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

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.

• 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. – nullUser Mar 10 '12 at 2:54
• (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. – André Nicolas Mar 10 '12 at 2:58

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.

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.

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 $0$.