Does the series $\sum_{n=0}^{\infty} \int_0^1 \cos(nt^2) dt$ converge or diverge?

Question

I know we can not write the primitive function of $\cos(nt^2)$ for any $n \in \mathbb{N}$, thus I could not calculate the term $ \int_0^1 \cos(nt^2) dt$, could someone help me? Thank you in advance.

Answer 1

Sustituting $u=\sqrt{n}t$, one has $$\int_0^1 \cos(nt^2)dt = \dfrac{1}{\sqrt{n}}\int_0^{\sqrt{n}} \cos(u^2) dt $$

Since $\displaystyle\int_0^{+\infty} \cos(u^2) du$ is well-known to be equal to $\dfrac{1}{2}\sqrt{\dfrac{\pi}{2}}$, one gets

$$\int_0^1 \cos(nt^2)dt \mathop{\sim}\limits_{+\infty} \dfrac{1}{2}\sqrt{\dfrac{\pi}{2n}}$$

so by comparison, the series diverges.

Answer 2

$$\int_0^1 \cos(nt^2)\mathrm{d}t=\sqrt{\frac{\pi}{2n}}C\left(\sqrt{\frac{2n}{\pi}}\right)\sim\frac{1}{2}\sqrt{\frac{\pi}{2n}}+\frac{\sin(n)}{2n}-\frac{\cos(n)}{4n^2}+O(n^{-2})$$ Where $C(z)$ is the Fresnel $C$ function.
$\sin(n)$ and $\cos(n)$ can be considered as constant $(s, c)$ since $-1\leq\sin(n),\cos(n)\leq 1$
In the series we have: $$\sum_{n=0}^{\infty}\int_0^1\cos(nt^2)\mathrm{d}t=1+\sum_{n=1}^{\infty}\int_0^1\cos(nt^2)\mathrm{d}t\sim1+\sum_{n=1}^{\infty}\left[\frac{1}{2}\sqrt{\frac{\pi}{2n}}+\frac{s}{2n}-\frac{c}{4n^2}\right]=\\ 1+\frac{\sqrt{\pi}}{2\sqrt{2}}\underbrace{\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}}_{\to \infty}+\frac{s}{2}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}}_{\to\infty}-\frac{c}{4}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^2}}_{\to\frac{\pi^2}{6}}\to\infty $$ I removed $n=0$ from the series because we have $n$ in the denominator.

Answer 3

$$S_p=1+\sum_{n=1}^{p} \int_0^1 \cos(nt^2)\, dt=1+\sqrt{\frac{\pi }{2}}\,\sum_{n=1}^{p} \frac 1 {\sqrt n} C\left(\sqrt{\frac{2n}{\pi }} \right)$$ $$S=1+\sum_{n=1}^{\infty} \int_0^1 \cos(nt^2)\, dt=1+\sqrt{\frac{\pi }{2}}\,\sum_{n=1}^{p} \frac 1 {\sqrt n} C\left(\sqrt{\frac{2n}{\pi }} \right)+$$ $$\sqrt{\frac{\pi }{2}}\,\sum_{n=p+1}^{\infty} \frac 1 {\sqrt n} C\left(\sqrt{\frac{2n}{\pi }} \right)$$

When $p$ is large (have a look at equation $(11)$ here) $$C\left(\sqrt{\frac{2n}{\pi }} \right)\simeq \frac{1}{2}+\frac{\sin (n)}{\sqrt{2 \pi n} }$$

$$\sqrt{\frac{\pi }{2}}\,\sum_{n=p+1}^{\infty} \frac 1 {\sqrt n} C\left(\sqrt{\frac{2n}{\pi }} \right)\simeq \frac 12 \sqrt{\frac{\pi }{2}}\color{red}{\sum_{n=p+1}^{\infty}{\frac 1 {\sqrt n} }}+\frac 12\sum_{n=p+1}^{\infty}\frac{\sin (n)}{n}$$

You could also compare $S_p$ to $$T_p=1+\sqrt{\frac{\pi }{2}}\,\int_{1}^{p} \frac 1 {\sqrt n} C\left(\sqrt{\frac{2n}{\pi }} \right)\,dn$$ $$T_p= \left(1+\sin (1)-\sqrt{2 \pi }\, C\left(\sqrt{\frac{2}{\pi }}\right)\right)+\left(\sqrt{2 p \pi } \, C\left(\sqrt{\frac{2p}{\pi }}\right)-\sin (p)\right)$$ which is an excellent approximation of $S_p$ (for example, $S_{1000}=40.1431$ while $T_{1000}=39.6654$).

Asymptotically $$T_p =\sqrt{\frac{p\,\pi }{2}}$$