How to prove that the improper integral $$\int_0^{\infty} \sin(x^3)\, dx$$ converges?
I wanted to use the Taylor series for $\sin(x)$ , and with that I get $\lim \sum \frac{(-1)^n a^{6n+4}}{(2n+1)!(6n+4)} $ as $a$ converges to infinity, but I don't know what to do afterwards. Any ideas?
Consider
$$
T(A) := \int_0^A \sin(x^3)\;dx .
$$
We want to show this converges as $A \to \infty$.
Change variables $y=x^3$,
$$
T(A) = \int_0^{A^3} \frac{\sin(y)}{3y^{2/3}}\;dy .
$$
So $T(A) = T_1+T_2(A)$, where
$$
T_1 = \int_0^1 \frac{\sin(y)}{3y^{2/3}}\;dy
\\
T_2(A) = \int_1^{A^3} \frac{\sin(y)}{3y^{2/3}}\;dy
$$
Now for $T_1$, note
$$
\left|\frac{\sin(y)}{3y^{2/3}}\right| \le
\frac{1}{3y^{2/3}}
$$
and
$$
\int_0^1 \frac{dy}{3y^{2/3}} = 1
$$
so $T_1$ conveges.
Next for $T_2$, integrate by parts
\begin{align}
T_2(A) &= \int_1^{A^3} \frac{\sin(y)}{3y^{2/3}}\;dy
= \left.-\frac{\cos(y)}{3y^{2/3}}\right|_{y=1}^{y=A^3} -
\frac{2}{9}\int_1^{A^3}\frac{\cos(y)}{y^{5/3}}\;dy
\\&
= \frac{-\cos(A^3)}{3A^2} +\frac{\cos(1)}{3}
-\frac{2}{9}\int_1^{A^3}\frac{\cos(y)}{y^{5/3}}\;dy
\end{align}
Note that
$$
\lim_{A \to\infty} \frac{-\cos(A^3)}{3A^2} = 0 .
$$
Also,
$$
\left|\frac{\cos(y)}{y^{5/3}}\right| \le \frac{1}{y^{5/3}}
$$
and
$$
\int_1^\infty \frac{dy}{y^{5/3}} = \frac53
$$
Thus $\lim_{A\to\infty}T_2(A)$ exists.
If you have seen the video link I mentioned then, Set $a=0,b=1,n=3,s=\frac{1}{3}$ Then it will follow that, $$\int_{0}^{\infty}\sin(x^{3})dx = \frac{\Gamma(1/3)}{6}$$
