Integral of 1/z using different paths


by jmcelve
Tags: 1 or z, integral, paths
jmcelve
jmcelve is offline
#1
Nov19-12, 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
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jackmell
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#2
Nov19-12, 06:19 AM
P: 1,666
Quote Quote by jmcelve View Post
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
Little ambiguous. The path (0,-i), (-i,a), (a,i), (0,i) does not represent a closed path. Did you mean, "and then back to (0,-i)"?

Suppose you did, then for the integral to be well-behaved, you'd had to indent around the singular point at the origin. If you indented into the right half-plane, then the contour encloses a region where 1/z is analytic and thus the integral is zero. If you indent into the left half-plane, then the contour includes the pole at the origin so the Residue Theorem applies. In either way, the integral represents a principal-valued integral since you're letting the indentation radius go to zero and taking the principal-valued integral along the imaginary axis. The integral is zero in the first case, or 2pi i in the other case right?
HallsofIvy
HallsofIvy is offline
#3
Nov19-12, 07:35 AM
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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].

jmcelve
jmcelve is offline
#4
Nov19-12, 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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