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?
Does $\int_{0}^{\infty} \sin(x) \cdot\sin(x^2)\,dx$ converge
calculusimproper-integrals
Best Answer
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.