Helpful Math

Group Properties:

Cylindrical Laplacian Solution

$\displaystyle \nu\neq 0; R = a\rho^{\nu} + b \rho^{-\nu}$ (168)
$\displaystyle \psi = A \cos{\nu\phi} + B \sin{\nu \phi}$    
$\displaystyle \nu=0; R=a_0 + b_0 \ln{\rho}$ (169)
$\displaystyle \psi = A_0 + B_0\phi$    
$\displaystyle \frac{\rho}{R}\frac{\partial}{\partial \rho}\big(\rho\frac{\partial R}{\partial \rho}\big) = \nu^2$ (170)
$\displaystyle \frac{1}{\psi}\frac{\partial^2\psi}{\partial\phi^2}$ (171)
$\displaystyle \nabla^2 \Phi = 0$ (172)

Spherical Laplacian Solution

$\displaystyle \nabla^2 \Phi = 0$    
$\displaystyle \Phi = \sum_{l,m}Y^m_l(\Omega)\bigg[A_lr^l + \frac{B_l}{r^{l+1}}\bigg]$ (173)
$\displaystyle = \sum_{l,m} P_l(\cos{\theta})e^{im\phi}\bigg[A_lr^l + \frac{B_l}{r^{l+1}}\bigg]$    

Legendre Polynomials

$\displaystyle P_0(\cos{\theta}) = 1$    
$\displaystyle P_1(\cos{\theta}) = \cos{\theta}$    
$\displaystyle P_2(\cos{\theta}) = \frac{1}{2}\big(3 \cos^2{\theta} - 1 \big)$    
Orthogonality relationship:    
$\displaystyle \int_{-1}^1 P_m(x) P_n(x) dx = \frac{2}{2n+1}\delta_{mn}$ (174)

Gaussian Integrals

$\displaystyle \boxed{\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}}$ (175)

Square the full integral over a second variable, pass this to polar coordinates, and perform the (now easy!) integral.

$\displaystyle \int_{-\infty}^{\infty}e^{-x^2}dx \int_{-\infty}^{\infty}e^{-y^2}dy = \int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy$    
$\displaystyle = \int_0^{\infty} dr \int_0^{2\pi}rd\theta e^{-r^2}$    
$\displaystyle u = r^2, du = 2rdr$    


$\displaystyle \boxed{\int_{\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}}$ (176)

Scale $x\rightarrow \sqrt{\alpha}x',dx=\sqrt{\alpha}dx'$. Easy.

$\displaystyle \boxed{\int_{-\infty}^{\infty}x^2 e^{-\alpha x^2}dx = \frac{\sqrt{\pi}}{2a^{3/2}}}$ (177)

The trick here is to differentiate under the integral sign (a professed favorite trick of a certain someone).

$\displaystyle \int_{-\infty}^{\infty}x^2 e^{-\alpha x^2}dx$    
$\displaystyle = - \int_{-\infty}^{\infty} \frac{d}{d\alpha} e^{-\alpha x^2} dx$    
$\displaystyle = - \frac{d}{d\alpha}\int_{-\infty}^{\infty} e^{-\alpha x^2} dx$    
$\displaystyle = - \frac{d}{d\alpha} \sqrt{\frac{\pi}{\alpha}}$    

Advanced Gaussian Integrals
Of the type used in paths, density matrices, etc.

$\displaystyle \boxed{\int_{-\infty}^{\infty}dx e^{-\frac{a}{2}x^2 + Jx} = \sqrt{\frac{2\pi}{\alpha}}e^{J^2/2\alpha} }$ (178)

$\displaystyle \int_{-\infty}^{\infty}dx e^{- \frac{\alpha}{2} + Jx}$    
complete the square in the exponent.    
$\displaystyle -\frac{\alpha}{2}x^2 + Jx = -\frac{\alpha}{2}\bigg(x^2 - \frac{2Jx}{\alpha}\bigg)$    
$\displaystyle = -\frac{\alpha}{2}\bigg(x - \frac{J}{\alpha}\bigg)^2 + \frac{J^2}{2\alpha}$    
$\displaystyle \rightarrow \int_{-\infty}^{\infty}e^{-\frac{\alpha}{2}(x-\frac{J}{\alpha})^2} dx e^{J^2/2\alpha}$    
$\displaystyle u = x-\frac{J}{\alpha}, du=dx$    

Now this is a Gaussian integral of the 'easy' type from the previous section.

$\displaystyle \boxed{ \int_{-\infty}^{\infty}dx e^{- \frac{\alpha}{2} + iJx}= \sqrt{\frac{2\pi}{\alpha}}e^{-J^2/2\alpha}}$ (179)

Just sub $J\rightarrow iJ$ in the previous solution.

$\displaystyle \boxed{ \int_{-\infty}^{\infty}dx e^{i\frac{\alpha}{2} + iJx}= \sqrt{\frac{2\pi i}{\alpha}}e^{-J^2/2\alpha}}$ (180)

Sub $\alpha \rightarrow -i\alpha$ in the previous solution.