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Contour Integration


Contour integration is the process of calculating the values of a contour integral around a given contour in the complex plane. As a result of a truly amazing property of holomorphic functions, such integrals can be computed easily simply by summing the values of the complex residues inside the contour.

ContourIntegral

Let P(x) and Q(x) be polynomials of polynomial degree n and m with coefficients b_n, ..., b_0 and c_m, ..., c_0. Take the contour in the upper half-plane, replace x by z, and write z=Re^(itheta). Then

 int_(-infty)^infty(P(z)dz)/(Q(z))=lim_(R->infty)int_(-R)^R(P(z)dz)/(Q(z)).
(1)

Define a path gamma_R which is straight along the real axis from -R to R and make a circular half-arc to connect the two ends in the upper half of the complex plane. The residue theorem then gives

lim_(R->infty)int_(gamma_R)(P(z)dz)/(Q(z))=lim_(R->infty)int_(-R)^R(P(z)dz)/(Q(z))
(2)
=lim_(R->infty)int_0^pi(P(Re^(itheta)))/(Q(Re^(itheta)))iRe^(itheta)dtheta
(3)
=2piisum_(I[z]>0)Res[(P(z))/(Q(z))],
(4)

where Res[z] denotes the complex residues. Solving,

 lim_(R->infty)int_(-R)^R(P(z)dz)/(Q(z))=2piisum_(I[z]>0)Res(P(z))/(Q(z))-lim_(R->infty)int_0^pi(P(Re^(itheta)))/(Q(Re^(itheta)))iRe^(itheta)dtheta.
(5)

Define

I_R=lim_(R->infty)int_0^pi(P(Re^(itheta)))/(Q(Re^(itheta)))iRe^(itheta)dtheta
(6)
=lim_(R->infty)int_0^pi(b_n(Re^(itheta))^n+b_(n-1)(Re^(itheta))^(n-1)+...+b_0)/(c_m(Re^(itheta))^m+c_(m-1)(Re^(itheta))^(m-1)+...+c_0)iRdtheta
(7)
=lim_(R->infty)int_0^pi(b_n)/(c_m)(Re^(itheta))^(n-m)iRdtheta
(8)
=lim_(R->infty)int_0^pi(b_n)/(c_m)R^(n+1-m)i(e^(itheta))^(n-m)dtheta
(9)

and set

 epsilon=-(n+1-m),
(10)

then equation (9) becomes

 I_R=lim_(R->infty)i/(R^epsilon)(b_n)/(c_m)int_0^pie^(i(n-m)theta)dtheta.
(11)

Now,

 lim_(R->infty)R^(-epsilon)=0
(12)

for epsilon>0. That means that for -n-1+m>=1, or m>=n+2, I_R=0, so

 int_(-infty)^infty(P(z)dz)/(Q(z))=2piisum_(I[z]>0)Res[(P(z))/(Q(z))]
(13)

for m>=n+2. Apply Jordan's lemma with f(x)=P(x)/Q(x). We must have

 lim_(x->infty)f(x)=0,
(14)

so we require m>=n+1.

Then

 int_(-infty)^infty(P(z))/(Q(z))e^(iaz)dz=2piisum_(I[z]>0)Res[(P(z))/(Q(z))e^(iaz)]
(15)

for m>=n+1 and a>0. Since this must hold separately for real and imaginary parts, this result can be extended to

 int_(-infty)^infty(P(x))/(Q(x))cos(ax)dx=2piR{sum_(I[z]>0)Res[(P(z))/(Q(z))e^(iaz)]}
(16)
 int_(-infty)^infty(P(x))/(Q(x))sin(ax)dx=2piI{sum_(I[z]>0)Res[(P(z))/(Q(z))e^(iaz)]}.
(17)

See also

Cauchy Integral Formula, Cauchy Integral Theorem, Contour, Contour Integral, Complex Residue, Inside-Outside Theorem, Jordan's Lemma, Sine Integral

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References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 406-409, 1985.Krantz, S. G. "Applications to the Calculation of Definite Integrals and Sums." §4.5 in Handbook of Complex Variables. Boston, MA: Birkhäuser, pp. 51-63, 1999.Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 353-356, 1953.Whittaker, E. T. and Watson, G. N. "The Evaluation of Certain Types of Integrals Taken Between the Limits -infty and +infty," "Certain Infinite Integrals Involving Sines and Cosines," and "Jordan's Lemma." §6.22-6.222 in A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, pp. 113-117, 1990.

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Contour Integration

Cite this as:

Weisstein, Eric W. "Contour Integration." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ContourIntegration.html

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