Integration with variables Notes_Part_18

Integration with variables Notes_Part_18 - as long as n...

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Unformatted text preview: as long as n negationslash = − 1. Therefore, we can use the Fundamental Theorem of Calculus (which works equally well for real integrals of complex-valued functions), to evaluate integraldisplay P z n dz = bracketleftbig t + i ( t 2 − 1) bracketrightbig n +1 n + 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 t = − 1 = , − 1 negationslash = n = 2 k + 1 odd , 2 n + 1 , n = 2 k even . Thus, when n ≥ 0 is a positive integer, we obtain the same result as before. Interestingly, in this case the complex integral is well-defined even when n is a negative integer because, unlike the real line segment, the parabolic path does not go through the singularity of z n at z = 0. The case n = − 1 needs to be done slightly differently, and integration of 1 /z along the parabolic path is left as an exercise for the reader — one that requires some care. We recommend trying the exercise now, and then verifying your answer once we have become a little more familiar with basic complex integration techniques. Finally, let us try integrating around a semi-circular arc, again with the same endpoints − 1 and 1. If we parametrize the semi-circle S + by z ( t ) = e i t , 0 ≤ t ≤ π , we find integraldisplay S + z n dz = integraldisplay π z n dz dt dt = integraldisplay π e i nt i e i t dt = integraldisplay π i e i ( n +1) t dt = e i ( n +1) t n + 1 vextendsingle vextendsingle vextendsingle vextendsingle π t = 0 = 1 − e i ( n +1) π n + 1 = , − 1 negationslash = n = 2 k + 1 odd , − 2 n + 1 , n = 2 k even . This value is the negative of the previous cases — but this can be explained by the fact that the circular arc is oriented to go from 1 to − 1 whereas the line segment and parabola both go from − 1 to 1. Just as with line integrals, the direction of the curve determines the sign of the complex integral; if we reverse direction, replacing t by − t , we end up with the same value as the preceding two complex integrals. Moreover — again providedsame value as the preceding two complex integrals....
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This note was uploaded on 02/10/2012 for the course MATH 5587 taught by Professor Olver during the Fall '10 term at University of Central Florida.

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Integration with variables Notes_Part_18 - as long as n...

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