contrattivo ha scritto:$ 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.
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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 \(\infty\).
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)