Recent content by Don'tKnowMuch

I think i have to use Rodrigues' formula

For clarity i should add that... P(sub L)(x) is a Legendre polynomial

I am doing a Laplace's equation in spherical coordinates and have come to a part of the problem that has the integral... ∫ P(sub L)*(x) * P(sub L')*(x) dx (-1<x<1) The answer to this integral is given by a Kronecker delta function (δ)... = 0 if L...

*High five*. Thanks again for the help! Have a nice weekend.

Awesome! Thanks for the extra set of eye's. It's like removing a splinter. Now i feel i can move on. Thanks Chopin! B.t.w., Chopin is a beast. Do you like Schumann? The piano/ string quintet's are this sh!+.

I do not think that completing the square is relevant at this point of the problem. It will be in the very next step. The author is trying to set up the integral in a way where completing the square is not necessary. By that i mean, he is cutting out the work of completing the square by...

Thank you for you response! Also sorry for the ambiguity of my query. I went over the integral from the page which is previous to the one that i linked and i understand it entirely. The integral on the page that i linked is essentially the same integral, only now t = 1 (instead of t = 0). So...

Here is a link to a course which i am studying, http://quantummechanics.ucsd.edu/ph130a/130_notes/node89.html#derive:timegauss My problem comes from the k' term attached to Vsub(g) (group velocity). I used the substitution k' = k - k(0), factored out all exponentials with no k'...

I"ve got it! Sheesh, it's like i removed splinter. Thank you so much. I love physics forums.

First off, thanks for the attention. I truly appreciate it! Sammy, yes you're right about the notation, I'm with you on that. dz should be s*(sec^2(theta))

I used the substitution that you suggested to get... (μ/4∏ε)(1/s) ∫ dz/[√1+√tan^2(θ)] I know the answer i should get is... (μ/4∏ε) Ln[ z+√s^2+√z^2] (evaluated at the bounds...i think i can do that part) I do not see the connection between your suggestion and the expression above...

the whole denominator is under the square root.

Homework Statement I have been hung up on this integral: (μ /4∏ε) ∫ dz/(√s^2 + z^2) Homework Equations The Attempt at a Solution i have tried a couple of different u-substitutions, and none are getting me anywhere. I do not that partial fractions, or by parts would help...

Dark energy = Vacuum energy? I am working on giving a presentation on dark energy and its possible relation to vacuum energy (i.e. one and the same). I have complied information from a couple of books, as well as websites (Scientific American, NASA, Astrophysical Journal). As i understand...

Looking at the sine term i can simplify the argument a little by setting... k' = (n∏/2a) making the sine term------> sin(k'x) Looking at the exponential, i can say that... p/h-bar = k Putting these in the integral looks something like... ζ(p) = 1/(2∏h-bar)^(1/2)*∫...