I've spent the last few hrs racking my head about how to solve this problem.
Show that
$$\int_0^\infty x\sin(x^3)\sin(tx)\,dx$$
converges uniformly for $t$ in any finite interval
So far I've managed to derive the following:
$$\int_0^\infty x\sin(x^3)\sin(tx)\,dx=\int_0^\infty\frac{x}{2}\big(\cos(tx-x^3)-\cos(tx+x^3)\big)\,dx$$
$$=\int_0^\infty\frac{x}{2}\left(\frac{e^{i(tx-x^3)}+e^{i(x^3-tx)}}{2}-\frac{e^{i(tx+x^3)}+e^{-i(tx+x^3)}}{2}\right)dx$$
From here, my strategy was to separate the integrals and then try to manipulate them into the gamma function as follows. On the first integral I had:
Let $u=-x^2\Rightarrow x=iu^{1/2}\Rightarrow i(tx-x^3)=-tu^{1/2}-u^{3/2}$
$$\frac{du}{-2x}=dx\Rightarrow\frac{1}{8}\int_{-\infty}^0e^{-tu^{1/2}-u^{3/2}}du$$
However, after this I don't know how to continue. I don't think this approach is fruitless, but I can't figure it out at this time. Any and all help is greatly appreciated.
Just integrate by parts repeatedly. For $a>0$ we have $$ I(a,t):=\int_a^\infty x\sin x^3\sin tx\,dx =-\frac13\int_a^\infty(\cos x^3)'\frac{\sin tx}{x}\,dx \\=\frac{\cos a^3\sin ta}{3a} -\frac13\int_a^\infty\frac{\cos x^3\sin tx}{x^2}\,dx +\frac{t}3J(a,t), \\ J(a,t):=\int_a^\infty\frac{\cos x^3\cos tx}{x}\,dx =\frac13\int_a^\infty(\sin x^3)'\frac{\cos tx}{x^3}\,dx \\=-\frac{\sin a^3\cos ta}{3a^3} +\frac{t}3\int_a^\infty\frac{\sin x^3\sin tx}{x^3}\,dx +\int_a^\infty\frac{\sin x^3\cos tx}{x^4}\,dx, \\ \therefore\qquad |I(a,t)|\leqslant\frac1{3a}+\frac1{3a}+\frac{|t|}3\left(\frac1{3a^3}+\frac{|t|}{6a^2}+\frac1{3a^3}\right), $$ and the estimate clearly converges to $0$ (as $a\to\infty$) uniformly for $t$ in any finite interval.