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In complex analysis , contour integration is a way to calculate an integral around a contour on the complex plane . In other words, it is a way of integrating along the complex plane.
More specifically, given a complex-valued function f {\displaystyle f} and a contour C {\displaystyle C} , the contour integral of f {\displaystyle f} along C {\displaystyle C} is written as ∫ C f ( z ) d z {\displaystyle \textstyle \int _{C}f(z)\,dz} or ∮ C f ( z ) d z {\displaystyle \textstyle \oint _{C}f(z)\,dz} .[ 1] [ 2]
For a standard contour integral, we can evaluate it by using the residue theorem . This theorem states that
∮ C f ( z ) d z = 2 π i ⋅ Res f ( z ) {\displaystyle \oint _{C}f(z)\,dz=2\pi i\cdot {\text{Res}}f(z)}
where Res {\displaystyle {\text{Res}}} is the residue of the function f ( z ) {\displaystyle f(z)} , C {\displaystyle C} is the contour located on the complex plane . Here, f ( z ) {\displaystyle f(z)} is the integrand of the function, or part of the integral to be integrated.
The following examples illustrate how contour integrals can be calculated using the residue theorem.
∮ C e z z 3 d z = 2 π i ⋅ Res ( e z z 3 ) = 2 π i ⋅ 1 2 = π i {\displaystyle {\begin{aligned}&\oint _{C}{\frac {e^{z}}{z^{3}}}\,dz\\&=2\pi i\cdot {\text{Res}}\left({\frac {e^{z}}{z^{3}}}\right)\\&=2\pi i\cdot {\frac {1}{2}}\\&=\pi i\end{aligned}}} ∮ C 1 z 3 d z = 2 π i Res f ( z ) = 2 π i Res 1 z 3 = 2 π i ⋅ 0 = 0 {\displaystyle {\begin{aligned}&\oint _{C}{\frac {1}{z^{3}}}\,dz\\&=2\pi i\,{\text{Res}}f(z)\\&=2\pi i\,{\text{Res}}{\frac {1}{z^{3}}}\\&=2\pi i\cdot 0\\&=0\end{aligned}}} To solve multivariable contour integrals (contour integrals on functions of several variables), such as surface integrals, complex volume integrals and higher order integrals , we must use the divergence theorem . For right now, let ∇ {\displaystyle \nabla } be interchangeable with Div {\displaystyle {\text{Div}}} . These will both serve as the divergence of the vector field written as F {\displaystyle \mathbf {F} } . This theorem states that:
∫ ⋯ ∫ U ⏟ n Div ( F ) d V = ∮ ⋯ ∮ ∂ U ⏟ n − 1 F ⋅ n d S {\displaystyle \underbrace {\int \cdots \int _{U}} _{n}{\text{Div}}(\mathbf {F} )\,dV=\underbrace {\oint \cdots \oint _{\partial U}} _{n-1}\mathbf {F} \cdot \mathbf {n} \,dS}
In addition, we also need to evaluate ∇ ⋅ F {\displaystyle \nabla \cdot \mathbf {F} } , where ∇ ⋅ F {\displaystyle \nabla \cdot \mathbf {F} } is an alternate notation of div F {\displaystyle {\text{div}}\,\mathbf {F} } . [ 1] The divergence of any dimension can be described as
Div F = ∇ ⋅ F = ( ∂ ∂ u , ∂ ∂ x , ∂ ∂ y , ∂ ∂ z , ⋯ ) ⋅ ( F u , F x , F y , F z ⋯ ) = ( ∂ F u ∂ u + ∂ F x ∂ x + ∂ F y ∂ y + ∂ F z ∂ z ⋯ ) {\displaystyle {\begin{aligned}&\operatorname {Div} {\mathbf {F} }\\\\&=\nabla \cdot {\textbf {F}}\\\\&=\left({\frac {\partial }{\partial u}},{\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}},\cdots \right)\cdot (F_{u},F_{x},F_{y},F_{z}\cdots )\\\\&=\left({\frac {\partial F_{u}}{\partial u}}+{\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}\cdots \right)\end{aligned}}} The following examples illustrate the use of divergence theorem in the calculation of multivariate contour integrals.
