There is apparently nothing wrong with zero,
but to me it is more logical to use small circles on a straight line (e.g. for integrating functions over the whole real axis) than to parametrize them themselve on a circle, which is our contour in this case.
Many thanks!
As mentioned before: I have now 2 complex poles.
When I take pi x i x Residu in both and add them I got zero, meaning the Principle value would be zero ...
That is just twice the same calculation, for the integration along the whole contour and for the poles on the contour. This seems strange...
When you do the typical substitution one gets
\frac{2}{i}\int_{C_1}\frac{dz}{z^2-2za+1}
which results in 2 complex poles (a is real)
z_\pm=a\pm i\sqrt{1-a^2}
which are indeed on the contour.
One could indeed avoid the poles by for example going around them in a small circle but with the...
Hi everyone,
I would like to calculate the following integral:
P\int_0^{\pi}\frac{1}{cos(x)-a}dx, with
|a|\leq 1.
The 'P' in front stands for the so-called Cauchy Principle value.
Whenever a is not in the specified domain, the integrand does not have a pole and one can do the integration...
Hey,
I was wondering in what way the detailed balance condition (DBC) has to do with thermodynamic equilibrium.
The DBC is defined as
W(i\to j)P(i) = W(j\to i)P(j)
with W the rates (probabilities per unit time) to jump from state i to j. P(i) is the equilibrium prob. density to have...
I still don't get it.
I think you mean this the following?
\frac{1}{2^{3N}}\int_0^\infty du_1\ldots \int_0^\infty du_{3N}\Theta\left(1-\sum_{j=1}^{3N}|u_j|\right)
with the substitution w=\sum_j u_j you get dw = \sum du_j. How exactly do you enter this differential in...
The substitution r_j^2=u_j
leads,
for the r-integral only,
to
\frac{1}{2^{3N}}\int_0^\infty du_1\ldots \int_0^\infty du_{3N}\Theta\left(1-\sum_{j=1}^{3N}|u_j|\right)
Playing with the limits... I don't see exactly what you mean.
\Theta is Heaviside's stepfunction, equals 1 when it's argument is larger than zero, and 0 otherwise.
That's why it would be the volume of a sphere with radius 1 when being factorized without the r_j in front
Yes, of course.
Is there maybe anothor way to factorize the original integral in x and y so that I can solve it an a closed form?
The change to polar coordinates seemed logical to me because I would get a product of two indepent integrals.
My bad...
I mean the r_j(instead of x_j) coming from the jacobian due to the coordinate transformation.
I wish they weren't there because the integral would be a volume of a 3N-sphere.
Is there more advance that can be made otherwise?
Hey,
I have the following integral:
I = \int_0^\infty dx_1\ldots \int_0^\infty dx_{3N} \int_0^\infty dy_1\ldots \int_0^\infty dy_{3N}\Theta\left(1-\sum_{j=1}^{3N}\left(|x_j|^2+|y_j|^2\right)\right)
Now I want to change to polar coordinates by the following substitution...