For which values $p$ does $\int_0^\infty x\sin(x^p) dx $ converge?

Question

In an old exercise sheet I have found the following question:

For which values $p\in \mathbb{R_{\geq 0}}$ does the following integral converge

$$\int_0^\infty x\sin(x^p) dx $$

In order to solve it, the paper suggests one should use the substitution $u = x^p$ (for $p > 0$) and use the Leibnitz-Criterium.

Applying this trick gives me $\int_0 ^\infty\frac{u^{\frac{1-p}{p}}}{p} \sin(u) u ^{\frac{1}{p}} $. ($\rightarrow$ using the substitution from above)

How does one proceed from here?

Another idea I have had was to use the taylor polynomial for $\sin(x) = x- \frac{x^2}{2!} \cdots$ and then approximate the integral from above and below $\rightarrow$ this however, seems to get very messy.

I am open for other ideas for solving this exercise.

Answer 1

You can determine what the answer is non-rigorously as follows. First of all:

$$I:=\int_0^\infty x \sin(x^p) dx = \sum_{k=1}^\infty \int_{r_{k-1}}^{r_k} x \sin(x^p) dx$$

where $r_k=(k\pi)^{1/p}$. This is of course rigorous (if we interpret a divergent integral and a divergent sum as "the same"). The point of doing this decomposition is to take each integrand to be of one sign.

Now the non-rigorous part starts. Think of the sine as if it just contributes a sign and a $\Theta(1)$ positive constant factor to the integral in between each pair of roots of $\sin(x^p)$. Similarly, think of the $x$ factor as not changing much between each pair of roots. This means that we are claiming

$$I=\sum_k (-1)^k a_k r_k (r_{k+1}-r_k)$$

for some positive numbers $a_k$. Now we deal with the difference:

$$(k+1)^{1/p}-k^{1/p}=k^{1/p}((1+1/k)^{1/p}-1) =b_k k^{1/p-1}$$

where $b_k>0,b_k=\Theta(1)$. So with $c_k=a_k b_k$ you claim

$$I=\sum_k c_k (-1)^k k^{2/p-1}.$$

If the $c_k$ doesn't matter for the convergence, then you get convergence iff $p>2$.

Cleaning this up to make it rigorous is probably more trouble than it's worth; you'd be better off following the hint or using a different technique entirely. Nevertheless this heuristic gives the correct answer.

Answer 2

By the Laplace transform

$$\lim_{M\to +\infty}\int_{0}^{M}z^{\frac{2}{p}-1}\sin(z)\,dz =\frac{1}{\Gamma\left(1-\frac{2}{p}\right)}\int_{0}^{+\infty}\frac{ds}{s^{2/p}(1+s^2)}\,ds $$ is a convergent integral for any $p>2$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\int_{0}^{\infty}x\sin\pars{x^{p}}\,\dd x \,\right\vert_{\ p\ \in\ \mathbb{R}_{\geq\ 0}} & \,\,\,\stackrel{x^{\large p}\ \mapsto\ x}{=}\,\,\, \int_{0}^{\infty}x^{1/p}\sin\pars{x}\pars{{1 \over p}\,x^{1/p - 1}}\dd x \\[5mm] & = {1 \over p}\int_{0}^{\infty}{\sin\pars{x} \over x^{1 - 2/p}}\dd x \qquad\pars{~\mbox{It}\ converges\ \mbox{whenever}\ p > 2 ~} \\ & = {1 \over p}\,\Im\int_{0}^{\infty}\expo{\ic x}\ \overbrace{{1 \over \Gamma\pars{1 - 2/p}} \int_{0}^{\infty}t^{-2/p}\expo{-xt}\dd t}^{\ds{1 \over x^{1 - 2/p}}}\ \dd x \\[5mm] & = {1 \over p}\,{\sin\pars{\pi\bracks{2/p}}\Gamma\pars{2/p} \over \pi} \,\Im\int_{0}^{\infty}t^{-2/p}\int_{0}^{\infty} \expo{-\pars{t - \ic}x}\dd x\,\dd t \\[5mm] & = {\sin\pars{2\pi/p}\Gamma\pars{2/p} \over \pi p} \,\Im\int_{0}^{\infty}{t^{-2/p} \over t - \ic}\,\dd t \\[5mm] & = {\sin\pars{2\pi/p}\Gamma\pars{2/p} \over \pi p} \int_{0}^{\infty}{t^{-2/p} \over t^{2} + 1}\,\dd t \\[1cm] & \stackrel{x\ =\ 1/\pars{t^{2} + 1}}{=}\,\,\, {\sin\pars{2\pi/p}\Gamma\pars{2/p} \over \pi p}\ \times \\ & \phantom{\stackrel{x\ =\ 1/\pars{t^{2} + 1}}{=}\,\,\,}\int_{1}^{0} {\bracks{\pars{1/x - 1}^{1/2}}^{-2/p} \over 1/x}\,{1 \over 2} \,\pars{{1 \over x} - 1}^{-1/2}\,\pars{-\,{\dd x \over x^{2}}} \\[1cm] & = {\sin\pars{2\pi/p}\Gamma\pars{2/p} \over 2\pi p} \int_{0}^{1}x^{1/p - 1/2}\pars{1 - x}^{-1/p - 1/2}\,\dd x \\[5mm] & = {\sin\pars{2\pi/p}\Gamma\pars{2/p} \over 2\pi p} \,{\Gamma\pars{1/p + 1/2}\Gamma\pars{-1/p + 1/2} \over \Gamma\pars{1}} \\[5mm] & = {\sin\pars{2\pi/p}\Gamma\pars{2/p} \over 2\pi p}\, {\pi \over \sin\pars{\pi\bracks{1/p + 1/2}}} \\[5mm] & = {\bracks{2\sin\pars{\pi/p}\cos\pars{\pi/p}}\Gamma\pars{2/p} \over 2p}\, {1 \over \sin\pars{\pi\bracks{1/p + 1/2}}} \\[5mm] & = {2\sin\pars{\pi/p}\cos\pars{\pi/p}\Gamma\pars{2/p} \over 2p}\, {1 \over \cos\pars{\pi/p}} = \bbx{\Gamma\pars{2/p}\sin\pars{\pi/p} \over p} \end{align}