Showing that $\int_0^{\infty}\sin t^2\,dt$ converges as an improper Riemann integral.

Question

I want to show that $\int_0^{\infty}\sin t^2\,dt$ exists.

I know that we define the above integral as $\lim_{b\to \infty}\int_0^b\sin t^2\,dt$, if it exists finitely.

Let $I:=\lim_{b\to \infty}\int_0^b\sin t^2\,dt=\lim_{b\to \infty}(\int_0^1\sin t^2\,dt+\int_1^b\sin t^2\,dt)$

We can do a variable substitution $(u=t^2)$ in the second integral inside parentheses to get: $\int_1^b\sin t^2\,dt=\dfrac{\cos 1}2-\dfrac {\cos b^2}{2b}-\frac 12\int_1^{b^2}\dfrac{\cos u}{u^{3/2}}\,du$ (we can do the substitution because $\sin t$ is continuous everywhere and hence on $[1,b]$ for any $b\gt 1$ and therefore the integral on LHS must be Riemann integrable.) $\tag 1$

Now let $A(b):=\int_1^{b^2}\dfrac{\cos u}{u^{3/2}}\,du$. Let $(x_n)$ be any sequence diverging to $\infty$ and define sequence $(y_n)$ as $y_n=A(x_n)=\int_1^{x_n^2}\dfrac{\cos u}{u^{3/2}}\,du$.

Given any $\epsilon\gt 0$, there must exist an $M\in \mathbb N$ such that for all $m\gt M$, it follows that $x_m\gt \Big(\dfrac 4\epsilon\Big)^{\dfrac 13}$.

It follows that for all $n\ge m\gt M$, we have $\begin{align} |y_n-y_m|=|\int_{x_m^2}^{x_n^2}\frac{\cos u}{u^{3/2}}\,du|& \leq |\int_{x_m^2}^{x_n^2}\frac{1}{u^{3/2}}\,du|\\&\leq 2\Big(\frac 1{x_m^3}+\frac 1{x_n^3}\Big)\\&\lt 2\frac 1{x_n^3}+\frac\epsilon 2\\&\lt \frac \epsilon 2+\frac\epsilon 2=\epsilon\end {align}\tag 2$

Thus, $(y_n)$ is a Cauchy sequence.

The only problem is: although we have shown $A(x_n)$ converges for every $(x_n)\to \infty$ but there is a possibility that for some two different sequences $x_n$ and $z_n$ both $\to \infty$, we may have $A(x_n)\to L_1$ and $A(z_n)\to L_2$ such that $L_1\ne L_2$.

A possible way to refute this possibility is: $\begin{align}|L_1-L_2|=&|L_1-A(x_n)+A(x_n)-A(z_n)+A(z_n)-L_2|\\&\leq |L_1-A(x_n)|+|A(x_n)-A(z_n)|+|A(z_n)-L_2|\end{align}$

For the second term in RHS, we proceed exactly as in case of $(2)$ so we can make it arbitrarily small and hence for large $n$, we can make RHS arbitrarily small and hence we must have $L_1=L_2$.

Hence, $\lim_{b\to \infty} A(b)$ exists finitely by sequential criteria of limits.

Hence, $\lim_{b\to \infty} \int_1^b\sin t^2\, dt$ exists (by $(1)$). Hence $I$ is finite.

Is my proof correct? Thanks.

the way of the proof is ok, but the nominator in your integral should be $2*\sqrt{u}$ — trula, Aug 21 '21 at 15:19
@trula: Thanks a lot for reviewing the post. I have made an edit. — Koro, Aug 21 '21 at 15:29
@trula: I have made the edit in $(1)$. I think there should be 4 and not 2 — Koro, Aug 21 '21 at 17:03
$u=x^2; x=\sqrt{u}, du=2xdx ;dx=du/(2\sqrt{u})$ and i don't see your edit, but never mind. — trula, Aug 21 '21 at 17:43
@trula: I’d wrongly written 4 instead of 2 in (1) and I have fixed it now. Thanks. — Koro, Aug 21 '21 at 18:37
Generally, if $\int_0^\infty |f(t)| dt$ is finite then $\int_0^\infty f(t) dt$ is said to “converge absolutely,” and that in particular implies that $\int_0^\infty |f(t)| dt$ exists. The proof is via Cauchy sequences and more or less what you did in this special case. — Martin R, Aug 21 '21 at 19:41
Does this answer your question? Prove: $\int_0^\infty \sin (x^2) , dx$ converges. — Mittens, Aug 21 '21 at 20:12
@OliverDiaz: My question was proof verification and I have understood now that my proof is correct. The answer below is an alternative to my approach. — Koro, Aug 21 '21 at 20:16

Showing that $\int_0^{\infty}\sin t^2\,dt$ converges as an improper Riemann integral.

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