Proving convergence of $\int_0^{\pi}\frac{\sin x}x\;dx+\int_\pi^{2\pi}\frac{\sin x}x\;dx+\dots$ 
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?
 A: 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.
A: 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.
A: 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
A: 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
$$\int_0^\infty \frac{\sin t}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty}  \int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a'_n $$
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 $1<\alpha <2$.

