en>Tom.Reding: Gen fixes (page/s, endash, , et al., unicodify, and/or concising wikilinks, etc.), & ref cleanup using AWB

2015-01-06T03:14:04Z

Gen fixes (page/s, endash, &nbsp, et al., unicodify, and/or concising wikilinks, etc.), & ref cleanup using AWB

200.104.245.226: /* Jeans mass */ rm promotion of particular views, excessive detail and unnecessary formatting

2013-11-13T04:41:09Z

Jeans mass: rm promotion of particular views, excessive detail and unnecessary formatting

← Older revision		Revision as of 06:41, 13 November 2013
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	~~Greetings! I am Marvella~~ and ~~I really feel comfortable~~ when ~~people use~~ the ~~full name~~. ~~To do aerobics~~ is a ~~factor that I~~'~~m completely addicted~~ to. ~~Hiring~~ is ~~my occupation~~. ~~Years~~ in the ~~past he moved~~ to ~~North Dakota~~ and ~~his family members enjoys it~~.<br><br>~~my web~~-~~site~~; ~~std home test~~ ([~~http~~://~~Www~~.~~pinaydiaries~~.~~com~~/~~user~~/~~LConsiden~~ i ~~thought about this~~])		In [[complex analysis]], '''Jordan's lemma''' is a result frequently used in conjunction with the [[residue theorem]] to evaluate [[contour integral]]s and [[improper integral]]s. It is named after the French mathematician [[Camille Jordan]].

			==Statement==
			Consider a [[complex number\|complex]]-valued, [[continuous function]] ''f'', defined on a semicircular contour

			:<math>C_R=\{z : z=R e^{i \theta}, \theta\in [0,\pi]\}</math>

			of radius ''R'' > 0 lying in the upper half-plane, centred at the origin. If the function ''f'' is of the form

			:<math>f(z)=e^{iaz} g(z)\,,\quad z\in C_R,</math>

			with a parameter ''a'' > 0, then Jordan's lemma states the following upper bound for the contour integral:

			:<math>\biggl\|\int_{C_R} f(z)\, dz\biggr\| \le \frac\pi{a}\max_{\theta\in [0,\pi]} \bigl\|g \bigl(R e^{i \theta}\bigr)\bigr\|\,.</math>

			where equal sign is when ''g(z)'' is identically zero. An analogous statement for a semicircular contour in the lower half-plane holds when ''a'' < 0.

			===Remarks===
			* If ''f'' is defined and continuous on the semicircular contour ''C<sub>R</sub>'' for all large ''R'' and

			::<math>M_R:=\max_{\theta\in [0,\pi]} \bigl\|g \bigl(R e^{i \theta}\bigr)\bigr\| \to 0\quad \mbox{as } R \to \infty\,,\qquad(*)</math>

			:then by Jordan's lemma

			::<math>\lim_{R \to \infty} \int_{C_R} f(z)\, dz = 0.</math>

			* For the case ''a'' = 0, see the [[estimation lemma]].

			* Compared to the estimation lemma, the upper bound in Jordan's lemma does not explicitly depend on the length of the contour ''C<sub>R</sub>''.

			==Application of Jordan's lemma==
			[[Image:Jordan's Lemma.svg\|right\|thumb\|300px\|The path ''C'' is the concatenation of the paths ''C''<sub>1</sub> and ''C''<sub>2</sub>.]]
			Jordan's lemma yields a simple way to calculate the integral along the real axis of functions {{nowrap\|''f'' (''z'') {{=}} ''e<sup>iaz</sup>g''(''z'')}} [[holomorphic]] on the upper half-plane and continuous on the closed upper half-plane, except possibly at a finite number of non-real points ''z''<sub>1</sub>, ''z''<sub>2</sub>, ..., ''z<sub>n</sub>''. Consider the closed contour ''C'', which is the concatenation of the paths ''C''<sub>1</sub> and ''C''<sub>2</sub> shown in the picture. By definition,

			:<math>\oint_{C} f(z)\, dz = \int_{C_1}f(z)\,dz + \int_{C_2} f(z)\,dz\,.</math>

			Since on ''C''<sub>2</sub> the variable ''z'' is real, the second integral is real:

			:<math>\int_{C_2} f(z)\,dz = \int_{-R}^{R} f(x)\,dx\,.</math>

			The left-hand side may be computed using the [[residue theorem]] to get, for all ''R'' larger than the maximum of \|''z''<sub>1</sub>\|, \|''z''<sub>2</sub>\|, ..., \|''z<sub>n</sub>''\|,

			:<math>\oint_{C} f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{Res}(f, z_k)\,,</math>

			where {{nowrap\|Res(''f'', ''z<sub>k</sub>'')}} denotes the [[Residue (complex analysis)\|residue]] of ''f'' at the singularity ''z<sub>k</sub>''. Hence, if ''f'' satisfies condition (*), then taking the limit as ''R''  tends to infinity, the contour integral over ''C''<sub>1</sub> vanishes by Jordan's lemma and we get the value of the improper integral

			:<math>\int_{-\infty}^{\infty} f(x)\,dx = 2\pi i \sum_{k=1}^n \operatorname{Res}(f, z_k)\,.</math>

