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Prove $g(x) = \sum_{n=1}^{∞}\int_{0}^{\frac{x}{\sqrt{n}}} \sin t^2 dt$ is well-defined and differentialbe on $(0,∞)$

Well-defined:

To prove that $g(x)$ is indeed well-defined, it should suffice to take any $x_0\in(0,∞)$ and show that the series of integrals converge. Of course the necessary condition for convergence is met, as $$\int_{0}^{\frac{x_0}{\sqrt{n}}} \sin t^2 dt \rightarrow0 \quad \text{as} \quad n\rightarrow ∞.$$ For any $x_0\in(0,∞)$, thare exists $n_0$ s.t. the integral above will be strictly greater than $0$. So I could use a comparison test for series, but I have trouble finding $a_n$ to bound the integral from above: $$ 0 < \int_{0}^{\frac{x_0}{\sqrt{n}}} \sin t^2 dt < a_n \quad (\sum a_n \text{converges})$$

Differentiability:

Let $f_n(x) = \int_{0}^{\frac{x}{\sqrt{n}}} \sin t^2 dt$. Now if $\sum ||f'_n(x)||_∞$ converges, then I fould use the fact that $$\frac{d}{dx}\sum f_n = \sum \frac{d}{dx} f_n$$ But $$||f'_n(x)||_∞ = \sup_{x\in (0,∞)}\left(\frac{1}{\sqrt{n}}\cdot \sin(\frac{x^2}{n})\right) = \frac{1}{\sqrt{n}} $$, so $\sum ||f'_n(x)||_∞$ doesn't converge. I've ran out of ideas, I will appreciate any hint.

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Hint: For both parts the inequality $\sin y<y$ which is true for positive $y$ is useful. For small enough $y$ you even have $0<\sin y<y$. As for the convergence of the series of derivatives, it is a common trick to prove convergence on $\left(0,M\right)$ for every $M>0$.

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