# Log(-i mu lambda)

by HACR
Tags: lambda, logi
 P: 37 1. The problem statement, all variables and given/known data Suppose a complex integral of the form $$\int_{-\infty}^{\infty}(\frac {1}{\alpha+i\lambda}-\frac{1}{\beta +i \lambda}) log(-i\mu\lambda) d\lambda$$ where $$\alpha,\beta$$ are arbitrary positive parameters and the sufficient condition for integrability (nonlatttice, etc) is assumed. The branch cut lies on the negative real axis i.e. $$C1; [-r,-R]U[re^{i\theta}:-\pi<θ<0]U[r,R]U[Re^{iθ}:-pi<θ<0]$$ 2. Relevant equations 3. The attempt at a solution Haven't learned how to get its residue first of all. Second of all the integrand is multiplied by log(-iμλ} which I haven't dealt with before. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
P: 1,666
 Quote by HACR 1. The problem statement, all variables and given/known data Suppose a complex integral of the form $$\int_{-\infty}^{\infty}(\frac {1}{\alpha+i\lambda}-\frac{1}{\beta +i \lambda}) log(-i\mu\lambda) d\lambda$$ where $$\alpha,\beta$$ are arbitrary positive parameters and the sufficient condition for integrability (nonlatttice, etc) is assumed. The branch cut lies on the negative real axis i.e. $$C1; [-r,-R]U[re^{i\theta}:-\pi<θ<0]U[r,R]U[Re^{iθ}:-pi<θ<0]$$ 2. Relevant equations 3. The attempt at a solution Haven't learned how to get its residue first of all. Second of all the integrand is multiplied by log(-iμλ} which I haven't dealt with before. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
You need to show some work. Tell you what, put that one up for the moment and first try and solve:

$$\text{P.V.}\int_{-\infty}^{\infty}\frac{\log(z)}{(z-2i)(z-3i)}dz$$

same dif. Notice it's a principal-valued integral. That's because the path of integration is going right through a singular point, the branch-point of log(z). Are you familiar with those types of integrals? I suspect you aren't from what you've said. That's a problem but not major. Just indent around it and let it's indentation radius go to zero. What's the value of the integral over such an indented path? Need to figure that out. Also need to clearly state where the branch-cut is. That's not too clear the way you stated it up there. Just move it down along the negative imaginary axis to get it out the way for now. Won't change the integral. Now everything in the upper half-plane is entirely analytic for the integrand so you can use the Residue Theorem but that means you'll have to figure out what that integral is over the large circular contour connecting the end points of your integration limits. Kinda' quite a bit of work. That's what it's gonna' take though.
 P: 37 two things: branch cuts and indentation to avoid singularities. I know indentation is used to avoid isolated singularities, but branch cuts are used to make log(z) single valued. Then "Just indent around it and let it's indentation radius go to zero" would mean making a branch cut at the same time avoiding isolated singularities.
 P: 37 Log(-i mu lambda) I had a branch cut on the positive imaginary axis; however, I didn't quite get what I was hoping. The two integrals on the real axis cancels. i.e. $$-\int_{\rho}^{R} \frac{(ln(r)+\pi i)e^{i\pi}}{(r-2i)(r-3i)} dr +\int_{C_\rho}f(z)dz+\int_{C_R}f(z)dz-\int_{\rho}^{R} \frac{(ln(r)+2\pi i)e^{2\pi i}}{(r-2i)(r-3i)}dr$$ Actually ln(r) cancels.
P: 37
The below is the contour considered.
Attached Images
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 P: 1,666 Ok, you tried. Fair enough. Consider though an upper-half washer contour right? That little half-hole in the washer corresponds to the indentation around the origin and the branch-cut is along the negative imaginary axis. Then the P.V. integral plus the indentation plus the large half-circle contour would, by the Residue Theorem, equal to $2\pi i(r_1+r_2)$ right? If you don't see that, you'll have to review what a Principal Value integral is. Basically the limit of that improper integral as the distance to the singular point approaches zero. $$\text{P.V.}\int_{-\infty}^{\infty} f(z)dz+\lim_{\rho\to 0}\int_{\pi}^0 f(z)dz+\lim_{R\to\infty}\mathop\int_{C} f(z)dz=2\pi i(r_1+r_2)$$ Alright, I tell you what. Just for now, assume the two limits tend to zero. That's a common and acceptable way to approach these types of problems as long as you go back and confirm or reject your assumption. Then the principal valued integral would just be that residue calculation. Ok, then that's not hard to check that assumption in Mathematica. Just numerically integrate it although you'll have to specify the "PrincipalValue" method: NIntegrate[f[z],{z,-Infinity,0,Infinity},Method->"PrincipalValue"] Get that answer, then compute the two residues to get that answer. Do they agree or not? Then further proceed depending on that answer.
 P: 37 NIntegrate[ f[z] = (log z)/((z - 2 i) (z - 3 i)), {z, -\[Infinity], 0, \[Infinity]}, Method -> "PrincipalValue"] I did type this in but didn't give me a numerical value. seems to be wrong in the interval -infty-> -1. I haven't used it before so not sure how to proceed from here. They should go to zero b/c as $$\rho$$ and R go to 0 and $$\infty$$ , they vanish.
 P: 37 the code didn't work although thanks. Well I tried this to get rid of two simple poles, 2i and 3i:NIntegrate[(Log z)/((z-2 i) (z-3 i)),{z,-∞,2 i,3 i,+∞},Method->PrincipalValue] . Note the 2i and 3i. but for some reason mathematica wouldn't recognize 2i and 3i.
 P: 37 "The integrand (Log\z)/((-3\i+z)\(-2\i+z)) has evaluated to non-numerical values for all sampling points in the region with boundaries {{-\[Infinity],0.}}." Yes this is what I received from the mathematica. But then the problem is that z can't be specified since it's an independent variable that needs to vary. One possible solution: The integral is computed without difficulty if the parameter has a numerical value:With[{c=1},NIntegrate[Cos[c x],{x,0,6}] The problem is that this is not true for the complex integrand considered.

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