Consider the improper integral $$ \int_{0}^{\infty} x \sin(x^3) \sin(tx) dx $$ which supposedly converges uniformly for any $t$ in a bounded set $E \subset [0, \infty)$.
I am having some trouble figuring out how to prove this. I have tried some ideas with the Abel-Dirichlet test, integrating by parts, bounding it from above, but nothing seems to work.
Any suggestions or hints?
If $t=0$, $I=0$.
If $t\neq 0$
Let $a>0$, such that $3a^2>|t|$
$$I=\int_{0}^{a} x \sin(x^3) \sin(tx) dx+\int_{a}^{\infty} x \sin(x^3) \sin(tx) dx$$
The first half is convergent. Let's look at the second half.
$$\int_{a}^{\infty} x \sin(x^3) \sin(tx) dx=\frac{1}2\int_a^\infty x\cos(x^3-tx)dx-\frac{1}2\int_a^\infty x\cos(x^3+tx)dx$$
Let $$I_1=\int_a^\infty x\cos(x^3-tx)dx,~~~I_2=\int_a^\infty x\cos(x^3+tx)dx$$
Next, we show $I_1$ is convergent. Integration by part, we get:
$$I_1=\int_a^\infty \frac{x}{3x^2-t}~d\sin(x^3-tx)=\int_a^\infty \sin(x^3-tx)~\frac{t + 3 x^2}{(t - 3 x^2)^2}dx$$
we have
$$|I_1|\le\int_a^\infty \left|\sin(x^3-tx)\frac{t + 3 x^2}{(t - 3 x^2)^2}\right|dx\le\int_a^\infty \left|\frac{t + 3 x^2}{(t - 3 x^2)^2}\right|dx\rightarrow \text{convergent}$$
Note that previously we choose $3a^2>|t|$, because we want to guarantee $3x^2\pm t\neq 0$. Also we can show $I_2$ is convergent by the similar way.