5

I want to show that the improper integral $\int_{0}^{\infty} x^{-a}\sin({x})~\text{d}x$ is convergent for all $0<a<2$. For $a=1$, there exist a few method. One of them is here : Improper integral of sin(x)/x from zero to infinity

I tried to apply the same method but I couldn't get the result. Which method should I use?

Integrand
  • 8,078
Tuğba Yesin
  • 299
  • https://en.wikipedia.org/wiki/Dirichlet%27s_test#Improper_integrals – metamorphy Jun 25 '20 at 20:07
  • 1
    The integral over $[1,\infty)$ can be dealt with in every case by cutting the integral into small pieces to generate an alternating series (or more easily for $1<a<2$ by using $|\sin(x)| \le 1$). The integral over $[0,1]$ is convergent either by using $|\sin x|\le 1$ if $a<1$ or $|\sin(x)| \le |x|$ if $1<a<2$ and using the result for $a<1$ . – nicomezi Jun 25 '20 at 20:12
  • 1
    For the word "integrable", you need to be careful because there is more than one definition and they are not equivalent. For example, if you are talking about Lebesgue integrable, it requires that $\int_0^\infty |f(x)| dx <\infty$. (i.e., absolutely integrable, the usual one). It is well-known that $\sin(x)/x$ is not Lebesgue integrable in that sense. However, it is conditionally integrable, in the sense that limit $\lim_{t\rightarrow \infty} \int_0^t \sin(x)/x dx$ exists (the well-known Dirichlet integral, which can be evaluated by Calculus of Residue). – Danny Pak-Keung Chan Jun 25 '20 at 20:25
  • 1
    I do not see the word "integrable" in the question anyway. – nicomezi Jun 25 '20 at 20:37
  • 1
    @ nicomezi It is rephrased as "convergence of imporper integral". – Danny Pak-Keung Chan Jun 25 '20 at 20:43
  • For improper integral, even in Riemann sense, one should distinguish between "absolutely integrable" and "conditionally integrable", which is an analog to absolute convergence and conditional convergence of infinite series $\sum_{n=1}^\infty a_n$. The "absolute" version has a lot of nice properties. – Danny Pak-Keung Chan Jun 25 '20 at 20:52
  • One can use other types of integration where $f_a(x)=x^{-a}\sin x$ is integrable, e.g. the Henstock–Kurzweil integral or Denjoy integral. But of course, $f_a$ is not Lebesgue integrable for $0<a\leq 1$. – Mittens Jun 25 '20 at 21:35

2 Answers2

4

Sketch of a proof:

Set $f_a(x)=x^{-a}\sin x$. Split the integral from $0$ to $\pi/2$ and then from $\pi/2$ say $(2m+1)\pi/2$.

If $0<a<1$ there is nothing to worry about $\int^{\pi/2}_0f(x)\,dx$ since then $|f_a(x)|\leq x^{-a}$ and $x^{-a}\in L_1(0,\pi/2)$. If $1\leq a$, then $|f_a(x)|=\Big|\frac{\sin x}{x}\Big|\Big|\frac{1}{x^{a-1}}\Big|\leq \Big|\frac{1}{x^{a-1}}\Big|\in L_1(0,\pi/2)$.

Thus, the only concern is $\lim_{b\rightarrow\infty}\int^b_{\pi/2} x^{-a}\sin x$. For $1<a<2$, there is nothing to worry about since $x^{-a}\in L_1(\pi/2,\infty)$. For $0<a\leq1$ you may proceed as follows.

It will be enough to consider $b$ growing along the sequence $b_n=(2n+1)\pi/2$.

$$ \int^{(2n+1)\pi/2}_{\pi/2} x^{-a}\sin x\,dx=\sum^{n-1}_{k=1}(-1)^k\int^{(2k+3)\pi/2}_{(2k+1)\pi/2}|f_a(x)|\,dx$$ Then it is not difficult to see that the sum on the right is in fact an alternating sum with terms decreasing to $0$.

Some details should be filled in, but nothing to complicated.

For $0<a<1$, the limit of the existence of limit $\lim_{b\rightarrow\infty}\int^b_{\pi/2}f_a(x)\,dx$ DOES NOT imply that $f_a\in L_1(\pi/2,\infty)$ (in the sense of Lebesgue) of course, in fact $f_a\notin L_1([\pi/2,\infty)$, as once can see by noticing that $\int^{(2m+3)\pi/2}_{(2m+1)\pi/2}|f_a|\geq c\frac{1}{m^a}$ for some constant.

Mittens
  • 46,352
  • There appears to be an issue. You write "Thus, the only concern is $\lim_{a\to \infty} \int^a_{\pi/2} x^{-a} \sin(x) dx$ ...." but we are only interested in $0 < a < 2$. I think it's a typo and it should be something like $\lim_{M \to \infty} \int^M_{\pi/2} x^{-a} \sin(x) dx$. – ccroth Jun 25 '20 at 21:57
  • 1
    @ccroth: Just fixed my bad choice of symbol. thanks – Mittens Jun 25 '20 at 22:01
2

Near $ 0^+$, $$\frac{\sin(x)}{x^a}\sim \frac{1}{x^{a-1}}$$

thus $$\int_0^1\frac{\sin(x)}{x^a}dx\; \text{converges} \; \iff $$ $$a-1<1 \; \iff \; \color{red}{a<2}$$

Near $ +\infty,$

Using by parts integration,

$$\int_1^X\frac{\sin(x)}{x^a}dx=$$ $$\Bigl[\frac{-\cos(x)}{x^a}\Bigr]_1^X-\int_1^X\frac{a\cos(x)}{x^{a+1}}dx$$

this shows that if $ a+1>1$, or $ \color{red}{a>0 }$, the integral $$\int_1^{+\infty}\frac{\sin(x)}{x^a}dx\; \text{converges}$$

MathFail
  • 21,529
hamam_Abdallah
  • 63,719