Computing the sine integral: Difference between revisions

Revision as of 23:10, 28 April 2008

This article studies the computation of the following improper real integral:

$P V \int_{- \infty}^{\infty} \frac{\sin x}{x} = π$

Here, the value at $0$ is assigned to be 1. (The function being integrated is termed the sinc function and its indefinite integral is termed the sine integral.

Computation

We first consider the function:

$z \mapsto \frac{e^{i z}}{z}$

This is a holomorphic function and its imaginary part (when restricted to the real axis) is $x \mapsto (\sin x) / (x)$ .

We now choose the "mousehole contour" method of integration (further information at mousehole contour integration method). The contour is given by semicircles centered at the origin and in the upper half-plane of radii $R_{1}$ and $R_{2}$ , where $R_{1}$ tends to zero and $R_{2}$ tends to $\infty$ .

By Jordan's lemma, the part of the integral on the outer circle tends to zero. The part of the integral on the inner circle tends to $- i π$ , hence the integral along the real axis is $i π$ . Thus, its imaginary part is precisely $π$ .

@@ Line 1: / Line 1: @@
 This article studies the computation of the following improper real integral:
-<math>\int_{-\infty}^\infty \frac{\sin x}{x} = \pi</math>
+<math>\operatorname{PV} \int_{-\infty}^\infty \frac{\sin x}{x} = \pi</math>
 Here, the value at <math>0</math> is assigned to be 1. (The function being integrated is termed the [[sinc function]] and its indefinite integral is termed the [[sine integral]].
@@ Line 11: / Line 11: @@
 <math>z \mapsto \frac{e^{iz}}{z}</math>
-This is a holomorphic function and its imaginary part is <math>(\sin x)/(x)</math>.
+This is a holomorphic function and its imaginary part (when restricted to the real axis) is <math>x \mapsto (\sin x)/(x)</math>.
-By the [[Jordan's lemma]], we see that the integral
+We now choose the "mousehole contour" method of integration (further information at [[mousehole contour integration method]]). The contour is  given by semicircles centered at the origin and in the upper half-plane of radii <math>R_1</math> and <math>R_2</math>, where <math>R_1</math> tends to zero and <math>R_2</math> tends to <math>\infty</math>.
+By [[Jordan's lemma]], the part of the integral on the outer circle tends to zero. The part of the integral on the inner circle tends to <math>-i\pi</math>, hence the integral along the real axis is <math>i\pi</math>. Thus, its imaginary part is precisely <math>\pi</math>.