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

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- Nov 30th 2007, 05:48 PMliyiSine integral
Evaluate $\displaystyle \int_{-\infty}^{+\infty}\frac{\sin x}x\,dx$

- Nov 30th 2007, 06:06 PMkalagota
$\displaystyle \int_{-\infty}^{+\infty}\frac{\sin x}x\,dx = \int_{-\infty}^0 \frac{\sin x}x\,dx + \int_0^{+\infty} \frac{\sin x}x\,dx$

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

now, integrate by parts.. continue.. - Nov 30th 2007, 06:22 PMgalactus
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 $\displaystyle \frac{e^{iz}}{z}$, because it has a simple pole at z=0 and has a residue $\displaystyle e^{i(0)}=1$

I have been studying a little CA when I have time, but don't have my sea legs yet. - Nov 30th 2007, 06:40 PMkalagota
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 - Nov 30th 2007, 07:00 PMKrizalid
$\displaystyle \frac1x=\int_0^\infty e^{-ux}\,du.$ so,

$\displaystyle \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 $\displaystyle {\int_0^\infty {e^{ - ux} \sin x\,dx} }.$ (Reverse integration order.)

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

$\displaystyle \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

$\displaystyle \int_{ - \infty }^{ + \infty } {\frac{{\sin x}}

{x}\,dx} = 2\int_0^\infty {\frac{1}

{{1 + u^2 }}\,du} = \pi .$ - Dec 1st 2007, 03:34 AMgalactus
Whoa, Kriz, you show off you. I am now bowing to the master. Very clever indeed.(Clapping)

- Dec 1st 2007, 02:20 PMThePerfectHacker
Let $\displaystyle f(z) = \frac{e^{iz}}{z}$. Consider the square contour $\displaystyle C$ (for $\displaystyle A,B>0$ large) from $\displaystyle -A$ to $\displaystyle B$ with $\displaystyle \gamma$ be a semicircular contour from $\displaystyle -r$ to $\displaystyle r$ (for $\displaystyle r>0$ small).

Then by Cauchy's theorem*:

$\displaystyle \int_{r}^{\infty}\frac{e^{ix}}{x}dx + \int_{-\infty}^{-r} \frac{e^{ix}}{x} dx - \int_{\gamma} f(z)dz = 0$

Now make $\displaystyle r\to 0^+$ to get:

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx = \pi i$

Equate real and imaginary parts.

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