# Math Help - covergence

1. ## covergence

show that int(sin^2(x)/ (x^2))dx from 0 to infinity converges and THEN use integration by parts and the identity sin(2x)=2sin(x)cos(x) to show that

int(sin^2(x)/ (x^2))dx from 0 to infinity = int(sin(x) / x) dx

2. Hello,

You can see here : http://www.mathhelpforum.com/math-he...-integral.html for something similar, proving it's convergent.

(similar method as (1-cos)/x²)

As for the integration by parts, integrate 1/x² and differentiate sin²(x) (which will give 2sin(x)cos(x), and this is where you'll use the identity)
And this would rather be int(sin(2x)/x), wouldn't it ?

3. Hello.
$\ 0 \leq sin^2 x \leq 1$ so $\int_1^{+\infty} \frac{sin^2 x}{x^2}dx$ converges and we can use $\lim_{x\to 0} \dfrac{\sin x}x = 1$ to show that
$\int_0^1 \frac{sin^2 x}{x^2}dx$ converges.