Computing the sine integral: Difference between revisions

Latest revision as of 19:11, 18 May 2008

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

$\int _{-\infty }^{\infty }{\frac {\sin x}{x}}=\pi$

We'll compute the Cauchy principal value of the integral, but this is the same as the actual integral since the integral is absolutely convergent.

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

Setting up the complex-valued function

We do not proceed to solve this problem by considering the sinc function itself as a complex-valued function, because that turns out not to have good decay properties at infinity (in other words, it doesn't become sufficiently small on large semicircles). Rather, we choose the function:

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

Some quick observations about this function:

It is holomorphic on $\mathbb {C} ^{*}$ . In particular, it is holomorphic on the upper half-plane
It has a simple pole at $0$
The imaginary part of its restriction to the real axis, is precisely the sinc function

Computing integrals over mousehole contours

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 $-i\pi$ , hence the integral along the real axis is $i\pi$ . Thus, its imaginary part is precisely $\pi$ .

@@ Line 1: / Line 1: @@
+{{contour integral computation|mousehole}}
 This article studies the computation of the following improper real integral:
-<math>\operatorname{PV} \int_{-\infty}^\infty \frac{\sin x}{x} = \pi</math>
+<math>\int_{-\infty}^\infty \frac{\sin x}{x} = \pi</math>
+We'll compute the [[Cauchy principal value]] of the integral, but this is the same as the actual integral since the integral is absolutely convergent.
 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 7: / Line 11: @@
 ==Computation==
-We first consider the function:
+===Setting up the complex-valued function===
+We do not proceed to solve this problem by considering the [[sinc function]] itself as a complex-valued function, because that turns out not to have good decay properties at infinity (in other words, it doesn't become sufficiently small on large semicircles). Rather, we choose the function:
 <math>z \mapsto \frac{e^{iz}}{z}</math>
-This is a holomorphic function and its imaginary part (when restricted to the real axis) is <math>x \mapsto (\sin x)/(x)</math>.
+Some quick observations about this function:
+* It is [[holomorphic function|holomorphic]] on <math>\mathbb{C}^*</math>. In particular, it is holomorphic on the [[upper half-plane]]
+* It has a [[simple pole]] at <math>0</math>
+* The imaginary part of its restriction to the real axis, is precisely the sinc function
+===Computing integrals over mousehole contours===
 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>.