Recent content by julian

I think this is the fourth edition: https://archive.org/details/landau-l.-d.-lifshitz-e.-m.-course-of-theoretical-vol-2/page/n3/mode/2up

I believe these notes cover Proposition 4.4.2 - in the author's notes it appears as Proposition 4.3.7. I spoke with the author about his notes, but that was likely over 10 years ago. Things are a bit hectic at the moment, so I won’t be able to revisit this right now.

The "stars and bars" thing: Consider the example where ##T_{ijk}## is symmetric under the interchange of any pair of indices, and the indices take the values ##1,2,3,4##. For example, all these are the same: \begin{align*} T_{112} = T_{121} = T_{211} . \end{align*} These can all be represented...

Say ##i=1,2,\cdots n##, and consider the object ##T_{i_i i_2 \cdots i_k}## that is symmetric under the interchange of any pair of indices. The number of independent components, ##N##, is equal to the number degree-##k## monomials in ##n## variables ##x_1, x_2, \dots, x_n##, i.e., expressions of...

The approach is thorough, though perhaps a bit more detailed than necessary. (1) This symmetry applies to ##i## and ##j##. For each fixed ##k##, ##T_{ijk}## is a symmetric 3×3 matrix in indices ##i,j##. The number of independent components in a symmetric 3×3 matrix is 6. Since ##k## can...

Just to clarify — the expression you gave for the integrating factor wasn’t correct, so I was pointing that out.

It should be ##\int \ln x dx = x \ln x -x +C##, so ##e^{\int \ln x dx} = K x^x e^{-x}##.

Alternatively, following the suggestion made by @Paul Colby. Note, \begin{align*} \Gamma(s)=\frac{1}{s}\Gamma(s+1)=\frac{1}{s(s+1)}\Gamma(s+2)=\frac{1}{s(s+1)(s+2) \cdots (s+n)}\Gamma(s+n+1) \end{align*} What does that tell us about the poles of ##\Gamma (s)^2##? Here are useful formula for...

My initial thought was to approach this using complex contour integration due to the structure of the inverse transform. However, there appears to be a possibly more straightforward way to derive the inverse Mellin transform. Consider \begin{align*} \Gamma (s)^2 = \int_0^\infty \left(...

They are mentioned briefly in Arfken, where it is noted that substituting ##t=\ln x## and ##i \omega = s -c## (possibly a typo — perhaps it should be ##i \omega = s -ic##?) into the Fourier transform and its inverse leads to \begin{align*} G(s) = \int_0^\infty x^{s-1} F(x) dx \qquad \text{the...

Your question could have been phrased more clearly. Try evaluating your integral by first expressing it as ##\sum_{n=0}^\infty \int_n^{n+1} [x] x^{s-1} dx## and then writing out the resulting series. That should make the connection to ##\zeta (-s)## clear—if that's what you're aiming for.

Indeed — as they say, "after a rather lengthy calculation."

Unfortunately, I don't have access to that part of the book. The Internet Archive version of L-L Volume 2 ends at page 183 for some reason.

Applying certain identities, the expression for ##R^{ik} - \frac{1}{2} g^{ik}R## in terms of ##g_{lp}## (using the quoted expression for ##R^{ik}##) just requires a few transformations to match the final form they provided. Below, I work backwards starting from their final expression. First...

Consider ##\vec{\nabla} \cdot (x \vec{P})##. Suppose ##\Omega## is a volume with a boundary ##\Gamma = \partial \Omega##. Integrating over ##\Omega## with respect to the volume ##dV##, and applying the divergence theorem, gives: \begin{align*} \int_\Gamma x \vec{P} \cdot \vec{n} d S & =...