The integral in question is: $$\int_1^{\infty}\frac{\sin^3(x)}{\sqrt{x}}dx$$ Graphically, I can see that the integral converges to about 0.5, but I've been unable to prove algebraically that the integral actually converges.
I've tried to do a comparison test, but I haven't been able to come up with a function that is larger and that I can show also converges.
How can I prove that this integral converges?
We first break the integral into portions on which the sign is constant,
$$\int_1^{\infty}\frac{\sin^3(x)}{\sqrt{x}}dx= \int_1^{\pi}\frac{\sin^3(x)}{\sqrt{x}}dx + \sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi}\frac{\sin^3(x)}{\sqrt{x}}dx.$$
Now, the terms on the right hand sum are certainly alternating in sign and decreasing in absolute value, since for each $x$ we have
$$\frac{|\sin^3(x)|}{\sqrt{x}}\geq \frac{|\sin^3(x+\pi)|}{\sqrt{x+\pi}}.$$
Moreover, they are converging in absolute value to $0$, since $$\frac{|\sin^3(x)|}{\sqrt{x}}<\frac{1}{\sqrt{x}},$$ and the latter's integral over $[n\pi,(n+1)\pi ]$ is at most $\frac{\pi}{\sqrt{n}}$. Therefore the integral converges by the alternating series test.
Strictly speaking, I was a little bit shady here, in that I implicitly suggested that if
$$\sum_{n=1}^\infty\int_{n\pi}^{(n+1)\pi} f(x)\,dx$$ converges, then so does
$$\int_{\pi}^\infty f(x)\,dx.$$
This is not necessarily true for general $f$ (just let $f\equiv n$ on $[n\pi,(n+\frac{1}{2})\pi)$ and $f\equiv -n$ on $[(n+\frac{1}{2})\pi,(n+1)\pi)$), but it can be easily verified to hold when $f$ does not change sign within the interval $[n\pi,(n+1)\pi]$, as is the case in our scenario. After all, in the case the sign does not change, we can trap the finite integrals
$$\int_{\pi}^\alpha f(x)\,dx.$$
between successive partial sums
$$\sum_{n=1}^M\int_{n\pi}^{(n+1)\pi} f(x)\,dx.$$
The answer is based on elementary trigonometric identities. We have $$\sin^3x={1\over 2}[1-\cos(2x)]\sin x\\ ={1\over 2}\sin x-{1\over 4}\sin( 3x) +{1\over 4}\sin x\\ ={3\over 4}\sin x-{1\over 4}\sin(3x)$$ The integrals $$\int\limits_1^\infty{1\over \sqrt{x}}\sin x \,dx\quad \int\limits_1^\infty{1\over \sqrt{x}}\sin (3x)\,dx$$ are convergent due to the Dirichlet test.
Remark The proof of the Dirichlet test in the particular cases above is pretty simple as it can be performed by integration by parts $$\int\limits_1^\infty{\sin x\over \sqrt{x}} \,dx=\cos 1-{1\over 2}\int\limits_1^\infty{\cos x\over x\sqrt{x}} \,dx $$
For your curiosity. $$I=\int\frac{\sin^3(x)}{\sqrt{x}}\,dx=2 \int\sin ^3\left(t^2\right)\,dt=\frac 32\int \sin(t^2)\,dt-\frac 12\int \sin(3t^2)\,dt$$ So, you face Fresnel sine integrals. $$I=\frac{1}{2} \sqrt{\frac{\pi }{6}} \left(3 \sqrt{3} S\left(\sqrt{\frac{2}{\pi }} t\right)-S\left(\sqrt{\frac{6}{\pi }} t\right)\right)$$
$$J=\int_1^\infty\frac{\sin^3(x)}{\sqrt{x}}\,dx=\frac{1}{4} \sqrt{\frac{\pi }{6}} \left(-6 \sqrt{3} S\left(\sqrt{\frac{2}{\pi }}\right)+2 S\left(\sqrt{\frac{6}{\pi }}\right)-1+3 \sqrt{3}\right)$$ which is $0.551171$.
$$K=\int_0^\infty\frac{\sin^3(x)}{\sqrt{x}}\,dx=\frac{1}{4} \left(3 \sqrt{3}-1\right) \sqrt{\frac{\pi }{6}}=0.759085$$
For T a multiple of $2\pi$, $T>0$, let's bound how much mass is added to the integral between a new period from $T$ and $T+2\pi$.
$$\int_{T}^{T+2\pi}\frac{sin^3(x)}{\sqrt{x}}dx \leq \int_{T}^{T+\pi}\frac{\sin^3(x)}{\sqrt{T}}dx - \int_{T+\pi}^{T+2\pi}\frac{|\sin^3(x)|}{\sqrt{T+2\pi}}dx$$ =$$\frac{4\pi}{3}\left(\frac{1}{\sqrt{T}} - \frac{1}{\sqrt{T+2\pi}}\right).$$
Now, telescoping series ...
