Improper integral

Nacho · #1 $Old$ September 28th, 2008, 05:58 PM

Diverge?

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

thanks

Krizalid · #2 $Old$ September 28th, 2008, 06:17 PM

It does converge. (Not absolutely, of course.)

Nacho · #3 $Old$ September 28th, 2008, 06:23 PM

Quote:

Originally Posted by Krizalid $View Post$

It does converge. (Not absolutely, of course.)

why?

Moo · #4 $Old$ September 29th, 2008, 11:59 AM

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 $x \to 0$

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

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

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

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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 :
$\frac{1-\cos(x)}{x^2} \le \frac{2}{x^2}$ .

Nacho · #5 $Old$ September 29th, 2008, 07:50 PM

thanks!!