Prove that this series converges. Its sum is $\frac\pi2$.

$$\int_0^{\pi}\frac{\sin x}x\;dx+\int_\pi^{2\pi}\frac{\sin x}x\;dx+\dots=\int_0^{\infty}\frac{\sin x}x\;dx$$

This is an exercise after a section on convergence tests. I tried to integrate but failed. Then I thought it should have something to do with the series. First, I just looked at the signs and thought all the terms were positive. But after plotting on Geogebra, I found some characteristics of the curve that make the series behave like an alternating series with decreasing terms (which makes it a sufficient condition for convergence): The curve crosses the x-axis at $\pi$ intervals, and every time it crosses it, it stays closer to it. But I just observed this on a graph. I still cannot show that the terms have alternating signs and are decreasing because they are very difficult to integrate. How can I prove it formally?

  • $\begingroup$ So you’re just trying to show that$$\int\limits_{0}^{\infty}\frac {\sin x}x\, dx=\frac {\pi}2$$? $\endgroup$ – Crescendo Oct 31 '17 at 15:35
  • $\begingroup$ See my answer Here . and replace by. $\endgroup$ – Guy Fsone Oct 31 '17 at 20:28

Use Leibinz criterium. The series is alternating: $$ \int_{2k\pi}^{(2k+1)\pi}\frac{\sin x}{x}\,dx>0,\quad\int_{(2k-1)\pi}^{2k\pi}\frac{\sin x}{x}\,dx<0. $$ Now you have to show that $$ \int_{k\pi}^{(k+1)\pi}\frac{\sin x}{x}\,dx=\pm\int_{0}^{\pi}\frac{\sin x}{x+k\,\pi}\,dx $$ is decreasing in absolute value.


Define $$a_k=\int_{k\pi}^{(k+1)\pi}\frac{\sin x}{x}\,dx$$ The series $$\sum_{k=0}^{\infty}a_k$$ is an alternating series, therefore we can use Leibniz criterion to prove that it converges.

$$\lim_{k\to\infty}a_k=0$$ Indeed thanks to the mean value theorem for any $k\in\mathbb{N}$ there exists $c_k\in[k\pi,(k+1)\pi]$ such that $$\int_{k\pi}^{(k+1)\pi}\frac{\sin x}{x}\,dx=\pi\frac{\sin c_k}{c_k}\to 0 \text { as }k\to\infty$$

Furthermore $|a_k|$ is monotonically decreasing

This is true because $$\left|\frac{\sin x}{x}\right|\leq \frac{1}{x}$$ therefore $$|a_n|=\left|\int_{k\pi}^{(k+1)\pi}\frac{\sin x}{x}\,dx\right|\leq \int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{x}\right|\,dx\leq\int_{k\pi}^{(k+1)\pi}\frac{1}{|x|}=\log\frac{k+1}{k}$$ which is decreasing for any $k\in\mathbb{N}$ because derivative $-\frac{1}{k^2+k}$ is negative for any $k$.

Thus for Leibniz criterion the series converges.

Hope this can be useful


All the zeroes of the sine function are of the form $n\pi$ with $n\in\mathbb{Z}$. Now consider the interval $\left[0;\pi\right]$. $\sin(x)=0$ at the endpoints of this interval. If there were $x_1,x_2$ in the interval $\left(0;\pi\right)$ such that $\sin(x_1)<0$ and $\sin(x_2)>0$, there would be a $c$ with $x_1<c<x_2$ such that $\sin(c)=0$ by the Intermediate Value Theorem since $\sin(x)$ is a continuous function. However, this is absurd since there is no number of the form $n\pi$ with $n\in\mathbb{Z}$ on the interval $(0;\pi)$. Thus, the function $\sin(x)$ does not change signs on the interval $(0;\pi)$. Now $\sin\left(\frac{\pi}{2}\right)=1$ and we conclude $\sin(x)>0$ for all $x\in(0;\pi)$. In a similar fashion we can show that $\sin(x)<0$ for all $x\in(\pi;2\pi)$. Since $\sin(x)$ is periodic with period $2\pi$ we conclude

