Integral

[contrattivo]

$ int_(0)^(+ oo)cos(x^2)dx $
if impose $ t=x^2rarr dx=dt/(2sqrt(t)) $
$ int_(0)^(+ oo)(cos(t)dt)/(2sqrt(t))<= int_(0)^(+ oo)1/(2sqrt(t))dt $
For this is divergent: $ (0, +oo) $ how can I prove that the integral exists?

[gugo82]

"contrattivo":$ int_(0)^(+ oo)cos(x^2)dx $
if impose $ t=x^2rarr dx=dt/(2sqrt(t)) $
$ int_(0)^(+ oo)(cos(t)dt)/(2sqrt(t))<= int_(0)^(+ oo)1/(2sqrt(t))dt $
For this is divergent: $ (0, +oo) $ how can I prove that the integral exists?

I'm going to explain fu^2's answer in full detail.

Just to fix our ideas, we choose \(M=\sqrt{\pi}\) (the reason for this choice will be evident in a moment); then, by the elementary properties of Riemann integral, we get:
\[
\int_0^\infty \cos x^2\ \text{d} t = \underbrace{\int_0^{\sqrt{\pi}} \cos x^2\ \text{d} x}_{=:I_1} + \underbrace{\int_{\sqrt{\pi}}^\infty \cos x^2\ \text{d} x}_{=:I_2}\; ;
\]
the integral \(I_1\) is finite, for \(\cos^2 x\) is a continuous function in \([0,\sqrt{\pi}]\), hence we have only to prove that \(I_2\) is finite.
In order to get the claim, let us consider the function:
\[
u(r) := \int_{\sqrt{\pi}}^r \cos x^2\ \text{d} x\; ,
\]
which rewrites:
\[
u(r) = \int_{\pi}^{r^2} \frac{\cos t}{\sqrt{t}}\ \text{d} t\; ;
\]
an integration by parts yields:
\[
u(r) = \frac{\sin t}{\sqrt{t}}\Big|_\pi^{r^2} + \frac{1}{2}\ \int_\pi^{r^2} \frac{\sin t}{t^{3/2}}\ \text{d} t
\]
therefore:
\[
u(r) = \frac{\sin r^2}{r} + \frac{1}{2}\ \int_\pi^{r^2} \frac{\sin t}{t^{3/2}}\ \text{d} t\; .
\]
Now we can pass to the limit to get:
\[
\begin{split}
I_2 &= \lim_{r\to \infty} u(r) \\
&= \lim_{r\to \infty} \frac{\sin r^2}{r} + \frac{1}{2}\ \int_\pi^{r^2} \frac{\sin t}{t^{3/2}}\ \text{d} t \\
&= \frac{1}{2}\ \int_\pi^\infty \frac{\sin t}{t^{3/2}}\ \text{d} t <\infty
\end{split}
\]
(the rightmost side is finite because \(\frac{\sin t}{t^{3/2}}\) decreases sufficiently fast at infinity).
It then follows that:
\[
\int_0^\infty \cos x^2\ \text{d} x = I_1+I_2<\infty
\]
i.e., that the improper Riemann integral of \(\cos x^2\) converges at \(\infty\).

[Paolo902]

"contrattivo":How can I prove that the integral exists?

The question is quite far from being trivial and the answer is not easy, indeed. The study of Fresnel integrals does require some tools from complex analysis.

[fu^2]

you can split it into two interval, a neighborhood of zero and infinity... Because your upper boud works in $(0, M)$, for some fixed $M<\infty$, so that you have to study only its existence in $(M, \infty)$.