Computing the sine integral - Revision history

Vipul: 7 revisions

2008-05-18T19:11:35Z

7 revisions

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(No difference)

Vipul at 17:52, 2 May 2008

2008-05-02T17:52:12Z

← Older revision		Revision as of 17:52, 2 May 2008
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	This article studies the computation of the following improper real integral:		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]].		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]].

Vipul: /* Setting up the complex-valued function */

2008-04-28T23:57:21Z

Setting up the complex-valued function

← Older revision		Revision as of 23:57, 28 April 2008
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	* It is [[holomorphic function\|holomorphic]] on <math>\mathbb{C}^*</math>. In particular, it is holomorphic on the [[upper half-plane]]		* It is [[holomorphic function\|holomorphic]] on <math>\mathbb{C}^*</math>. In particular, it is holomorphic on the [[upper half-plane]]
	* It has ~~an essential singularity~~ at <math>0</math>		* It has a [[simple pole]] at <math>0</math>
	* The imaginary part of its restriction to the real axis, is precisely the sinc function		* The imaginary part of its restriction to the real axis, is precisely the sinc function

Vipul at 23:23, 28 April 2008

2008-04-28T23:23:22Z

← Older revision		Revision as of 23:23, 28 April 2008
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			{{contour integral computation\|mousehole}}

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

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	==Computation==		==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>		<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 an essential singularity 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>.		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>.		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>.

Vipul at 23:10, 28 April 2008

2008-04-28T23:10:59Z

← Older revision		Revision as of 23:10, 28 April 2008
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	This article studies the computation of the following improper real integral:		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]].		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]].
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	<math>z \mapsto \frac{e^{iz}}{z}</math>		<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>.

Vipul at 20:42, 28 April 2008

2008-04-28T20:42:02Z

← Older revision		Revision as of 20:42, 28 April 2008
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	This is a holomorphic function and its imaginary part is <math>(\sin x)/(x)</math>.		This is a holomorphic function and its imaginary part is <math>(\sin x)/(x)</math>.

			By the [[Jordan's lemma]], we see that the integral

Vipul at 20:32, 28 April 2008

2008-04-28T20:32:31Z

← Older revision		Revision as of 20:32, 28 April 2008
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	This article studies the computation of the following improper real integral:		This article studies the computation of the following improper real integral:

	<math>\int_{\-infty}^\infty \frac{\sin x}{x} = \pi</math>		<math>\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]].		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]].

Vipul: New page: This article studies the computation of the following improper real integral: $\int_{\-infty}^\infty \frac{\sin x}{x} = \pi$ Here, the value at $0$ is assigned to b...

2008-04-28T20:32:05Z

New page: This article studies the computation of the following improper real integral: <math>\int_{\-infty}^\infty \frac{\sin x}{x} = \pi</math> Here, the value at <math>0</math> is assigned to b...

New page

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

<math>\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]].

==Computation==

We first consider the function:

<math>z \mapsto \frac{e^{iz}}{z}</math>

This is a holomorphic function and its imaginary part is <math>(\sin x)/(x)</math>.