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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?

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  • $\begingroup$ It is $$\sqrt{\frac{\pi }{2}} \left(C\left(\frac{1}{\sqrt{2 \pi }}\right) \cos \left(\frac{1}{4}\right)+S\left(\frac{1}{\sqrt{2 \pi }}\right) \sin \left(\frac{1}{4}\right)\right)$$ $\endgroup$ – Dr. Sonnhard Graubner Jun 16 at 11:51
  • $\begingroup$ @Dr.SonnhardGraubner I checked it in wolfram alpha, but I don't know what $C$ and $S$ are and I assume it is not elementary to find that antiderivative. I'm looking for something elementary or for another solution (without Dirichlet's test) $\endgroup$ – Omer Jun 16 at 11:53
  • $\begingroup$ @Omer en.wikipedia.org/wiki/Trigonometric_integral $\endgroup$ – J.G. Jun 16 at 11:54
  • $\begingroup$ @J.G. I see, but I still prefer something else, I was not taught about that and this is a question from previous calculus 2 test, so there should be something. $\endgroup$ – Omer Jun 16 at 11:57
  • $\begingroup$ While $\sin t$ goes through a complete cycle, $\sin\sqrt t/\sqrt t$ is nearly constant. Can you find the leading order of an integral for $t$ from $2n\pi$ to $2(n+1)\pi$ ? $\endgroup$ – Empy2 Jun 16 at 12:11
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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.

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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.

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    $\begingroup$ $\sin t \frac{\sin\sqrt{t}}{\sqrt{t}}< \sin t \cdot \varepsilon/2$ is not true for $\sin t<0$. $\endgroup$ – A.Γ. Jun 16 at 12:46
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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.

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