User - MathOverflow

Gram series for more general integrals

2013-04-08T13:06:27Z

given the Gram series

$$ \pi (x) \sim \sum_{n=1}^{\infty} \frac{1}{n}\frac{log^{n}(x)}{n!\zeta (n+1)} $$

can this result be generalized to more integral transforms ?

$$ g(x)= \int_{0}^{\infty}dyK(xy)f(y)dy $$

for exapmle in Gramm series for the prime counting function the integral equation is

$$ log\zeta(s)= s\int_{0}^{\infty}\frac{1}{e^{xy}-1}\pi(e^{y})dy $$

Numerical evaluation of a triple integral can be made ?

2013-04-08T11:14:56Z

let be the integral

$$ \int_{0}^{\infty}dx\int_{0}^{\infty}dy\int_{0}^{\infty}dz \frac{f(x,y,z)}{(1+x^{2}+y^{2}+z^{2})^{s}} $$

here $ s $ is a parameter so the integral converges

now let us make the change to polar coordinates , however the function $ f(x,y,z) $ will not be invariant under rotations so there will be an extra integration over the angular variables

$$ \int_{\Omega} d \Omega \int_{0}^{\infty}r^{2}f(r, \Omega )(1+r^{2})^{-s} $$

my question is can we ALWAYS integrate with Numerical (MOnte carlo method) inside the angular variables $ \int_{\Omega}d \Omega $ so we are left only with a sum of one dimensional integrals ??

$$ \sum_{i} \int_{0}^{\infty}dr f(r, \Omega _{i})r^{2}(1+r^{2})^{-s}. $$

Bessel function with complex argument and index

2012-12-18T10:03:32Z

let be a Bessel function $ J_{u} (x) $

if both the index 'u' and the argument 'x' are complex

$$ J_{ia}(ib) $$

for real 'a' and 'b' what is then the name for this function ??

if only the index is complex $ J_{ia}(b) $ or the argumentis complex $ J_{a}(ib) $

what is the name of this function ??

how can i solve a boundary value numerically on an infinite interval ??

2012-08-04T09:36:31Z

let be the differential equation $ -y''(x)+x^{4}y(x)-E_{n}y(x)=0 $ with the boundary conditions $ y(0)=0=y(\infty) $

how could i use the shooting method or other numerical method to solve this equation ? , my only idea is to set $ R=10000 $ for example and solve $ y(0)=0=y(R) $

of course i also could made the substitution $ u= \frac{x}{x-1} $ so the new boundary conditions could become $ y(0)=0=y(1) $ but know the differential equation would be singular at the point $ u=1 $

Comment by

2013-04-11T11:37:13Z

oh sorry tehn Ben if possible erase teh questio please :)

Comment by

2013-04-10T11:00:13Z

my idea is to change to polar coordinates :) and then to apply numerical integration ONLY to the angular variables so we are left with a set of ONE dimensional integral in the form $ \int_{0}^{\infty}drr^{n-1}f(r, \Omega _{i} ) $

Comment by

2013-04-08T19:01:55Z

my question was about if my numerical method is correct a) change to polar coordinates b) numerical integration ONLY over the angular variables.

Comment by

2013-01-27T13:47:03Z

mathdl.maa.org/images/upload_library/22/Ford/… inverse problems in physics.. see form equation (6.1) we can get the inverse of the potentia $ f^{-1}(x) $ from the eigenvalue staircase in this case the staricase is the Riemann zeta function

Comment by

2013-01-27T13:41:02Z

Hua Wu and D. W. L. Sprung, "Riemann zeta and a fractal potential", Physical Review E 48 (1993) 2595. Fractal fits to Riemann zeros P B Slater, Canadian Journal of Physics, 2007, 85(4): 345–357, 10.1139/p07-050

Comment by

2012-11-12T13:06:15Z

as far as i know an operator of the form $ -D^{2}+f(x) $ has only Real Eigenvalues doesn't it ? , of course as a phsyicist i could be wrong :D , look on the web WU-SPRUNG potential PHYSICS OF THE RIEMANN HYPOTHESIS and similar questions or simply use a computer to get the eigenvalues to see i am right

Comment by

2012-11-11T14:48:38Z

the idea is that using WKB method , or semiclassical theory , look at this en.wikipedia.org/wiki/Old_quantum_theory one could get a Hamiltonian with a density of states given by the Riemann Weil formula and whose energies are the Riemann Zeros the potential is given implicitly of course

Comment by

2012-11-08T16:07:33Z

given any 'x' positive i would like to find a method to sum the divergent series with $ a=1/2$ or with a a real number on the interval $ (0,1) $