# Tag Info

0

You incorrectly found the poles. See the output of ToRadicals[Solve[((w^2 - \[CapitalOmega]^2)^2 + (\[Gamma] w)^2) == 0, w, Assumptions -> \[CapitalOmega] > 0 && \[Gamma] > 0], Assumptions -> \[CapitalOmega] > 0 && \[Gamma] > 0] Here is its part $$w\to \fbox{i \left( \begin{array}{cc} \{ & \begin{array}{cc} \... 1 Probably there is no solution in the range you are looking for. To show this we can visualize solution of two equations with using ContourPlot[] as follows d = 3; b = 0.1; tb = 10.103; zmcritical = 13.128; zh = 10; f[z_] := 1 - (z/zh)^(d + 1) Sv[z_?NumericQ, zm_?NumericQ] := zm^d/(z^d Sqrt[zm^(2 d) f[z] - z^(2 d) f[zm]]) SA[z_?NumericQ, zs_?NumericQ] := ... 1 You get the right analytical result quite fast, if you impose the addidional condition b<1 . Maybe Integrate then finds an other branch. expr = (b (b - 2 d + Sqrt[4 + b^2 - 4 b d]) (c + I x) x (3 + b^2 c^2 + 2 I b^2 c x - (-6 + b^2) x^2 + 3 x^4) (-1 + a x (-2 I c + a x)))/((2 + b^2 + 2 d (d - Sqrt[4 + b^2 - 4 b d]) + b (-4 d + Sqrt[4 + b^2 ... 4 Phew, this is quite a long and dense code, so I can give only some general hints. First you should look where the actual bottlenecks are. Please do not expect this from members of this forum. Anyways, here are some spots to look: The code throws many General::munfl errors because you feed the exponential functions with many negative values of large magniture.... 2 Separate x-independent prefactor: u = (b (b - 2 d + Sqrt[4 + b^2 - 4 b d]))/((2 + b^2 + 2 d (d - Sqrt[4 + b^2 - 4 b d]) + b (-4 d + Sqrt[4 + b^2 - 4 b d])) (\[Pi]^3) ); y = ((c + I x) x (-I + a x (2 c + I a x)) (b^2 (c + I x)^2 + 3 (1 + x^2)^2))/((c + I a x) (1 + a^2 x^2) (-b^2 (c + I x)^2 + (1 + x^2)^2)^2); FullSimplify[expr - u y] (* 0 *) Use an ... 0 Too long for a comment. The command result1 = Integrate[expr, {x, -Infinity, Infinity}, Assumptions -> c > -1 && c < 1 && d > 1 && a >= 0, GenerateConditions -> True] performs a huge output under the conditions Im[Sqrt[-4 - b^2 + 4 b c]] < Re[b] && Re[b] < Im[Sqrt[-4 - b^2 - 4 b c]] && Im[... 3 Update: Typo in the original code made it work more easily. Compute one integral, not two. Don't use "LocalAdaptive" -- it's usually slower except when it isn't (and sometimes it isn't). Use MaxRecursion -> 0 in Plot to experiment with possible solutions. (It showed rather quickly there was a discontinuity around 0.8. Suddenly, the plot jumped ... 4 You forgot the derivative (see Residue of monomial here): NIntegrate[func[Exp[I \[CurlyPhi]]]/(2 Pi I)*I*Exp[I \[CurlyPhi]], {\[CurlyPhi], 0, 2 Pi}] 1.99993 - 1.11022*10^-16 I 4 final version(corrected) Perhaps you could try to integrate along a path which excludes the singular point p= 0+I 0 (quarter-circle radius eps) : int[eps_?NumericQ] := NIntegrate[func[ p], {p, I Pi/2, I eps}] + NIntegrate[func[eps (Sin[\[CurlyPhi]] + I Cos[\[CurlyPhi]])] D[ eps (Sin[\[CurlyPhi]] +I Cos[\[CurlyPhi]]), \[CurlyPhi]], {\[CurlyPhi], 0, Pi/2}] ... 1 Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global*"] function[z_] := 1/(Sinh[z/2] Sqrt[Cosh[z]]) Add assumuptions integ[a_] = Assuming[0 < a < Pi/2, Integrate[function[z], {z, a I, -a I + Pi I/2}]] (* 2 (ArcTanh[Cos[a/2] Sqrt[Sec[a]]] - ArcTanh[Sin[1/4 (2 a + π)]/Sqrt[Sin[a]]]) *) Verifying ... 1 The integral divergence in the range [0,\infty] d = 2; func[p_] := 1/(Cosh[p/2]^(2/d) Tanh[p/2] Sqrt[ 1 - (Cosh[p/2]^(4 - 4/d) Tanh[p/2]^2)/(-0.419602)]) // Rationalize[#, 0] &; Integrate[func[p], {p, r, ∞}, Assumptions -> r > 0] Limit[%, r -> 0, Direction -> "FromAbove"]$$\log \left(\frac{\sqrt{81193660402}-2 \...

