Recent content by jack5322

Oh, it comes from jordans lemma, thank you! I guess i was caught up on this problem because the book problems in analytic number theory by murty doesn't thoroughly explain it, at least to my satisfaction. I think its mistaken, though. If you have it, could you please take a look at it and...

Yes, but can you show me how the semicircle goes to zero?

no, s is the variable we are integrating. The integrand, including the dummy variable, is x^s/sds.

prove: x^s/s integrated from c-infinity*i to c+infinity*i, where c is any constant bigger than zero, is equal to 1 if x is bigger than 1. The outline of the proof is that we use a semicircle contour enclosing the origin, whose line segment parallel to the imaginary axis is intersecting the...

the limits are from c - infinity*i to c + infinity*i where c is a constant bigger than zero.

can someone prove this for me using a semicircle contour as opposed to a rectangle? I always get nowhere with the semicircle.

x^s/s integrated on the semicircular contour with radius R and center c>0, where x>1, s is the complex variable, and R is meant to go to infinity. please help.

following this logic, wouldn't the 8pi/3 be on the left of the cut for the MB contour, the one going out to the left? If not, why?

ok, why is it that the 2pi is on the top and the zero on the bottom for the example of something like (1-z)^-1/2 for -1<x<1 but the 2pi is on the bottom for the sqrtz with 0<x<infinity and zeroon the top?

Actually that doesn't work. My question is this: Why is it that we calculate the residues from the negative square root if the 2pi is on top?

I think I know now, we take the residues from the integral in the positive orientation, i.e. the one going from zero to infinity! That makes sense now.

I know I don't have to, but I need to see that It's possible that we can do this. It will make sense If I can do this, because If the choice from 2pi can be shifted to the bottom then Ill see why we can manipulate how the cuts phases can be changed. Also If we shift the 2pi to the top, the...

why is it that we choose the negative sqrt if the 2pi is on top?

I still don't see why the residue are negatives for the example with the 2pi on top. I get negative 2I = 2pi*i residues, where the residues are from the positive sqrt z/z^2+1.

how do we know which part of the multifunction to compute for the residues?