			==Example==
			The function

			:<math>f(z)=\frac{e^{iz}}{1+z^2},\qquad z\in{\mathbb C}\setminus\{i,-i\},</math>

			satisfies the condition of Jordan's lemma with ''a'' = 1 for all ''R'' > 0 with ''R'' ≠ 1. Note that, for ''R'' > 1,

			:<math>M_R=\max_{\theta\in[0,\pi]}\frac1{\|1+R^2e^{2i\theta}\|}=\frac1{R^2-1}\,,</math>

			hence (*) holds. Since the only singularity of ''f'' in the upper half plane is at ''z'' = ''i'', the above application yields

			:<math>\int_{-\infty}^\infty \frac{e^{ix}}{1+x^2}\,dx=2\pi i\,\operatorname{Res}(f,i)\,.</math>

			Since ''z'' = ''i'' is a [[simple pole]] of ''f'' and {{nowrap\|1 + ''z''<sup>2</sup>}} = {{nowrap\|(''z'' + ''i'')(''z'' - ''i'')}}, we obtain
			:<math>\operatorname{Res}(f,i)=\lim_{z\to i}(z-i)f(z)
			=\lim_{z\to i}\frac{e^{iz}}{z+i}=\frac{e^{-1}}{2i}</math>

			so that

			:<math>\int_{-\infty}^\infty \frac{\cos x}{1+x^2}\,dx=\operatorname{Re}\int_{-\infty}^\infty \frac{e^{ix}}{1+x^2}\,dx=\frac{\pi}{e}\,.</math>

			This result exemplifies how some integrals difficult to compute with classical tools are easily tackled with the help of complex analysis.

			==Proof of Jordan's lemma==
			By definition of the [[Line_integral#Complex_line_integral\|complex line integral]],

			:<math>\begin{align}
			\int_{C_R} f(z)\, dz
			&=\int_0^\pi g(Re^{i\theta})\,e^{iaR(\cos\theta+i \sin\theta)}\,i Re^{i\theta}\,d\theta\\
			&=R\int_0^\pi g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta}\,d\theta\,.
			\end{align}</math>

			Now the inequality

			:<math>\biggl\|\int_a^b f(x)\,dx\biggr\|\le\int_a^b \|f(x)\|\,dx</math>

			yields

			:<math>\begin{align}
			I_R:=\biggl\|\int_{C_R} f(z)\, dz\biggr\|
			&\le R\int_0^\pi\bigl\|g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta} \bigr\|\,d\theta\\
			&=R\int_0^\pi \bigl\|g(Re^{i\theta})\bigr\|\,e^{-aR\sin\theta}\,d\theta\,.
			\end{align}</math>

			Using ''M<sub>R</sub>'' as defined in (*) and the symmetry {{nowrap\|sin ''θ''}} = {{nowrap\|sin(''π'' – ''θ'')}}, we obtain

			:<math> I_R \le RM_R\int_0^\pi e^{-aR\sin\theta}\,d\theta = 2RM_R\int_0^{\pi/2} e^{-aR\sin\theta}\,d\theta\,.</math>

			Since the graph of {{nowrap\|sin ''θ''}} is [[concave function\|concave]] on the interval {{nowrap\|''θ'' ∈ <nowiki>[</nowiki>0,''π'' /2<nowiki>]</nowiki>}}, the graph of {{nowrap\|sin ''θ''}} lies above the straight line connecting its endpoints, hence

			:<math>\sin\theta\ge \frac{2\theta}{\pi}\quad</math>

			for all {{nowrap\|''θ'' ∈ <nowiki>[</nowiki>0,''π'' /2<nowiki>]</nowiki>}}, which further implies

			:<math>I_R
			\le 2RM_R \int_0^{\pi/2} e^{-2aR\theta/\pi}\,d\theta
			=\frac{\pi}{a} (1-e^{-a R}) M_R\le\frac\pi{a}M_R\,.</math>

			==See also==
			*[[Estimation lemma]]

			==References==
			* {{Cite book\| last1=Brown\| first1=James W.\| last2=Churchill\| first2=Ruel V.\| date=2004\| title=Complex Variables and Applications\| edition=7th\|place=New York \| publisher=McGraw Hill\| pages=262–265\| isbn = 0-07-287252-7}}

			{{DEFAULTSORT:Jordan's Lemma}}
			[[Category:Complex analysis]]
			[[Category:Articles containing proofs]]
			[[Category:Lemmas]]

en>Ego White Tray: Delete merge tag, merge done

2012-05-26T03:47:33Z

Delete merge tag, merge done

New page

Greetings! I am Marvella and I really feel comfortable when people use the full name. To do aerobics is a factor that I'm completely addicted to. Hiring is my occupation. Years in the past he moved to North Dakota and his family members enjoys it.<br><br>my web-site; std home test ([http://Www.pinaydiaries.com/user/LConsiden i thought about this])

Jeans instability - Revision history

en>Tom.Reding: Gen fixes (page/s, endash, , et al., unicodify, and/or concising wikilinks, etc.), & ref cleanup using AWB

200.104.245.226: /* Jeans mass */ rm promotion of particular views, excessive detail and unnecessary formatting

en>Ego White Tray: Delete merge tag, merge done