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If $p >0$, determine if the following integral converges: $$\int_1^{\infty} \frac{\sin{(ax+b)}}{x^p} \,\mathrm dx$$

What i did so far is:

  1. If $p>1$, then $$f(x)= \frac{|\sin{(ax+b)}|}{x^p}<\frac{1}{x^p}$$ hence the integral converges.
  2. If $0<p\le1$, The above is false, And then what can i do?
Lord_Farin
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StationaryTraveller
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3 Answers3

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Since you know that $$\frac{|\sin{(ax+b)}|}{x^p}<\frac{1}{x^p}$$ you can just consider $\frac{1}{x^p}$. If $\frac{1}{x^p}$ converges, so will $\frac{|\sin{(ax+b)}|}{x^p}$. So, lets integrate $\frac{1}{x^p}$:

$$\int_1^{\infty} \frac{1}{x^p} \,\mathrm dx=\lim_{a \to +\infty}(\int_1^{\infty} \frac{1}{x^p}dx)=\lim_{a \to +\infty}\Big(\frac{1}{1-p}x^{1-p}|_1^a\Big)=\lim_{a \to +\infty}\Big(\frac{1}{1-p}(a^{1-p}-1^{1-p})\Big)=$$

So, limit of one to any power is one. The only tricky thing here is that there is a negative sign in front of one, but we can get rid of it. As a result the last expression above reduces to:

$$=\lim_{a \to +\infty}\Big(\frac{1}{1-p}(a^{1-p})\Big)+\frac{1}{p-1}$$

Forgot to mention that $\frac{|\sin{(ax+b)}|}{x^p}<\frac{1}{x^p}$ holds regardless of whether p is greater or smaller than 1 because $|\sin{(ax+b)}|<1$ always. So, now you have to work with this limit only. If $p>1$, then the limit equals zero and we are left with $\frac{1}{p-1}$ which means the integral converges (to $\frac{1}{p-1})$. If $p<1$ then the limit goes to infinity (diverges) and the integral diverges.

To see why the integral in question converges for $p<1$ read Baby Dragon's comment.

Koba
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    How do you know that if $\frac{1}{x^p}$ diverges then $\frac{\sin(ax+b)}{x^p}$ also diverges? – Quimey Apr 11 '13 at 16:42
  • Oh yeah I dont. If a series $b_k$, $b_k>a_k$, diverges it does not imply that $a_k$ diverges. So, what if picked and integrated $\frac{-1}{x^p}$ because $\frac{-1}{x^p}<\frac{|\sin{(ax+b)}|}{x^p}$? – Koba Apr 11 '13 at 17:05
  • I think you still have the same problem. – Quimey Apr 11 '13 at 17:10
  • But $\frac{-1}{x^p}$ is smaller than the other series and if a smaller series diverges this means a greater series diverges as well. Yeah I think it would work. Cause we would get the same limit, but with a negtive sign in front. So, for p<1 that limit diverges. – Koba Apr 11 '13 at 17:19
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    By that reasoning (in the edit), the integral $\int_1^\infty 0dx$ also diverges since $-1<0$, and $\int_1^\infty -1dx$ diverges. – Baby Dragon Apr 11 '13 at 17:40
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    It still may be that $\Sigma_0^\infty \frac{\sin(ax+b)}{x^p}$ converges if $0<p\leq 1$. The reason I think that they could converge is that $\sin(ax+b)$ behaves like $(-1)^n$, the integral behaves like a sum, and the sums $\Sigma_{n=0}^\infty\frac{(-1)^n}{n}$ converges. – Baby Dragon Apr 11 '13 at 17:42
  • For $a=1$, $b=0$ and $p=1/2$ the integral converges: http://www.wolframalpha.com/input/?i=Integrate[Sin[x]%2FSqrt[x]%2C{x%2C1%2Cinfinity}] – Quimey Apr 11 '13 at 17:51
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    Wolfram alpha says that this integral converges for any $p$. Ok I take it back. my edit was incorrect. But how would I show that the integral in question converges for p smaller than 1? I guess babydragon's explanations is the best we can get. – Koba Apr 11 '13 at 18:06
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Let us consider the integral, $\int_0^\infty\frac{\sin x}{x^p}dx$ (I changed the lower limit of integration. Note that the function $\frac{\sin x}{x^p}$ has a limit at zero if $0<p\leq1$). We may rewrite this integral as $$\sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx.$$ Note that this is an alternating series so the alternating series test applies. We must show that

1 $|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx|>|\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx|$

2 $\lim_{n\to\infty}\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx=0$

Let us start with (1). Let us assume that $\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx>0$ (the other case is similar). Note that in this case that $\frac{\sin x}{x^p}\geq 0$. This also means that $\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx<0$, so $$|\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx|=-\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx$$ We will now make a change of variables, $u=x-\pi$, which makes $$-\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx=-\int_{(n)\pi}^{(n+1)\pi}\frac{\sin (u+\pi)}{(u+\pi)^p}du=\int_{(n)\pi}^{(n+1)\pi}\frac{\sin x}{(x+\pi)^p}dx$$ (note that $\int_a^bf(x)dx=\int_a^bf(u)du$). Again note that $\frac{\sin (x+\pi)}{(x+\pi)^p}>0$ in the interval $[n\pi,(n+1)\pi]$. We must now compare the integrals, $$\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx\mbox{ and }\int_{n\pi}^{(n+1)\pi}\frac{\sin (x+\pi)}{(x+\pi)^p}dx. $$ Since $x^p<(x+\pi)^p$, we get that $$\frac{\sin x}{x^p}>\frac{\sin (x)}{(x+\pi)^p}. $$ Therefore $$\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx>\int_{n\pi}^{(n+1)\pi}\frac{\sin (x+\pi)}{(x+\pi)^p}. $$, which shows (1).

Now to show (2) simply note that $|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx|\leq\int_{n\pi}^{(n+1)\pi}\frac{1}{x^p}dx$ and that $0<\int_{n\pi}^{(n+1)\pi}\frac{1}{x^p}dx$. An elementary calculation will $\lim_{n\to\infty}\int_{n\pi}^{(n+1)\pi}\frac{1}{x^p}dx=0$

hbghlyj
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Baby Dragon
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For simplicity, we can just set $a=1$ and $b=0$.

If $0<p\le1$, we can use Dirichlet's test.

Because $x^{-p}$ is monotonically decreasing and tends to $0$, and $|\int_s^t \sin x\ dx|\le 2$ for arbitrary $t>s\ge1$, then $\int_1^{\infty}x^{-p}\sin x\ dx$ converges.

Moreover, we can prove that this integral is conditionally convergent.

$\int_1^{\infty}x^{-p}|\sin x|\ dx \ge \int_1^{\infty}x^{-p} (\sin x)^2\ dx =\frac{1}{2}\int_1^{\infty}x^{-p}[1-\cos (2x)]\ dx$

Using Dirichlet's test again, we know that $\int_1^{\infty}x^{-p}\cos (2x)\ dx$ is convergent.

However, $\int_1^{\infty}x^{-p}\ dx$ is divergent, making the whole integral $\int_1^{\infty}x^{-p}|\sin x|\ dx$ is divergent.

Therefore, we can conclude that:

  1. If $p>1$, the integral is absolutely convergent.
  2. If $0<p\le 1$, the integral is conditionally convergent.
Cheng Li
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