Let the vector field F = sin ( 2 x ) + sin ( 2 y ) + sin ( 2 z ) {\displaystyle \mathbf {F} =\sin(2x)+\sin(2y)+\sin(2z)} be bounded by the following conditions
0 ≤ x ≤ 1 0 ≤ y ≤ 3 π − 1 ≤ z ≤ 4 {\displaystyle {0\leq x\leq 1}\quad {0\leq y\leq 3\pi }\quad {-1\leq z\leq 4}} The corresponding double contour integral would be set up as such:
{\displaystyle } S {\displaystyle {\scriptstyle S}} F ⋅ n d S {\displaystyle {\mathbf {F} \cdot n}\,{\rm {d}}\,S} We now evaluate ∇ ⋅ F {\displaystyle \nabla \cdot \mathbf {F} } by setting up the corresponding triple integral:
= ∭ V ( ∂ F x ∂ x + ∂ F y ∂ y + ∂ F z ∂ z ) d V = ∭ V ( ∂ sin ( 2 x ) ∂ x + ∂ sin ( 2 y ) ∂ y + ∂ sin ( 2 z ) ∂ z ) d V = ∭ V = 2 ( cos ( 2 x ) + cos ( 2 y ) + cos ( 2 z ) ) d V {\displaystyle {\begin{aligned}&=\iiint _{V}\left({\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}\right)\,dV\\\\&=\iiint _{V}\left({\frac {\partial \sin(2x)}{\partial x}}+{\frac {\partial \sin(2y)}{\partial y}}+{\frac {\partial \sin(2z)}{\partial z}}\right)\,dV\\\\&=\iiint _{V}{=2(\cos(2x)+\cos(2y)+\cos(2z))}\,dV\end{aligned}}} From this, we can now evaluate the integral as follows:
∫ 0 1 ∫ 0 3 ∫ − 1 4 2 ( cos ( 2 x ) + cos ( 2 y ) + cos ( 2 z ) ) d x d y d z = ∫ 0 1 ∫ 0 3 ( 10 cos ( 2 y ) + sin ( 8 ) + sin ( 2 ) + 10 cos ( z ) ) d y d z = ∫ 0 1 ( 30 cos ( 2 z ) + 3 sin ( 2 ) + 3 sin ( 8 ) + 5 sin ( 6 ) ) d z = 18 sin ( 2 ) + 3 sin ( 8 ) + 5 sin ( 6 ) {\displaystyle {\begin{aligned}&\int _{0}^{1}\int _{0}^{3}\int _{-1}^{4}2(\cos(2x)+\cos(2y)+\cos(2z))\,dx\,dy\,dz\\\\&=\int _{0}^{1}\int _{0}^{3}(10\cos(2y)+\sin(8)+\sin(2)+10\cos(z))\,dy\,dz\\\\&=\int _{0}^{1}(30\cos(2z)+3\sin(2)+3\sin(8)+5\sin(6))\,dz\\\\&=18\sin(2)+3\sin(8)+5\sin(6)\end{aligned}}}
Given the vector field F = u 4 + x 5 + y 6 + z − 3 {\displaystyle \mathbf {F} =u^{4}+x^{5}+y^{6}+z^{-3}} and n {\displaystyle n} being the fourth dimension. Let this vector field be bounded by the following:
0 ≤ x ≤ 1 − 10 ≤ y ≤ 2 π 4 ≤ z ≤ 5 − 1 ≤ u ≤ 3 {\displaystyle {0\leq x\leq 1}\quad {-10\leq y\leq 2\pi }\quad {4\leq z\leq 5}\quad {-1\leq u\leq 3}} To evaluate this, we use the divergence theorem as stated before, and evaluate ∇ ⋅ F {\displaystyle \nabla \cdot \mathbf {F} } afterwards. Let d V = d x d y d z d u {\displaystyle \,dV=\,dx\,dy\,dz\,du} , then:
