Does $\int_{0}^{\infty} \sin(x) \cdot\sin(x^2)\,dx$ converge? My try: Let $x = \sqrt{t}$, then $dx = \frac{1}{2\sqrt{t}}dt$. We get the following integral: $\int_{0}^{\infty}\frac{\sin(t)\sin(\sqrt{t})}{2\sqrt{t}}dt$. Now I tried to use Dirichlet's test: The function $g(x) = 1/2\sqrt{t}$ has limit $0$ at infinity and it is monotonically decreasing. Now if I could show that the function $F(b) = \int_{0}^{b} \sin(t)\sin(\sqrt{t})$ is bounded, that would mean the integral converges by Dirichlet's test, but I don't know how to prove it. Suggestions?
 A: I think I found an answer: We will prove it by Cauchy's test: That is, we will prove that for all $\varepsilon > 0$ there exists $N > 0$ such that for all $a>b>N$ we have $|\int_{a}^{b} \sin t \frac{\sin\sqrt{t}}{\sqrt{t}}dt| < \varepsilon. $  Take $\varepsilon > 0$. Then  since $\lim_{t \to \infty} \frac{\sin\sqrt{t}}{\sqrt{t}} = 0$ we can take $N>0$ such that for all $t > N$, $|\frac{\sin\sqrt{t}}{\sqrt{t}}|<\varepsilon/2$. Then, if $a >b >N$ we have: $\int_{a}^{b} \sin t \frac{\sin\sqrt{t}}{\sqrt{t}}dt< \int_{a}^{b} \sin t \cdot \varepsilon/2 \leq 2 \cdot \varepsilon/2=\varepsilon$, and similarlly we can get $\int_{a}^{b} \sin t \frac{\sin\sqrt{t}}{\sqrt{t}} > -\varepsilon$ which is what we wanted.
A: As usual, trig identities are the answer:
\begin{align}
\int_0^\infty \sin(x^2)\sin(x)dx =& \frac{1}{2}\int_0^\infty\left[ \cos(x^2-x)-\cos(x^2+x)\right]dx \\ =& \frac{1}{2}\int_0^\infty\left( \cos\left[\left(x-\frac{1}{2}\right)^2 -\frac{1}{4}\right]-\cos\left[\left(x+\frac{1}{2}\right)^2 -\frac{1}{4}\right]\right)dx
\\ = &\frac{1}{2}\left[\int_{-1/2}^\infty\cos\left(x^2 -\frac{1}{4}\right)dx-\int_{1/2}^\infty\cos\left(x^2 -\frac{1}{4}\right)dx\right]
\end{align}
and these integrals both converge, since the integrals of $\sin(x^2)$ and $\cos(x^2)$ both converge, and $\cos(x^2-1/4) = \cos(1/4)\cos(x^2)+\sin(1/4)\sin(x^2)$. Thus, the original integral converges as well.
The difference of integrals can be simplified to
\begin{align}
\int_0^\infty \sin(x^2)\sin(x)dx = &\int_0^{1/2}\cos\left(x^2-\frac{1}{4}\right)dx 
\\= &\cos\left(\frac{1}{4}\right)\int_0^{1/2}\cos(x^2)dx + \sin\left(\frac{1}{4}\right)\int_0^{1/2}\sin(x^2)dx
\\ = &\sqrt{\frac{\pi}{2}}\left[\cos\left(\frac{1}{4}\right)C\left(\frac{1}{\sqrt{2\pi}}\right) + \sin\left(\frac{1}{4}\right)S\left(\frac{1}{\sqrt{2\pi}}\right)\right],
\end{align}
where $S$ and $C$ are the Fresnel sine and cosine integrals, respectively.
A: Dirichlet test is proved by integration by parts. One can apply it directly to the original integral. The convergence on $[0,+\infty)$ is equivalent to the convergence on $[1,+\infty)$. Now integrate by parts
$$
\int_1^{+\infty}\sin x\sin x^2\,dx=\int_1^{+\infty}\frac{\sin x}{2x}\,d(-\cos x^2)=\\=\left[
-\frac{\sin x}{2x}\cos x^2\right]_1^{+\infty}+\int_1^{+\infty}\left(\frac{\cos x\cos x^2}{2x}-\frac{\sin x\cos x^2}{2x^2}\right)\,dx.
$$
The only non-obviouos part is the integral
$$
\int_1^{+\infty}\frac{\cos x\cos x^2}{2x}\,dx=\int_1^{+\infty}\frac{\cos x}{4x^2}\,d\sin(x^2),
$$
however, a similar integration by part will do the job.
