Tag Info

Hot answers tagged

5

To illustrate whats going on, your function is 0 at x=0, rises to a max and becomes essentially zero very quickly. With[{a = .9}, Plot[x Exp[-(a^2+.001^2) x^2], {x, 0, 3}]] Now look at the values NIntegrate computes: res = Reap[With[{a = .9}, NIntegrate[y = x Exp[-(a^2 + 0.001^2) x^2], {x, 0, 8000}, EvaluationMonitor :> Sow[{x, ...


3

First note that, $$ \oint_{E(\vec{x})=\epsilon}\frac{dS}{|\nabla E(\vec{x})|} =\oint_{E(\vec{x})=\epsilon}\frac{\nabla E(\vec{x})\cdot d\vec{S}}{|\nabla E(\vec{x})|^2} = \int_{E(\vec{x})\leq\epsilon} \nabla\cdot\left( \frac{\nabla E(\vec{x})}{|\nabla E(\vec{x})|^2}\right) dV$$ By now, the volume integral can be evaluated by Integrate[f ...


3

Treat the maximum machine number as a singularity: ListPlot[Table[{a, NIntegrate[x Exp[-(a^2 + 0.001^2) x^2], {x, 0, 0.5 Log[$MaxMachineNumber/(a^2 + 0.001^2)], 8000}]}, {a, 0.001, 1, 1/100}], Joined -> True Update [Forgive me, I actually have a job, and, while I could solve the problem quickly over breakfast, I could not compose a complete ...


2

The code in the Question runs slowly, because it evaluates n^2 integrals. However, all the integrands are of the form Dc Exp[-Ax (x - a1)^2] Exp[-Ay (y - b1)^2] Exp[-Ax (x - a2)^2] Exp[-Ay (y - b2)^2]; This generic term can be integrated symbolically in several seconds Integrate[Laplacian[%, {x, y}], {x, -a, a}, {y, -b, b}] (* Dc (E^((-2*a^2 - a1^2 - ...


2

Trace[Integrate[ Integrate[ Integrate[ E^(-v1 - v2 - v3) (v1 + v2 + v3)/3, {v3, 2 v1 - 5/2 v2, v2}], {v2, 4 v1/7, 5 v1/7}], {v1, 0, Infinity}]] Check one line before the result. I think this is what you want.



Only top voted, non community-wiki answers of a minimum length are eligible