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


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: $$ 0 < \int_{0}^{\frac{x_0}{\sqrt{n}}} \sin t^2 dt < a_n \quad (\sum a_n \text{converges})$$


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.


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 $M>0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.