12
$\begingroup$

This question already has an answer here:

Hi I am trying to find out for what values of the real parameter does the integral $$ I=\int_0^\infty \frac{\sin x}{x^s}dx $$ (a) convergent and (b) absolutely convergent.

I know that the integral is convergent if $s=1$ since $$ \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}. $$ For $s=0$ it is easy to see divergent integral since $\int_0^\infty \sin x\, dx$ is divergent. However I am stuck on figuring out when it is convergent AND or absolutely convergent.

I know to check for absolute convergence I can determine for an arbitrary series $\sum_{n=0}^\infty a_n$ by considering $$ \sum_{n=0}^\infty |a_n|. $$ If it helps also $$\sin x=\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{(2n+1)!} {x^{2n+1}}$$. Thank you all

$\endgroup$

marked as duplicate by user, Lee David Chung Lin, Alex Provost, YiFan, Eevee Trainer Mar 20 at 6:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

12
$\begingroup$

This is a good problem to analyze. We can solve it by just series methods and careful thought.

Given the following integral \begin{equation} \int_{0}^\infty \frac{\sin x}{x^s}dx, \tag1 \end{equation} for what values of the real parameter s is the integral convergent and absolutely convergent.

(a) In order to solve this problem we break (1) into two pieces \begin{equation} \int_{0}^\infty \frac{\sin x}{x^s}dx=\int_{0}^1 \frac{\sin x}{x^s}dx + \int_{1}^\infty \frac{\sin x}{x^s}dx \tag2 \end{equation} We can analyze each term separately. It is easy to see that the term $$ \int_{1}^\infty \frac{\sin x}{x^s}dx $$ is divergent for $ s \leq 0$ since integral is proportional to $x^s$ which diverges as $x \to \infty$. For $ s > 0$, the series is convergent since $x^{-s} \downarrow 0 \ \text{as}\ x \to \infty$. We now consider the other term in (2) and write it explicitly in terms of a sum $$ \int_{0}^1 \frac{\sin x}{x^s}dx=\int_{0}^1 \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}{x^{-s}}dx= \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}dx. $$ We can evaluate if this integral is convergent by analyzing the series inside which is \begin{equation} \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}\equiv \xi \end{equation} Using the ratio test on $\xi$, we have $$ \lim_{n\to \infty}\bigg| \frac{(-1)^{n+1} x^{2n+3-s} \cdot (2n+1)}{(2n+3)! \cdot (-1)^n x^{2n+1-s}} \bigg|=\lim_{n\to \infty} \frac{x^2}{4n^2+10n+6}=0. $$ By the definition of the ratio test, this series is absolutely convergent since $$ \lim_{n\to \infty} \bigg|\frac{\xi_{n+1}}{\xi_n}\bigg| =0 <1. $$ We now check for uniform convergence by swapping the order of summation and integration, that is doing the integral first which yields $$ \sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)!}\int_{0}^{1} x^{2n+1-s} dx=\sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)! \cdot (2n+2-s)}. $$ Note, the $(2n+2-s) >0$ to be defined. Computing the sum for $n=0$ we have the condition $2 -s > 0$, or $ 2>s$. Evaluating the integral at $n=0, s=2$ we have $$ \int_{0}^{1} x^{2n+1-s} dx=\int_{0}^{1} {x^{-1}} dx $$ which diverges as the logarithm.

We can conclude that (1) is convergent for $s \in (0,2)$.

(b):For absolute convergence we check the convergence of $$ \int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx. $$ Once again, we break the integral into two parts $$ \int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx=\int_{0}^{1} \bigg|\frac{\sin x}{x^s}\bigg| dx + \int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx. $$ The second term on the right converges for $s > 1$ and is seen easily since $$ \int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx < \int_{1}^{\infty} \bigg|\frac{1}{x^s}\bigg| dx $$ which is convergent for $s > 1$. We check the other term for convergence by noting that $$ \bigg|\frac{\sin x}{x^s}\bigg|=\frac{\sin x}{x^s} $$ for $ x \in [0,1]$. Thus we conclude that $$ \int_0^1 \frac{\sin x}{x^s} $$ is absolutely convergent for $s \in (0,2)$.

