How would one prove the convergence of $$ \int_0^{\infty} \cos\left(x^2\right) \,\mathrm dx $$
I tried using the integral test for convergence by noting that making the substitution $u = x^2$ means you can write the integral in the form:
$$ \int_0^{\infty} \frac{1}{2\sqrt{u}}\cos(u) \space \mathrm{du}$$
I'm not sure what to do here - I though of rewriting the cosine as a Taylor series, and then applying the integral test, but I don't think that would work.
The convergence of such integral follows from the continuous analogue of the Dirichlet criterion for the convergence of a series.
Dirichlet criterion: If $\{a_n\}_{n\in\mathbb{N}}$ is a sequence with the property that the partial sums $\sum_{k=0}^n a_k$ are bounded, and $\{b_n\}_{n\in\mathbb{N}}$ is a decreasing sequence converging towards zero, the series $$\sum_{n=0}^{+\infty} a_n b_n $$ is convergent. The proof relies on summation by parts.
Dirichlet criterion (integral version): If $f(x)$ is a Riemann-integrable function with the property that $\int_{0}^{x}f(t)\,dt$ is bounded for any $x\geq 0$, and $g(x)$ is a continuous decreasing function over $\mathbb{R}^+$ such that $\lim_{x\to +\infty}g(x)=0$, then $f\cdot g$ is a Riemann-integrable function over $\mathbb{R}^+$.
The proof relies on integration by parts.
In order to apply the last criterion, it is sufficient to take $f(x)=\cos x$ and $g(x)=\frac{1}{2\sqrt{x}}$.
This is a brutal way to prove the convergence of the given integral. We have the following formula
\begin{equation} \int_0^\infty \cos ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}+\sin\frac{b^2}{a}\right) \end{equation}
The detail proof can be seen here: Duo Fresnel-like integrals $(??)$
Therefore, setting $a=1$ and $b=0$, we get \begin{equation} \int_0^\infty \cos x^2\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}=\sqrt{\frac{\pi}{8}} \end{equation} Addendum of the proof in the cited link, we can use the fact \begin{equation} \int_{-\infty}^\infty e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}} \end{equation}
I can give you a fast way of treating this question:
First you notice that your function is continuous on $[0,+\inf[ $
So your integral is defined on [0,1] for instance:
I(X) = $\int_{[0,X]} cos(t^2)dt = \int_{[0,1]} cos(t^2)dt + \int_{[1,X]} cos(t^2)dt = A + I_1(X)$
$ I_1(X) = \int_{[1,X]} 2t*\frac{cos(t^2)}{2t}dt $ = $[\frac{sin(t^2)}{2t}]_1^X + \int_{[1,X]} \frac{sin(t^2)}{2t^2}dt $
The first term is a constant and a term that ->0 when x->inf. The second verifies:
$|g(t)| = |\frac{sin(t^2)}{2t^2}| \leq \frac{1}{2t^2} $ => g integrable on [1,+inf[ , because it is dominated by an integrable function on [1, +inf[
Hence $I_1(X)$ has a limit when X->inf, and so does your integral
