore $\int_1^\infty |F(y)|y^{\sigma - 1} \,dy < \infty$ for all $\sigma \in \mathbb{R}$.Therefore if $\Omega = \{ s \in \mathbb{C} : \Re(s) \le \sigma \}$ then the integral $\int_1^k F(y)y^{s-1}\,dy$ converges uniformly on $\Omega$ as $k \to \infty$. Therefore holomorphic, and so on and so on. – 2017-02-23
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    @james.nixon Well $F'(s) = \int_1^\infty (\sum_n f(ny)) y^{s-1}\ln y\, dy$ so $F(s)$ is holomorphic – 2017-02-23

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