Sine integral

**liyi** · November 30th 2007, 05:48 PM

Evaluate $\int_{-\infty}^{+\infty}\frac{\sin x}x\,dx$

**kalagota** · November 30th 2007, 06:06 PM

Originally Posted by liyi

Evaluate $\int_{-\infty}^{+\infty}\frac{\sin x}x\,dx$

$\int_{-\infty}^{+\infty}\frac{\sin x}x\,dx = \int_{-\infty}^0 \frac{\sin x}x\,dx + \int_0^{+\infty} \frac{\sin x}x\,dx$

$= \lim_{a \rightarrow -\infty} \int_a^0 \frac{\sin x}x\,dx + \lim_{b \rightarrow +\infty} \int_0^b \frac{\sin x}x\,dx$

now, integrate by parts.. continue..

**galactus** · November 30th 2007, 06:22 PM

I don't think sin(x)/x is readily done by parts or any other elementary operation. This is a case for PH's forte, complex analysis.

You can start by using $\frac{e^{iz}}{z}$ , because it has a simple pole at z=0 and has a residue $e^{i(0)}=1$

I have been studying a little CA when I have time, but don't have my sea legs yet.

**kalagota** · November 30th 2007, 06:40 PM

yeah, i was solving it using ibp but i don't get anything useful..
about using CA concepts, it is possible if s/he has taken the course.. personally, i haven't taken it so i can't even use it.. Ü

**Krizalid** · November 30th 2007, 07:00 PM

Originally Posted by liyi

Evaluate $\int_{-\infty}^{+\infty}\frac{\sin x}x\,dx$

$\frac1x=\int_0^\infty e^{-ux}\,du.$ so,

$\int_{ - \infty }^{ + \infty } {\frac{{\sin x}} {x}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin x\,du} \,dx} .$

Now the remaining challenge is to compute ${\int_0^\infty {e^{ - ux} \sin x\,dx} }.$ (Reverse integration order.)

This is nasty applyin' integration by parts, but it's quickly considering the following:

$\int_0^\infty {e^{ - ux} \sin x\,dx} = \text{Im} \int_0^\infty {e^{ - (u - i)x} \,dx} = \text{Im} \frac{1} {{u - i}} = \frac{1} {{u^2 + 1}}.$

And finally

$\int_{ - \infty }^{ + \infty } {\frac{{\sin x}} {x}\,dx} = 2\int_0^\infty {\frac{1} {{1 + u^2 }}\,du} = \pi .$

**galactus** · December 1st 2007, 03:34 AM

Whoa, Kriz, you show off you. I am now bowing to the master. Very clever indeed.

**ThePerfectHacker** · December 1st 2007, 02:20 PM

Let $f(z) = \frac{e^{iz}}{z}$ . Consider the square contour $C$ (for $A,B>0$ large) from $-A$ to $B$ with $\gamma$ be a semicircular contour from $-r$ to $r$ (for $r>0$ small).
Then by Cauchy's theorem*:
$\int_{r}^{\infty}\frac{e^{ix}}{x}dx + \int_{-\infty}^{-r} \frac{e^{ix}}{x} dx - \int_{\gamma} f(z)dz = 0$
Now make $r\to 0^+$ to get:
$\int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx = \pi i$
Equate real and imaginary parts.

*)The issue of convergence is covered by Jordan's lemma.

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