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$$I(k) = \int_{0}^{∞} \frac{\sin(kx)}{x} dx$$

I have no idea how to evaluate this.... but I'm guessing the method of steepest descents (saddle point method) may work.... but not sure.

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You can use double dollar signs to get displayed equations, which look better especially for things like integrals with limits. By the way, your integral is missing the integration variable. – joriki Oct 10 '12 at 0:07
Are you talking about $I(k)$ instead of $I(x)$? – Jack Oct 10 '12 at 0:18
thank you..... didn't notice that. – Squirtle Oct 10 '12 at 0:31
The integration variable is still missing. – joriki Oct 10 '12 at 0:37
i thought that was there ..... i put it now – Squirtle Oct 10 '12 at 0:45
up vote 1 down vote accepted


$$\forall\,0<M\in\Bbb R\;\;\;,\;\,\Gamma_M:=\{z\in\Bbb C\;:\;|z|=M\,,\,\Im(z)>0\Longrightarrow z=Me^{it}\,,\,0\leq t\leq\pi\}$$

we define the complex path

$$\Gamma:=[-R,-\epsilon]\cup\Gamma_\epsilon\cup[\epsilon,R]\cup\Gamma_R\,\,\,,\,\,\epsilon,R\in\Bbb R^+\,\,,\,\epsilon<<R$$

Defining now $\,\displaystyle{f(z):=\frac{e^{ikz}}{z}}\,$ and using the corollary to the lemma in this answer , we get (all the integrals are evaluated following the positive orientation):

$$Res_{z=0}(f)=\lim_{z\to 0}e^{ikz}=1\Longrightarrow\int_{\Gamma_\epsilon}f(z)\,dz=\pi i$$

And since Jordan's Lemma gives us at once $\,\displaystyle{\int_{\Gamma_R}f(z)\,dz\xrightarrow [R\to\infty]{}0}\,$ -- or directly:

$$\left|\int_{\Gamma_R}\frac{e^{ikz}}{z}dz\right|\leq\max_{|z|=R}\frac{e^{-Rk}}{R}\pi R\xrightarrow[R\to\infty]{}0\;\;\text{(as long as}\,k>0)--$$

we get (since $\,f(z)\,$ is holomorphic within $\,\Gamma\,$):

$$0=\oint_\Gamma f(z)\,dz\xrightarrow[R\to\infty]{}\int_{-\infty}^\infty\frac{e^{ikx}}{x}dx-\pi i$$

and taking the imaginary part above we get

$$\int_{-\infty}^\infty\frac{i\sin kx}{x}dx=\pi i\Longrightarrow \int_0^\infty\frac{\sin kx}{x}dx=\frac{\pi}{2}$$

since the integrand is an even function.

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You can use the fact that the sinc function is the Fourier transform of a rectangle and then apply the Plancherel theorem.

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Very roughly since the actual solution is long to type out:

You can calculate this definite integral by using a method called "contour integration," which informally I think you can describe as line integration for complex variables.

The method behind contour integration is to take a complex-valued function, integrate it along some closed curve, and then extract the real-variable integral you want from that integration. For example, you can say:

$$ I(k) = \Im \displaystyle \int_{-\infty}^{\infty} \dfrac{e^{ikx}}{x} \ dx $$

Now you can evaluate instead $\displaystyle \int_{\gamma} \dfrac{e^{ikz}}{z} \ dz$, where $\gamma$ is the contour that runs from $-R$ to $R$ along the real axis, with a small concave down infinitesimal semi-circle at the origin, and then along the semi-circle from $R$ to $-R$.

Notice that in the limit, along the real-axis you get $\displaystyle \int_{-\infty}^{\infty} \dfrac{e^{ikz}}{z} \ dz $. By use of Jordan's lemma (see Wikipedia), you can show that the integral along the semi-circle goes to zero.

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Shouldn't that be the Imaginary part of I(k)? Also why are the limits of integration changing? – Squirtle Oct 10 '12 at 5:11
Ah, yes that should be the imaginary part not the real part. The limits of integration are changed for symmetry sake. – Michael Zhao Oct 10 '12 at 14:08

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