Therefore, the integral in (1) is absolutely convergent for $s \in (1,2)$.

$\endgroup$
  • $\begingroup$ Hi is there a reason for the down vote? Thank you. $\endgroup$ – Jeff Faraci May 13 '14 at 19:45
  • $\begingroup$ Thank you!!!! I would have missed the evaluating the integral at n=0,s=2. I did not down vote though. Check as answer and +1 $\endgroup$ – user143444 May 13 '14 at 19:52
  • $\begingroup$ Yes that is a subtle point. Glad it helped. $\endgroup$ – Jeff Faraci May 13 '14 at 19:57
  • 2
    $\begingroup$ 2 downvoters? It is quite interesting you down vote the correct solution. Thank you $\endgroup$ – Jeff Faraci May 13 '14 at 20:14
  • $\begingroup$ Nice answer - literally (+1) $\endgroup$ – Ron Gordon May 14 '14 at 18:12
6
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{\infty}{\sin\pars{x} \over x^{s}}\,\dd x:\ {\large ?}}$

\begin{align} I&=\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty} {1 \over x^{s - 1}}{\sin\pars{x} \over x}\,\dd x =\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{1 \over x^{s - 1}}\bracks{% \half\Re\int_{-1}^{1}\expo{\ic\verts{k}x}\,\dd k}\,\dd x \\[3mm]&=\half\Re\int_{-1}^{1}\bracks{\color{blue}{% \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty} {\expo{\ic\verts{k}x} \over x^{s - 1}}\,\dd x}}\,\dd k\tag{1} \end{align}

\begin{align} &\overbrace{\color{blue}{% \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{\expo{\ic\verts{k}x} \over x^{s - 1}} \,\dd x}}^{\ds{\ic\verts{k}x = -t\ \imp\ x = {\ic \over \verts{k}}\,t}} =\lim_{\epsilon \to 0^{+}}\int_{-\epsilon\ic}^{-\infty\ic}\pars{\expo{\ic\pi/2}t \over \verts{k}}^{1 - s} \expo{-t}\,{\ic \over \verts{k}}\,\dd t \\[3mm]&=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}} \lim_{\epsilon \to 0^{+}}\int_{-\epsilon\ic}^{-\infty\ic}t^{1 - s}\expo{-t}\,\dd t \\[3mm]&=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}\times \\[3mm]&\lim_{\epsilon \to 0^{+}}\bracks{% -\int^{\epsilon}_{\infty}t^{1 - s}\expo{-t}\,\dd t -\lim_{R \to \infty}\int_{-\pi/2}^{0} R^{1 - s}\expo{\ic\pars{1 - s}\theta}\exp\pars{-R\expo{\ic\theta}}R \expo{\ic\theta}\ic\,\dd\theta}\qquad\pars{2} \end{align}

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \mbox{When}\quad\epsilon \to 0^{+},\ \mbox{the first integral converges when}\ \Re\pars{1 - s} > -1\ \imp\ \Re\pars{s} < 2\tag{3} $$

Let's study the second integral in the limit $\ds{R \to \infty}$: \begin{align} &\verts{\int_{-\pi/2}^{0} R^{1 - s}\expo{\ic\pars{1 - s}\theta}\exp\pars{-R\expo{\ic\theta}}R \expo{\ic\theta}\ic\,\dd\theta} \leq R^{2 - s}\int_{-\pi/2}^{0}\exp\pars{-R\cos\pars{\theta}}\,\dd\theta \\[3mm]&=R^{2 - s}\int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}\,\dd\theta <R^{2 - s}\int_{0}^{\pi/2}\exp\pars{-R\,{2\theta \over \pi}}\,\dd\theta \\[3mm]&={\pi \over 2}\pars{R^{1 - s} - R^{1 - s}\expo{-R}} \to 0\ \mbox{when}\ \Re\pars{1 - s} < 0\ \imp\ \Re\pars{s} > 1\tag{4} \end{align}

