does $\intop_{1}^{\infty}x\sin(x^{3})dx$ really converge?

Question

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:

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

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

3. is the claim actually true (there exists a function which has an improper integral but isn't bounded)?

Many thanks!

Answer 1

1. 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.

Answer 2

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.

Answer 3

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: $$\int x^m\sin(x^n)dx = \frac{x^{m+n+1}}{m+n+1} \,_1F_2\left(\begin{array}{c}\frac{1}{2}+\frac{m+1}{2n}\\ \frac{3}{2}+\frac{m+1}{2n},\frac{3}{2}\end{array}\mid -\frac{x^{2n}}{4}\right)$$

Answer 4

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 $b < a$.