1.a 1 & lt； three

Because x & gt=0, sinx

A 1 = integer (0, pi/2)sinx/x dx+ integer (pi/2, pi) sinx /x dx.

& lt= 1 * pi/2+ 1/(pi/2)* pi/2 = pi/2+ 1 & lt； three

2. In any integral interval of, the sign of the integrand does not change.

3.|An| >|A(n+ 1)|

| an |-| a (n+1) | = integral ((n-1) pi, NPI) | sinx | (1/x-1(x+pi) dx > 0 because of the integrand function. 0

4.{An} is a positive and negative staggered sequence.

So let 0

| integral (a, b) sinx/xdx | < =| am+a (m+1)+...+an |

& lt= | Am | & lt= a 1 & lt； three

Let's explain that "by properly enlarging or narrowing the interval, the end point of the interval is npi hours"

There is n 1

Assumption: integral (a, b) sinx/xdx > 0, Sina & gt0, sinb & lt0

Then because the integral is >: 0,

Take the interval where m=n 1 and has positive amplification.

Take the interval where n=n2- 1 minus the negative function. .

| integral (a，b)sinx/xdx | < = | Integral((n 1- 1)pi，(N2- 1) pi) sinx/xdx |

= |Am+...+An||

In other cases, the treatment is similar.