$$\sin(x)\ge0\,\forall x\in[2n\pi;(2n+1)\pi]\qquad,n\in\mathbb{N}$$ $$\sin(x)\le0\,\forall x\in[(2n+1)\pi;(2n+2)\pi]\qquad,n\in\mathbb{N}$$ This allows us to rewrite the series in the following way: \begin{align*} \sum\limits_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{\sin(x)}{x}\text{d}x&=\sum\limits_{n=0}^\infty\int_{2n\pi}^{(2n+1)\pi}\frac{\sin(x)}{x}\text{d}x+\sum\limits_{n=0}^\infty\int_{(2n+1)\pi}^{(2n+2)\pi}\frac{\sin(x)}{x}\text{d}x\\&=\sum\limits_{n=0}^\infty\int_{2n\pi}^{(2n+1)\pi}\left|\frac{\sin(x)}{x}\right|\text{d}x-\sum\limits_{n=0}^\infty\int_{(2n+1)\pi}^{(2n+2)\pi}\left|\frac{\sin(x)}{x}\right |\text{d}x\\&=\sum\limits_{n=0}^\infty(-1)^n\int_{n\pi}^{(n+1)\pi}\left|\frac{\sin(x)}{x}\right|\text{d}x \end{align*} Consider the sequence $a_n=\int_{n\pi}^{(n+1)\pi}\left|\frac{\sin(x)}{x}\right|\text{d}x$. We notice the inequality $\int_a^b\left|f(x)\right|\text{d}x\le(b-a)\max_{x\in[a;b]}(f(x))$ and write \begin{equation} \int_{n\pi}^{(n+1)\pi}\left|\frac{\sin(x)}{x}\right|\text{d}x\le((n+1)\pi-n\pi)\max_{x\in[n\pi;(n+1)\pi]}\left(\frac{\sin(x)}{x}\right)\le\pi\max_{x\in[n\pi;(n+1)\pi]}\left(\frac{1}{x}\right)=\frac{1}{n} \end{equation} $0\le\left|\sin(x)/x\right|$ yields $0\le a_n\le1/n$ and thus $\lim_{n\rightarrow\infty}a_n=0$. We now want to show $a_n\ge a_{n+1}$: \begin{equation} \int_{n\pi}^{(n+1)\pi}\left|\frac{\sin(x)}{x}\right|\text{d}x\ge\int_{(n+1)\pi}^{(n+2)\pi}\left|\frac{\sin(x)}{x}\right|\text{d}x \end{equation} Use the substitution $\varphi(y)=y+n\pi$ for the left and $\varphi(y)=y+(n+1)\pi$ for the right integral to achieve \begin{equation} \int_0^\pi\left|\frac{\sin(y+n\pi)}{y+n\pi}\right|\text{d}y\ge\int_0^\pi\left|\frac{\sin(y+(n+1)\pi)}{y+(n+1)\pi}\right|\text{d}y \end{equation} Now for all natural $n$ we have $\sin(y+n\pi)=(-1)^n\sin(x)$ and since $\sin(y)\ge0$ on the interval $[0;\pi]$ this is equivalent to \begin{equation} \int_0^\pi\frac{\sin(y)}{y+n\pi}\text{d}x\ge\int_0^\pi\frac{\sin(y)}{y+(n+1)\pi}\text{d}x \end{equation} Since $f(x)\ge g(x)$ for all $x\in I$ implies $\int_If(x)\text{d}x\ge\int_Ig(x)\text{d}x$, it suffices to prove \begin{equation} \frac{\sin(y)}{y+n\pi}\ge\frac{\sin(y)}{y+(n+1)\pi}\Leftrightarrow\frac{1}{y+n\pi}\ge\frac{1}{y+(n+1)\pi}\Leftrightarrow y+(n+1)\pi\ge y+n\pi\Leftrightarrow\pi\ge0 \end{equation} , which is true. Thus, $a_n$ is a monotonically decreasing sequence with $\lim_{n\rightarrow\infty}a_n=0$ and - by the Alternating Series Test - the series in question converges.


More General result(for your Question take $\alpha = 1$)

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

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

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


\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 $1<\alpha <2$.


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