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Jordan's Lemma


Jordan's lemma shows the value of the integral

 I=int_(-infty)^inftyf(x)e^(iax)dx
(1)

along the infinite upper semicircle and with a>0 is 0 for "nice" functions which satisfy lim_(R->infty)|f(Re^(itheta))|=0. Thus, the integral along the real axis is just the sum of complex residues in the contour.

The lemma can be established using a contour integral I_R that satisfies

 lim_(R->infty)|I_R|<=pi/alim_(R->infty)epsilon=0.
(2)

To derive the lemma, write

x=Re^(itheta)
(3)
=R(costheta+isintheta)
(4)
dx=iRe^(itheta)dtheta,
(5)

and define the contour integral

 I_R=int_0^pif(Re^(itheta))e^(iaRcostheta-aRsintheta)iRe^(itheta)dtheta
(6)

Then

|I_R|<=Rint_0^pi|f(Re^(itheta))||e^(iaRcostheta)||e^(-aRsintheta)||i||e^(itheta)|dtheta
(7)
=Rint_0^pi|f(Re^(itheta))|e^(-aRsintheta)dtheta
(8)
=2Rint_0^(pi/2)|f(Re^(itheta))|e^(-aRsintheta)dtheta.
(9)

Now, if lim_(R->infty)|f(Re^(itheta))|=0, choose an epsilon such that |f(Re^(itheta))|<=epsilon, so

 |I_R|<=2Repsilonint_0^(pi/2)e^(-aRsintheta)dtheta.
(10)

But, for theta in [0,pi/2],

 2/pitheta<=sintheta,
(11)

so

|I_R|<=2Repsilonint_0^(pi/2)e^(-2aRtheta/pi)dtheta
(12)
=2epsilonR(1-e^(-aR))/((2aR)/pi)
(13)
=(piepsilon)/a(1-e^(-aR)).
(14)

As long as lim_(R->infty)|f(z)|=0, Jordan's lemma

 lim_(R->infty)|I_R|<=pi/alim_(R->infty)epsilon=0
(15)

then follows.


See also

Contour Integration

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References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 406-408, 1985.Jordan, C. Cours d'Analyse de l'Ecole polytechnique, Tome 2, 3. éd., rev. et corrigé. Paris: Gauthier-Villars, pp. 285-86, 1909-1915.Whittaker, E. T. and Watson, G. N. "Jordan's Lemma." §6.222 in A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, pp. 115-117, 1990.

Referenced on Wolfram|Alpha

Jordan's Lemma

Cite this as:

Weisstein, Eric W. "Jordan's Lemma." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/JordansLemma.html

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