S {\displaystyle {\scriptstyle S}} F ⋅ n d S {\displaystyle \mathbf {F} \cdot n\,{\rm {d}}{\mathbf {S}}} = ⨌ V ( ∂ F u ∂ u + ∂ F x ∂ x + ∂ F y ∂ y + ∂ F z ∂ z ) d V = ⨌ V ( ∂ u 4 ∂ u + ∂ x 5 ∂ x + ∂ y 6 ∂ y + ∂ z − 3 ∂ z ) d V = ⨌ V 4 u 3 z 4 + 5 x 4 z 4 + 5 y 4 z 4 − 3 z 4 d V {\displaystyle {\begin{aligned}&=\iiiint _{V}\left({\frac {\partial F_{u}}{\partial u}}+{\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}\right)\,dV\\\\&=\iiiint _{V}\left({\frac {\partial u^{4}}{\partial u}}+{\frac {\partial x^{5}}{\partial x}}+{\frac {\partial y^{6}}{\partial y}}+{\frac {\partial z^{-3}}{\partial z}}\right)\,dV\\\\&=\iiiint _{V}{\frac {4u^{3}z^{4}+5x^{4}z^{4}+5y^{4}z^{4}-3}{z^{4}}}\,dV\end{aligned}}} From this, we now can evaluate the integral:
⨌ V 4 u 3 z 4 + 5 x 4 z 4 + 5 y 4 z 4 − 3 z 4 d V = ∫ 0 1 ∫ − 10 2 π ∫ 4 5 ∫ − 1 3 4 u 3 z 4 + 5 x 4 z 4 + 5 y 4 z 4 − 3 z 4 d V = ∫ 0 1 ∫ − 10 2 π ∫ 4 5 ( 4 ( 3 u 4 z 3 + 3 y 6 + 91 z 3 + 3 ) 3 z 3 ) d y d z d u = ∫ 0 1 ∫ − 10 2 π ( 4 u 4 + 743440 21 + 4 z 3 ) d z d u = ∫ 0 1 ( − 1 2 π 2 + 1486880 π 21 + 8 π u 4 + 40 u 4 + 371720021 1050 ) d u = 371728421 1050 + 14869136 π 3 − 105 210 π 2 ≈ 576468.77 {\displaystyle {\begin{aligned}&\iiiint _{V}{\frac {4u^{3}z^{4}+5x^{4}z^{4}+5y^{4}z^{4}-3}{z^{4}}}\,dV\\\\&=\int _{0}^{1}\int _{-10}^{2\pi }\int _{4}^{5}\int _{-1}^{3}{\frac {4u^{3}z^{4}+5x^{4}z^{4}+5y^{4}z^{4}-3}{z^{4}}}\,dV\\\\&=\int _{0}^{1}\int _{-10}^{2\pi }\int _{4}^{5}\left({\frac {4(3u^{4}z^{3}+3y^{6}+91z^{3}+3)}{3z^{3}}}\right)\,dy\,dz\,du\\\\&=\int _{0}^{1}\int _{-10}^{2\pi }\left(4u^{4}+{\frac {743440}{21}}+{\frac {4}{z^{3}}}\right)\,dz\,du\\\\&=\int _{0}^{1}\left(-{\frac {1}{2\pi ^{2}}}+{\frac {1486880\pi }{21}}+8\pi u^{4}+40u^{4}+{\frac {371720021}{1050}}\right)\,du\\\\&={\frac {371728421}{1050}}+{\frac {14869136\pi ^{3}-105}{210\pi ^{2}}}\\\\&\,\approx {576468.77}\end{aligned}}} Thus, we can evaluate a contour integral of the fourth dimension.
↑ 1.0 1.1 "List of Calculus and Analysis Symbols" . Math Vault . 2020-05-11. Retrieved 2020-09-18 . ↑ "Contour Integration | Brilliant Math & Science Wiki" . brilliant.org . Retrieved 2020-09-18 .