does $\intop_{1}^{\infty}x\sin(x^{3})dx$ really converge? I'm trying to find a continuous function $f(x)$ on $[0,\infty)$ such that: 
$\intop_{1}^{\infty}f(x)dx$ converges while $f(x)$ isn't bounded.
I came up with  $f(x)=x\sin(x^{3})dx$, as a function which oscillates like crazy when x tends to infinity, and much faster than x, which is the direction IMO.
Wolfram says it converges, and plugging big numbers shows Cauchy's criterion holds, but I wasn't able to rigorously prove the convergence.
A few questions:


*

*Is there a "nice" way of showing this integral converges?

*(general question) is Wolfram's numeric approximation always positive? 

*is the claim actually true (there exists a function which has an improper integral but isn't bounded)?
Many thanks!
 A: *

*You can integrate by parts with $u=\dfrac{1}{x}$ and $dv = x^2\sin(x^3)dx$.  You'll get $\frac13\cos(1)$ plus an obviously absolutely convergent integral.

A: If you make your life easier by allowing for functions that aren't given by explicit formulae then you can easily convince yourself such $f$ exist. For example, define $f$ so that at $x=n$, the function has a spike of height $n$ with width $\frac{1}{n^{3}}$ and is otherwise zero. This might cause problems for $n=1$ so start at $n=2$ if you like. This contributes less than $\frac{1}{n^{2}}$ to the integral, and summing over $n$ shows that this would converge, but is clearly unbounded. You can even smooth out the spike and make $f$ smooth. 
A: In general, if $a >0$,
$$ \begin{align} \int_{1}^{\infty} x^{b-1} \sin(x^{a}) \ dx &= \int_{1}^{\infty} (u^{1/a})^{b-1} \sin (u) \frac{1}{a} u^{1/a-1} \ du \\ &= \frac{1}{a} \int_{1}^{\infty} u^{b/a-1} \sin (u) \ du \end{align}$$
which by Dirichlet's convergence test converges if $\frac{b}{a} -1 < 0$. That is, if $b < a$.
A: Recall the Fresnel diffraction in physics. I guess this is a type of generalised Fresnel integral. I think that could be perhaps the most rigorous approach.
The integral $$\int x^m \exp(ix^n)dx = \int\sum_{l=0}^\infty\frac{i^lx^{m+nl}}{l!}dx
 = \sum_{l=0}^\infty \frac{i^l}{(m+nl+1)}\frac{x^{m+nl+1}}{l!}$$
which reduces to Fresnel integrals if real or imaginary parts are taken:
$$\int x^m\sin(x^n)dx = \frac{x^{m+n+1}}{m+n+1}
\,_1F_2\left(\begin{array}{c}\frac{1}{2}+\frac{m+1}{2n}\\
\frac{3}{2}+\frac{m+1}{2n},\frac{3}{2}\end{array}\mid -\frac{x^{2n}}{4}\right)$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{1}^{\infty}x\sin\pars{x^{3}}\,\dd x:\ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{1}^{\infty}x\sin\pars{x^{3}}\,\dd x}
=\int_{1}^{\infty}x^{1/3}\sin\pars{x}\,{1 \over 3}\,x^{-2/3}\,\dd x
={1 \over 3}\int_{1}^{\infty}x^{-1/3}\sin\pars{x}\,\dd x
\\[3mm]&={1 \over 3}\,\Im\
\int_{1}^{\infty}{\expo{\ic x} \over x^{1/3}}\,\dd x
=\color{#66f}{\large{1 \over 3}\,\Im\pars{{\rm E_{1/3}}\pars{-\ic}}}
\approx 0.2056
\end{align}

where $\ds{{\rm E_{n}}\pars{x}}$ is the
Exponential Integral of Order $\ds{\rm n}$.
