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Consider the $\displaystyle \int_0^\infty $sin$ (x^p) dx$. Does it converge when $p<0$? Does it converge when $p>1$?

My Work:

Let $x^p=y$, then $\displaystyle \int_0^\infty $sin$ \displaystyle (x^p) dx=\frac{1}{p}\sum_{n=1}^\infty \int_{(n-1)\pi}^{n\pi} \frac{\text{sin} \; y}{y^r} dy$ , where $r=\frac{p-1}{p}>1 $when $p<0$ and $0<r=\frac{p-1}{p}\leq 1 $when $p>1$. I know that $\displaystyle \int_0^\infty $sin$ (\frac{1}{x}) dx$ diverges which is a special case of first case. But failed to show it generally. Can anyone please give me a hint to preceed?

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  • $\begingroup$ When $p<0$ you have $\sin(x^p) \sim x^p$ as $x \to \infty$, so use this to investigate convergence near $\infty$. And near $x=0$, despite wild oscillations, the integrand is bounded, so there is no problem. $\endgroup$ – GEdgar Jan 2 '15 at 22:47
  • $\begingroup$ See Riemann-Lebesgue lemma. $\endgroup$ – Lucian Jan 2 '15 at 23:42
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    $\begingroup$ Which "series"? $\endgroup$ – Did Jan 2 '15 at 23:48
  • $\begingroup$ Combining the integral expression for the $\Gamma$ function with Euler's formula, for $p>1$ we have $$\int_0^\infty\sin(x^p)~dx~=~\sin\frac\pi{2p}\cdot\Gamma\bigg(1+\frac1p\bigg)$$ and $$\int_0^\infty\cos(x^p)~dx~=~\cos\frac\pi{2p}\cdot\Gamma\bigg(1+\frac1p\bigg)$$ where $\Gamma(1+n)=n!$ $\endgroup$ – Lucian Jan 2 '15 at 23:51
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If $p<0$, then $\sin(x^p)$ behaves like $x^p$ for $x\to +\infty$, so the integral is converging iff $p<-1$.

On the other hand, if $p>0$ then: $$\int_{0}^{+\infty}\sin(x^p)\,dx = \frac{1}{p}\int_{0}^{+\infty}u^{\frac{1}{p}-1}\sin(u)\,du $$ is converging iff $p>1$ in virtue of the integral version of Dirichlet's test - in such a case $\sin u$ is a function with a bounded primitive and $u^{\frac{1}{p}-1}$ is a continuous function decreasing towards zero.

By putting together these two cases, we have that $\sin(x^p)$ is Riemann integrable over $\mathbb{R}^+$ iff $|p|>1$.

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  • $\begingroup$ But $u^{\frac{1}{p}-1}$ is not defined at $0$. So, how did you use Dirichlet test? $\endgroup$ – Extremal Jan 2 '15 at 23:25
  • $\begingroup$ And why cannot we use Dirichlet test when $p<0$? $\endgroup$ – Extremal Jan 2 '15 at 23:30
  • $\begingroup$ @Mathi: because we need a function with a bounded derivative and a decreasing function converging to zero. $\endgroup$ – Jack D'Aurizio Jan 2 '15 at 23:55
  • $\begingroup$ But even in case where $p<0$ we have a such functions. For eg: when $p=-\frac{1}{2},$ we have $\frac{1}{u^3}$ which is decreasing to $0$. $\endgroup$ – Extremal Jan 3 '15 at 0:07
  • $\begingroup$ @Mathi: but in such a case the integrand function is not integrable in a right neighbourhood of zero. $\endgroup$ – Jack D'Aurizio Jan 3 '15 at 11:19
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We can establish convergence using Cauchy's Theorem. For $p \gt 0$, consider the integral

$$\oint_C dz \, e^{i z^p} $$

where $C$ is the boundary of a circular sector of radius $R$ and angle $\pi/(2 p)$,in the upper right of the complex plane, with one segment along the positive real axis. The contour integral is then

$$\int_0^R dx \, e^{i x^p} + i R \int_0^{\pi/(2 p)} d\theta \, e^{i \theta} \, e^{i R^p e^{i p \theta}} + e^{i \pi/(2 p)} \int_R^0 dt \, e^{-t^p}$$

The second integral vanishes in the limit as $R \to \infty$ when $p \gt 1$. We can see this by bounding the magnitude of this integral, which is less than or equal to

$$R \int_0^{\pi/(2 p)} d\theta \, e^{-r^p \sin{p \theta}} = \frac{R}{p} \int_0^{\pi/2} d\phi \, e^{R^p \sin{\phi}} \le \frac{R}{p} \int_0^{\pi/2} d\phi \, e^{2 R^p \phi/\pi} \le \frac{\pi}{2 p} \frac1{R^{p-1}}$$

Thus, when $p \gt 1$, we have in this limit

$$\int_0^{\infty} dx \, e^{i x^p} = e^{i \pi/(2 p)} \int_0^{\infty} dx \, e^{-x^p} $$

The integral on the RHS clearly converges, so that the integral on the LHS converges and its imaginary part. Thus, when $p \gt 1$, the integral in question converges and is equal to

$$\int_0^{\infty} dx \, \sin{x^p} = \sin{\left (\frac{\pi}{2 p}\right )} \int_0^{\infty} dx \, e^{-x^p} = \frac1{p} \Gamma \left ( \frac1{p} \right ) \sin{\left (\frac{\pi}{2 p}\right )} $$

The integral does not converge for $0 \lt p \lt 1$. Cauchy's Theorem is unable to say anything about convergence when $p=1$.

Note that the question of convergence for $p \lt 0$ may be answered by subbing $u=1/x$ in the original integral to get

$$\int_0^{\infty} du \frac{\sin{u^{|p|}}}{u^2} $$

From the analysis above and the behavior at $u=0$, we may establish that the integral converges when $|p| \gt 1$.

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The integral does converge when $p>1$. Try this:

Let $$S_n = \int_{(n\pi)^{\frac{1}{p}}}^{((n+1)\pi)^{\frac{1}{p}}}\text{sin}(x^p)\, \mathrm{d}x $$ This sequence changes signs every time, so we can apply leibniz criterion. So we want to prove the following:

  • $S_n$ is decreasing
  • Its limit equals zero. This should get you going.
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