Integration with variables Notes_Part_18

# Integration with variables Notes_Part_18 - as long as n =-1...

This preview shows pages 1–2. Sign up to view the full content.

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 = 0 , 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 0 z n dz dt dt = integraldisplay π 0 e i nt i e i t dt = integraldisplay π 0 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 = 0 , 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 provided n negationslash = 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern