1. ## Improper integral

Diverge?

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

thanks

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

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

4. Hello,

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

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

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

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

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

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

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

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

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

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See $\displaystyle \int_0^1 \frac{1-\cos(x)}{x^2} ~dx$
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 $\displaystyle I$
(actually, you have to check this explanation...it's the sort of parts in which I am always wrong)

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

See $\displaystyle \lim_{b \to + \infty} \int_1^b \frac{1-\cos(x)}{x^2} ~dx$
This converges by comparison with the Riemann integral, since :
$\displaystyle \frac{1-\cos(x)}{x^2} \le \frac{2}{x^2}$.

5. thanks!!