# Chapter4 - Chapter 4 Integration Everybody knows that...

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Chapter 4 Integration Everybody knows that mathematics is about miracles, only mathematicians have a name for them: theorems. Roger Howe 4.1 Deﬁnition and Basic Properties At ﬁrst sight, complex integration is not really anything diﬀerent from real integration. For a continuous complex-valued function φ : [ a,b ] R C , we deﬁne Z b a φ ( t ) dt = Z b a Re φ ( t ) dt + i Z b a Im φ ( t ) dt. (4.1) For a function which takes complex numbers as arguments, we integrate over a curve γ (instead of a real interval). Suppose this curve is parametrized by γ ( t ) , a t b . If one meditates about the substitution rule for real integrals, the following deﬁnition, which is based on (4.1) should come as no surprise. Deﬁnition 4.1. Suppose γ is a smooth curve parametrized by γ ( t ) , a t b , and f is a complex function which is continuous on γ . Then we deﬁne the integral of f on γ as Z γ f = Z γ f ( z ) dz = Z b a f ( γ ( t )) γ 0 ( t ) dt. This deﬁnition can be naturally extended to piecewise smooth curves, that is, those curves γ whose parametrization γ ( t ), a t b , is only piecewise diﬀerentiable, say γ ( t ) is diﬀerentiable on the intervals [ a,c 1 ] , [ c 1 ,c 2 ] ,..., [ c n - 1 ,c n ] , [ c n ,b ]. In this case we simply deﬁne Z γ f = Z c 1 a f ( γ ( t )) γ 0 ( t ) dt + Z c 2 c 1 f ( γ ( t )) γ 0 ( t ) dt + ··· + Z b c n f ( γ ( t )) γ 0 ( t ) dt. In what follows, we’ll usually state our results for smooth curves, bearing in mind that practically all can be extended to piecewise smooth curves. Example 4.2. As our ﬁrst example of the application of this deﬁnition we will compute the integral of the function f ( z ) = z 2 = ( x 2 - y 2 ) - i (2 xy ) over several curves from the point z = 0 to the point z = 1 + i . 42

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CHAPTER 4. INTEGRATION 43 (a) Let γ be the line segment from z = 0 to z = 1 + i . A parametrization of this curve is γ ( t ) = t + it, 0 t 1. We have γ 0 ( t ) = 1 + i and f ( γ ( t )) = ( t - it ) 2 , and hence Z γ f = Z 1 0 ( t - it ) 2 (1 + i ) dt = (1 + i ) Z 1 0 t 2 - 2 it 2 - t 2 dt = - 2 i (1 + i ) / 3 = 2 3 (1 - i ) . (b) Let γ be the arc of the parabola y = x 2 from z = 0 to z = 1 + i . A parametrization of this curve is γ ( t ) = t + it 2 , 0 t 1. Now we have γ 0 ( t ) = 1 + 2 it and f ( γ ( t )) = ± t 2 - ( t 2 ) 2 ² - i 2 t · t 2 = t 2 - t 4 - 2 it 3 , whence Z γ f = Z 1 0 ( t 2 - t 4 - 2 it 3 ) (1 + 2 it ) dt = Z 1 0 t 2 + 3 t 4 - 2 it 5 dt = 1 3 + 3 1 5 - 2 i 1 6 = 14 15 - i 3 . (c) Let γ be the union of the two line segments γ 1 from z = 0 to z = 1 and γ 2 from z = 1 to z = 1 + i . Parameterizations are γ 1 ( t ) = t, 0 t 1 and γ 2 ( t ) = 1 + it, 0 t 1. Hence Z γ f = Z γ 1 f + Z γ 2 f = Z 1 0 t 2 · 1 dt + Z 1 0 (1 - it ) 2 idt = 1 3 + i Z 1 0 1 - 2 it - t 2 dt = 1 3 + i ³ 1 - 2 i 1 2 - 1 3 ´ = 4 3 + 2 3 i.
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## This note was uploaded on 06/18/2011 for the course MATH 375 taught by Professor Marchesi during the Spring '10 term at Binghamton University.

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Chapter4 - Chapter 4 Integration Everybody knows that...

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