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Integral of 1/z using different paths 
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#1
Nov1912, 01:00 AM

P: 52

Hi everyone,
I'm trying to work out the integral of 1/z over the path (0, i) to (i, a) to (a, i) to (0, i) for a any real number greater than 0. I'm having trouble trying to determine what to do at z=0 since the integral doesn't exist here. Any ideas as far as how to push forward? My impression is that I'll have to exclude the pole at z=0 using z=ρ*exp[iθ] and taking the limit as ρ goes to 0 which will give me a closed contour. Presumably I'd have to calculate the residue as well, but I want to make sure I have this done rigorously. Would I have to use the principal value as well? Many thanks, jmcelve2 


#2
Nov1912, 06:19 AM

P: 1,666

Suppose you did, then for the integral to be wellbehaved, you'd had to indent around the singular point at the origin. If you indented into the right halfplane, then the contour encloses a region where 1/z is analytic and thus the integral is zero. If you indent into the left halfplane, then the contour includes the pole at the origin so the Residue Theorem applies. In either way, the integral represents a principalvalued integral since you're letting the indentation radius go to zero and taking the principalvalued integral along the imaginary axis. The integral is zero in the first case, or 2pi i in the other case right? 


#3
Nov1912, 07:35 AM

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PF Gold
P: 39,564

If, as I would read this, you are integrating on the straight line path from (0, i) to (a, i) to (a, i) to (0, i), you would not have to do anything about z= 0 because you never go anywhere near z= 0.
On the path from (0, i) to (a, i) you can take z= x i so that dz= dx and the integral is [itex]\int_0^a (x i)^{1}dx[/itex]. On the path from (a, i) to (a, i) you can take z= a+ iy so that dz= idy and the integral is [itex]\int_{1}^1 (a+iy)^{1}dy[/itex]. Finally, on the path from (a, i) to (0, i), you can take z= x+ i so that dz= dx and the integral is [itex]\int_1^0 (x+ i)^{1}dx[/itex]. 


#4
Nov1912, 11:50 AM

P: 52

Integral of 1/z using different paths
Yeah, that's it HallsofIvy. For some reason, I had the impression I was supposed to do a full contour integral. On a second glance, I realized I had completely misread the problem.
Shows the value of reading comprehension in mathematics! Thanks for the help. 


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