$\pars{3}$ and $\pars{4}$ show that both terms in $\pars{2}$ converge whenever $\ds{1 < \Re\pars{s} < 2}$: $$ \color{blue}{\lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{\infty}{\expo{\ic\verts{k}x} \over x^{s - 1}}\,\dd x} =-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}\,\Gamma\pars{2 - s}\,,\qquad\qquad 1 < \Re\pars{s} < 2 $$ where $\ds{\Gamma\pars{z}}$ is the Gamma Function. This result is replaced in $\pars{1}$ to find: \begin{align} I&=-\,\half\,\cos\pars{\pi s \over 2}\Gamma\pars{2 - s} \int_{-1}^{1}\verts{k}^{s - 2}\,\dd k =-\,\half\,\cos\pars{\pi s \over 2}\Gamma\pars{2 - s}\,{2 \over s - 1} \end{align}

$$\color{#00f}{\large% I = \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty} {\sin\pars{x} \over x^{s}}\,\dd x = \cos\pars{\pi s \over 2}\Gamma\pars{1 - s}}\,,\qquad 1 < \Re\pars{s} < 2 $$ where we used the Gamma Recurrence Formula ${\bf\mbox{6.1.15}}$.

$\endgroup$
  • $\begingroup$ This one is quite similar. Above I included the convergence along an arc ( Jordan Lemma ) which yields $\large\Re\left(s\right) > 1$ as one of the conditions. $\endgroup$ – Felix Marin May 14 '14 at 18:43
  • $\begingroup$ @WillieWong Thanks a lot. You're very kind. $\endgroup$ – Felix Marin May 15 '14 at 8:08
  • $\begingroup$ @FelixMarin: since you are around, I hope you don't mind that I edited your first comment above to remove the reference to the erasure. I find it very worthwhile to keep your comment pointing out the other answer. $\endgroup$ – Willie Wong May 15 '14 at 8:13
0
$\begingroup$

Note that when x close to $0$ the integrand behaves as

$$ \frac{x}{x^s} .$$

On the other hand at infinity behaves as

$$ \frac{1}{x^s} .$$

Now check the integrability of the above funtions and see the conditions on $s$.

$\endgroup$
  • 2
    $\begingroup$ -1 For not providing a solution, please post as a comment. $\endgroup$ – user143444 May 13 '14 at 19:50
  • 2
    $\begingroup$ @Gregory: This is the main idea (in fact the solution ) and what's left is for the OP to do. Gettingdetailed answers is not a good strategy for learning. $\endgroup$ – Mhenni Benghorbal May 13 '14 at 20:02
  • 2
    $\begingroup$ This is not the main idea however. You miss a lot of key points. Please post as a comment next time thanks :) Greg $\endgroup$ – user143444 May 13 '14 at 20:03
  • 2
    $\begingroup$ Even as a hint, this is false. "On the other hand at infinity behaves as 1/x^s" squarely misses the point and yields the (wrong) answer that s should be in (1,2) although the integral converges for every s in (0,2). @upvoters Can you explain? $\endgroup$ – Did May 14 '14 at 6:46
  • 7
    $\begingroup$ For the nth time, Mhenni, this is how the site is meant to work: everybody nosing in everybody's posts (if this unsettles you that your answers are read and critiqued, don't post). About the four upvotes: I agree, these upvotes are amazing seeing the defective mathematical content of this answer--but once again this is how the site works. $\endgroup$ – Did May 14 '14 at 7:23
0
$\begingroup$

$$\varphi_1(\alpha) =\int_0^\infty \frac{\sin t}{t^\alpha}\,dt\tag{I}$$

case $\alpha\gt 0$

Near $t=0$, $\sin t\approx t.$ Which yields, $\frac{\sin t}{t^{\alpha}}\approx \frac{1}{t^{\alpha -1}}$ and the convergence of the integral in (I) holds nearby $t=0$ if and only if $\alpha<2 $.

Now let take into play the case where $t $ is large.

case $\alpha\leq 0$

Employing integration by part, \begin{eqnarray*} \Big| \int_{\frac{\pi}{2}}^\infty \frac{\sin t}{t^\alpha}\,dt\Big| &= & \Big| -\alpha \int_{\frac{\pi}{2}}^\infty \frac{\cos t}{t^{\alpha+1}}\,dt\Big|\\ % &\leq & \alpha \int_{\frac{\pi}{2}}^\infty \frac{ 1 }{t^{\alpha+1}}\,dt< \infty \qquad\text{since} \qquad \alpha +1>1~~\text{with} ~~\alpha >0. \end{eqnarray*} Thus for $\alpha>0 $

$\varphi_1(\alpha)$ exists if and only if $0<\alpha<2$.

We will later these are the only values of $\alpha$ which guarantee the existence of $\varphi_1$. For now let have a look on the integrability of functions under (I). In other to see that, one can quickly check the following

$$ \mathbb{R}_+ = \bigcup_{n\in\mathbb{N}} [n\pi, (n+1)\pi).$$

Then, $$\int_0^\infty \frac{|\sin t|}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty} \int_{n\pi}^{(n+1)\pi} \frac{|\sin t|}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a_n$$

With suitable change of variable ($u = t-n\pi$) we get

\begin{eqnarray*} a_n &=& \int_{0}^{\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\qquad\text{since } \sin(t+n\pi)= (-1)^n\sin t \end{eqnarray*} On the oder hand, it is also easy to check

\begin{eqnarray} \frac{2}{(n+1\pi)^\alpha} \leq a_n \leq \frac{2}{(n\pi)^\alpha}. % \end{eqnarray} These inequality together with the Riemann sums show that the series of general terms $(a_n)_n$ and $(b_n)_n$ converge if and only if $\alpha>1.$ Moreover we have seen from the foregoing that

$$\int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt$$ converges only for $\alpha <2$

Taking profite of the tricks above, we get the result for the case $\alpha \leq 0$ as follows

$$\int_0^\infty \frac{\sin t}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty} \int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a'_n $$

With

\begin{eqnarray*} |a'_n| &=&\left|\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\right|= \left|\int_{0}^{\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\right| \geq \frac{2}{(\pi+n\pi)^\alpha} \qquad\qquad\text{since } \sin(t+n\pi) = (-1)^n\sin t . \end{eqnarray*}
and the equalities hold in both cases when $\alpha = 0.$ Therefore, $$\lim |a'_n|= \begin{cases} 2 &~~if ~~\alpha = 0 \nonumber\\ \infty & ~~if ~~\alpha <0. \nonumber \end{cases}$$ What prove that the divergence of the series $\sum\limits_{n=0}^{\infty} a'_n$ since $a_n'\not\to 0$. Consequently the left hand side of the previous relations always diverge since $\int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt $ converges for $\alpha\leq 0.$

Conclusion$ \frac{\sin t}{t^\alpha} $ converges for $0<\alpha<2$ and converges absolutely for $1<\alpha <2$.

$\endgroup$
-3
$\begingroup$

It is absolutely convergent for $1<s<2$. First, write the integral as

$\int_{0}^{\infty} \left|{\frac{\sin(x)}{x^s}}\right| \;dx = \int_{0}^{1} \left|\frac{\sin(x)}{x^s}\right| \;dx + \int_{1}^{\infty} \left|\frac{\sin(x)}{x^s} \right|\;dx$.

Then $ \int_{1}^{\infty} \left|\frac{\sin(x)}{x^s} \right|\;dx$ converges for any $s>1$ since $\int_{1}^{\infty} \left|\frac{1}{x^s} \right|\;dx$ converges for such $s$ (using the obvious fact that $|\sin(x)| \leq 1$). Moreover, this integral diverges for $s\leq 1$.

For the other summand, recall that $\frac{\sin(x)}{x} < 1$ on $(0,1]$, so we bound the integrand:

$\left|\frac{sin(x)}{x^s}\right| \leq \frac{1}{x^{s-1}}$

for $x\in (0,1].$

It is well known that the integral $\int_0^1 \frac{1}{x^p}\;dx$ converges for $0<p<1$, so $\int_0^1 \frac{1}{x^{s-1}}\;dx$ converges for $1<s<2$. It remains to show that this integral diverges for $s\geq 2$, but this could be accomplished by using your taylor series for $\sin(x)$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.