# 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:

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!

• If this function is just an example you made up, you can make your life easier by using instead $f(x)=x\sin(x^3)+\cos(x^3)/3x^2$. The second term is obviously negligible for large $x$, so it won't affect the asymptotic behavior you are looking for, but the antiderivative is simple: if $F(x)=-\cos(x^3)/3x$, then $F'=f$. (This is also another way of proving that your original $f$ is convergent.) Jun 8, 2013 at 7:41
• You're right, but I'm looking for a function which is defined and continuous also at $x=0$.
– Paz
Jun 8, 2013 at 7:48
• What about $g(x):=f(x+1)$? That pushes the inconvenient behavior to $x=-1$. Jun 8, 2013 at 8:01
• Actually, a much easier function to work with is $F(x)=e^{-x/2}\sin(e^x)$ and $f(x)=e^{x/2}\cos(e^x)-\frac12e^{-x/2}\sin(e^x)$. Jun 8, 2013 at 8:06

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.
• Fixed thanks to @Lord_Farin's comment. Jun 8, 2013 at 7:32
• That absolutely convergent point was exactly what I missed. Thanks!
– Paz
Jun 8, 2013 at 7:38

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.

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 $b < a$.

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

• Thanks, but I've never heard of nor seen the Fresnel diffraction.
– Paz
Jun 8, 2013 at 7:49
• just google "Fresnel Integrals" you will get endless... Jun 8, 2013 at 7:56

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{1}^{\infty}x\sin\pars{x^{3}}\,\dd x:\ {\large ?}}$

\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 $\ds{\rm n}$.