Prove the inequality$\int_0^{+\infty} {\sin x \over x}dx<\int_0^\pi {\sin x \over x}dx$ Prove:
$$\int_0^{+\infty} {\sin x \over x}dx<\int_0^\pi {\sin x \over x}dx$$
Here is my answer,but I want a different way to prove it.
\begin{aligned}
\int_0^{+\infty} {\sin x \over x}dx&=\lim_{x \to +\infty} \int_0^x {\sin t \over t} dt\\
&=\lim_{n\to +\infty} \int_0^{n\pi} {\sin t \over t} dt\\
&=\sum_{i=1}^{+\infty} \int_{(i-1)\pi}^{i\pi } {\sin t \over t} dt
\end{aligned}
let $a_n=\int_{(n-1)\pi}^{n\pi } \mid {\sin t \over t} \mid dt$ 
then $$\sum (-1)^n a_n=\sum_{i=1}^{+\infty} \int_{(i-1)\pi}^{i\pi } {\sin t \over t} dt$$
easy to prove $\sum (-1)^n a_n$ converges
s.t$\mid r_0\mid=\mid S-S_0\mid<a_1$
s.t$$\mid\sum (-1)^n a_n\mid<\int_0^\pi {\sin x \over x}dx$$
s.t
$$\mid\int_0^{+\infty} {\sin x \over x}dx\mid<\int_0^\pi {\sin x \over x}dx$$
 A: For every $x\geqslant0$, consider $$I(x)=\int_0^x\frac{\sin t}t\mathrm dt,$$ then the variations of $I(x)$ are given by the sign of $\sin(x)$, that is, the function $I$ increases on $[0,\pi]$ then decreases on $[\pi,2\pi]$, ..., then increases on each $[2n\pi,(2n+1)\pi]$ then decreases  on each $[(2n+1)\pi,(2n+2)\pi]$. Thus, everything becomes obvious once one can show that the sequence $$x_n=(-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\sin t}t\mathrm dt,$$ is decreasing. But $$x_n=\int_{0}^{\pi}\frac{\sin t}{n\pi+t}\mathrm dt,$$ hence the conclusion follows, for example, $$\int_0^\infty\frac{\sin t}t\mathrm dt=\lim_{x\to\infty}I(x)=\sum_{n\geqslant0}(-1)^nx_n\lt x_0=\int_0^\pi\frac{\sin t}t\mathrm dt.$$
Edit: Another presentation of the same idea: note that $$\int_0^\infty\frac{\sin t}t\mathrm dt=\int_0^\pi S(t)\sin t\, \mathrm dt,\qquad S(t)=\sum_{n\geqslant0}\frac{(-1)^n}{n\pi+t}.$$ The series $S(t)$ is alternating hence (it converges and), for every $t$ in $(0,\pi)$, $$S(t)\lt\frac1t.$$ In particular, $$\int_0^\pi S(t)\sin t\, \mathrm dt\lt\int_0^\pi\frac1t\sin t\, \mathrm dt.$$
A: 
area of sinx/x from 0 to pi is more than area of that function from pi to 2pi  . and so on , area between 2pi to 3pi is less than area of pi to 2pi .so : if we name them to s1 , s2 , s3 , s4 ,...
s1 = area between 0 to pi 
s2=area between pi tp 2pi ...
s1>s2>s3>s4>s5>...
integral from 0 to npi = s1-s2 +s3 -s4+ s5 -s6 ...=s1- (s2-s3 +s4-s5 ...)
(s2-s3 +s4-s5 ...) >0
s1>s1-(s2-s3 +s4-s5...)
A: Not using series is not easy, since the inner reason why this inequality is true is because of properties of alternate series with a decreasing term ->0, however one could try to prove that:
For every N>1 : $\int_\pi^{N{\pi}}\frac{\sin t}t\mathrm dt  <0 $ using induction.
First case: n is odd (ie sin(t) < 0 on $]N\pi,(N+1)\pi[$ ) is obvious since you add a negative term.
To treat the case : n is even, you have to use an induction on all the previous term (two is enough actually) and write :
$\int_\pi^{(N+1){\pi}}\frac{\sin t}t\mathrm dt  = \int_\pi^{(N-1){\pi}}\frac{\sin t}t\mathrm dt  + \int_{(N-1)\pi}^{N{\pi}}\frac{\sin t}t\mathrm dt + \int_{N\pi}^{(N+1){\pi}}\frac{\sin t}t\mathrm dt  $
The term that goes from 1 to N-1 is <0 with the hypothesis , and all you have to prove is that the sum of the two other terms is <0, which is not very complicated using the change of variable $t= u+ N*\pi$
Now you can affirm that: $\int_\pi^{N{\pi}}\frac{\sin t}t\mathrm dt  <0$  for every N>1, so knowing that the integral has a limit you can write N->+∞ and you get :
$\int_\pi^{+∞}\frac{\sin t}t\mathrm dt  <0$ which is your result
