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Show that $\int_0^\infty x^{-\alpha}\sin x dx$ exists for $\alpha \in (0,2)$.

This is a real analysis class question. I am not quite sure how to show this. I tried a whole bunch of things like integration by parts, bounding sine, ... and nothing worked. Any hints, ideas, or solutions would be appreciated.

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    $\begingroup$ One thing that sticks out to me. Near 0, $\sin x \approx x$ so the integrand is like $x^{-\alpha+1}$ which requires $-\alpha+1 > -1$ for the integral to converge on $[0,1]$, i.e. $\alpha < 2$. But also we need $\alpha >0$ because the integral is like an alternating series on $[1,\infty)$. This isn't a proof of course, but I think the intuition is there. $\endgroup$ – nullUser Mar 10 '12 at 2:54
  • $\begingroup$ (1) Split into $0$ to $1$, and the rest. (2) There is no trouble at $0$, because $\sin x \le x$ near $0$. That's where $\alpha<2$ is useful. (3) For $1$ to infinity, look at integral from $1$ to $M$, integrate by parts, letting $u=x^{-\alpha}$, $dv=\sin x\,dx$. Get something nice, plus an integral with $\cos x$ on top, $x^{1+\alpha}$ at bottom. This behaves nicely for large $M$, by comparing absolute value if $\frac{1}{x^{1+\alpha}}$. That's where $\alpha>0$ is handy. $\endgroup$ – André Nicolas Mar 10 '12 at 2:58
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Consider separately $$\int_0^1 x^{-\alpha}\sin x\, dx \qquad\text{and}\qquad \int_1^\infty x^{-\alpha}\sin x\, dx.$$

For the first integral, use the fact that for $0<x\le 1$ we have $0\le \sin x \le x$. Now prove the convergence of the first integral by comparing our function with $\dfrac{1}{x^{\alpha-1}}$, noting that $\alpha-1<1$.

For the second integral, let $$I(M)=\int_1^M x^{-\alpha} \sin x\,dx$$ and examine the behaviour of $I(M)$ as $M\to\infty$.

There are various ways to proceed. One is to integrate by parts, letting $u=x^{-\alpha}$ and $dv=\sin x\,dx$. We get something which behaves nicely as $M$ gets large, and an integral of something that is a constant times $\dfrac{\cos x}{x^{1+\alpha}}$. This integral behaves nicely as $M\to\infty$, by comparison of $\dfrac{|\cos x|}{x^{1+\alpha}}$ with $\dfrac{1}{x^{1+\alpha}}$.

The reason for the integration by parts was to increase the power of $x$ at the bottom, in order to make the function decrease faster, fast enough for obvious convergence of the integral.

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Kb100 has exactly the right idea. We can split the proof into three parts:

  1. Show that $\int_0^\epsilon x^{-\alpha} \sin x dx$ exists only if $\alpha<2$ by utilizing $\sin x = x +\mathcal{O}(x^3)$.

  2. Observe that the improper integral diverges for $\alpha\le0$ because $\lim\limits_{x\to\infty} x^{-\alpha}\sin x \ne 0$.

  3. For $\alpha\in(0,2)$ split the integral into an alternating series (note $\sin$ is nonnegative on $[0,\pi]$), $$\int_0^\pi x^{-\alpha}\sin xdx - \int_0^\pi (x+\pi)^{-\alpha}\sin x dx+\int_0^\pi (x+2\pi)^{-\alpha}\sin x dx -\cdots $$ Prove that the terms above are decreasing in magnitude by first proving that $x^{-\alpha}>(x+\pi)^{-\alpha}$ for all $x>0$ and positive $\alpha\;$ (hint: $(1+\epsilon)^\alpha>1$). Now just invoke the alternating series test.

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For the integral $\int_1^\infty x^{-\alpha}\sin x\, dx$ we may apply Dirichlet's Convergence Test for Improper Integrals, witch states:

Let

  1. function $f(x)$ is integrable in closed interval $[a,A]$ $(A>a)$, and

$$|\int_a^Af(x)dx|\leq K \ \ \ \ \ \ (K=\text{const},a\leq A<\infty)$$.

  1. function $g(x)$ is monotone and

$$\lim_{x\to\infty}g(x)=0.$$

Then $\int_a^{\infty}f(x)g(x)dx$ converges.


In our case,

$$ |\int_a^A\sin(x)dx|=|\cos(a)-\cos(A)|\leq 2 $$,

and, $\frac{1}{x^{\alpha}}$ is monotonic decreasing to $0$.

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