Contour Integration

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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)
👁 Image	👁 =	👁 lim_(R->infty)int_0^pi(P(Re^(itheta)))/(Q(Re^(itheta)))iRe^(itheta)dtheta	(3)
👁 Image	👁 =	👁 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)
👁 Image	👁 =	👁 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)
👁 Image	👁 =	👁 lim_(R->infty)int_0^pi(b_n)/(c_m)(Re^(itheta))^(n-m)iRdtheta	(8)
👁 Image	👁 =	👁 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)

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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 Resource. https://mathworld.wolfram.com/ContourIntegration.html

URL: https://mathworld.wolfram.com/ContourIntegration.html