# Improper integral

• Sep 28th 2008, 04:58 PM
Nacho
Improper integral
Diverge?

$
\int_0^\infty {\frac{{\sin x}}
{x}dx}

$

thanks (Rock)
• Sep 28th 2008, 05:17 PM
Krizalid
It does converge. (Not absolutely, of course.)
• Sep 28th 2008, 05:23 PM
Nacho
Quote:

Originally Posted by Krizalid
It does converge. (Not absolutely, of course.)

why? (Thinking)
• Sep 29th 2008, 10:59 AM
Moo
Hello,

I'll try the proof I read a time ago...

Let $I_{a,b}=\int_a^b \frac{\sin(x)}{x} ~dx$

Integrate by parts with :
$u(x)=\frac 1x$ and $v'(x)=\sin(x)$. So we have $u'(x)=-\frac 1{x^2}$ and $v(x)=-\cos(x)+c$, and we'll take c=1. So $v(x)=1-\cos(x)$

$I_{a,b}=\left[\frac{1-\cos(x)}{x}\right]_a^b+\int_a^b \frac{1-\cos(x)}{x^2} ~dx$

But we know by Taylor series that $1-\cos(x) \approx \frac{x^2}{2}$ when x is near from 0.

So we can make a=0 since $\frac{1-\cos(x)}{x}$ and $\frac{1-\cos(x)}{x^2}$ have finite limits when $x \to 0$

$I_b=\left[\frac{1-\cos(x)}{x}\right]_a^b+\int_0^b \frac{1-\cos(x)}{x^2} ~dx$

$I_b=\frac{1-\cos(b)}{b}+\int_0^1 \frac{1-\cos(x)}{x^2} ~dx+\int_1^b \frac{1-\cos(x)}{x^2} ~dx$

$I=\lim_{b \to + \infty} I_b=\lim_{b \to + \infty} \frac{1-\cos(b)}{b}+\int_0^1 \frac{1-\cos(x)}{x^2} ~dx+\lim_{b \to + \infty} \int_1^b \frac{1-\cos(x)}{x^2} ~dx$

_________________________________________________
See $\int_0^1 \frac{1-\cos(x)}{x^2} ~dx$
By extension, the integrand is continuous for all x in [0,1] and finite. Thus the integral exists and it is finite value. So it doesn't intervene in the convergence or divergence of $I$
(actually, you have to check this explanation...it's the sort of parts in which I am always wrong)

See $\lim_{b \to + \infty} \frac{1-\cos(b)}{b}$
This obviously converges.

See $\lim_{b \to + \infty} \int_1^b \frac{1-\cos(x)}{x^2} ~dx$
This converges by comparison with the Riemann integral, since :
$\frac{1-\cos(x)}{x^2} \le \frac{2}{x^2}$.
• Sep 29th 2008, 06:50 PM
Nacho
thanks!! (Clapping)