- Residue theorem
The residue theorem in
complex analysis is a powerful tool to evaluateline integral s ofanalytic function s over closed curves and can often be used to compute real integrals as well. It generalizes theCauchy integral theorem andCauchy's integral formula .The statement is as follows. Suppose "U" is a
simply connected open subset of thecomplex plane , and "a"1,...,"a""n" are finitely many points of "U" and "f" is a function which is defined and holomorphic on "U" {"a"1,...,"a""n"}. If γ is arectifiable curve in "U" which bounds the "a""k", but does not meet any and whose start point equals its endpoint, then:oint_gamma f(z), dz =2pi i sum_{k=1}^n operatorname{I}(gamma, a_k) operatorname{Res}( f, a_k ).
If γ is a
Jordan curve , I(γ, "a""k") = 1 and so :oint_gamma f(z), dz =2pi i sum_{k=1}^n operatorname{Res}( f, a_k ).Here, Res("f", "a""k") denotes the residue of "f" at "a""k", and I(γ, "a""k") is the
winding number of the curve γ about the point "a""k". This winding number is aninteger which intuitively measures how often the curve γ winds around the point "a""k"; it is positive if γ moves in a counter clockwise ("mathematically positive") manner around "a""k" and 0 if γ doesn't move around "a""k" at all.In order to evaluate real integrals, the residue theorem is used in the following manner: the integrand is extended to the complex plane and its residues are computed (which is usually easy), and a part of the real axis is extended to a closed curve by attaching a half-circle in the upper or lower half-plane. The integral over this curve can then be computed using the residue theorem. Often, the half-circle part of the integral will tend towards zero as the radius of the half-circle grows, leaving only the real-axis part of the integral, the one we were originally interested in.
Example
The integral
:int_{-infty}^infty {e^{itx} over x^2+1},dx
arises in
probability theory when calculating the characteristic function of theCauchy distribution , and it resists the techniques of elementarycalculus . We willevaluate it by expressing it as a limit of contour integralsalong the contour "C" that goes along the realline from −"a" to "a" and then counterclockwise alonga semicircle centered at 0 from "a" to −"a". Take"a" to be greater than 1, so that the imaginaryunit "i" is enclosed within the curve. The contour integral is:int_C {f(z)},dz =int_C {e^{itz} over z^2+1},dz.
Since "e""itz" is an
entire function (having no singularitiesat any point in the complex plane), this function hassingularities only where the denominator"z"2 + 1 is zero. Since"z"2 + 1 = ("z" + "i")("z" − "i"),that happens only where "z" = "i" or "z" = −"i".Only one of those points is in the region bounded by thiscontour. Because "f"("z") is:
the residue of"f"("z") at "z" = "i" is
:operatorname{Res}_{z=i}f(z)={e^{-t}over 2i}.
According to the residue theorem, then, we have
:int_C f(z),dz=2pi icdotoperatorname{Res}_{z=i}f(z)=2pi i{e^{-t} over 2i}=pi e^{-t}.
The contour "C" may be split into a "straight"part and a curved arc, so that
:int_{mbox{straight+int_{mbox{arc=pi e^{-t},
and thus
:int_{-a}^a =pi e^{-t}-int_{mbox{arc.
It can be shown that if "t" > 0 then
:int_{mbox{arc{e^{itz} over z^2+1},dz ightarrow 0 mbox{as} a ightarrowinfty.
Therefore if "t" > 0 then
:int_{-infty}^infty{e^{itz} over z^2+1},dz=pi e^{-t}.
A similar argument with an arc that winds around −"i"rather than "i" shows that if "t" < 0 then
:int_{-infty}^infty{e^{itz} over z^2+1},dz=pi e^t,
and finally we have
:int_{-infty}^infty{e^{itz} over z^2+1},dz=pi e^{-left|t ight.
(If "t" = 0 then the integral yields immediately to elementary calculus methods and its value is π.)
ee also
*
Methods of contour integration
*Morera's theorem
*Nachbin's theorem References
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*External links
* [http://mathworld.wolfram.com/ResidueTheorem.html Residue theorem] in
MathWorld
* [http://math.fullerton.edu/mathews/c2003/ResidueCalcMod.html Residue Theorem Module by John H. Mathews]
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