1

There are several messages when using DSolve[] In:= DSolve[{H'[u] == -a*H[u], H == HMax, X'[u] == X[u]*b*H[u]/HMax - X[u]*d*X[u]/K, X == K}, {H[u], X[u]}, u] During evaluation of In:= Solve::incnst: Inconsistent or redundant transcendental equation. After reduction, the bad equation is -2000+Subscript[\[ConstantC], 1] == 0. During ...

1

If I correctly understand the question, you ask about the asymptotic of the integral $$\int_0^\infty \cosh (x) \exp (-z \cosh (x))\,dx$$ as $z\to 0$ and/or $z->\infty$. Unfortunately, Mathematica 12.3 fails with that integral: Integrate[Cosh[x] Exp[-z Cosh[x]], {x, 0, \[Infinity]}, Assumptions -> z > 0] returns the input. The integral under ...

6

Change the function is easy to change the region. p = 2*Pi; f[x_, y_, z_] = Sin[x + y] - Cos[y - z] - Sin[x + z]; NIntegrate[ f[x, y, z]^2*Boole[f[x, y, z] >= 0], {x, 0, p}, {y, 0, p}, {z, 0, p}, Method -> "LocalAdaptive"] 239.804

0

You should try plotting your functions before assuming they're oscillatory. Some are not. Some have numerics issues. You're asking a lot if you want people to help you with 211 slow integrals. A MWE should be minimal. Here's an idea to help with a few of the 211 integrals: Figure out where the support of the integral lies, and make sure it's sampled. rat =...

0

Clear["Global*"] τ = 4; d = 1/2; (* use exact values for known constants *) w = 5; i[d_, τ_, g_, w_] = -(1/8) (-1 + Sqrt[1 - d^2]) (2 g^2 + w^2 - w^2 Cos[2 g*τ]) // Simplify; a[τ_, g_, d_] = ArcCos[1/4 Sqrt[ 1 - d] (Sqrt[1 - d] + Sqrt[ 1 + d] + (Sqrt[1 - d] - Sqrt[1 + d]) Cos[g*τ]) + 1/4 Sqrt[ 1 + ...

0

Use ImplicitRegion: region = With[{a = 1, b = 1/2, c = 1, d = .03, Ef = .5}, With[{sq = Sqrt[(b - x^2 - y^2)^2 + d (x^2 + y^2)]}, ImplicitRegion[ Abs[a*x - sq] < Ef && Abs[a*x + sq] < Ef, {x, y} ] ] ] (* test it works and we get a point *) RandomPoint[region] (* NIntegrate[f[x,y], {x,y} ∈ region] *)

-1

This is not a general answer. But it works for the problem at hand. Since the integrand is simple enough (a Gaussian) it can be integrated over the variable the analytically. The integral over r is done afterwards numerically, with the default NIntegrate configuration. Clear[PDFTheta] PDFTheta[the_] := PDF[ NormalDistribution [0, DeltaTheta/2], the] theLimUp[...

4

There is two ways you can go about this. As Anton noted, NDSolve will return Indeterimate for solution function evaluations that are outside of the region. This is the correct behavior for the finite element method as generally speaking no information is available beyond the boundary condition. That behavior, however, can be changed. This is explained in the ...

3

FindMinimum seems able to handle this. I had to increase precision to deal with machine underflow in intermediate steps. I gave decreasing initial values mostly to avoid issues from the fact that all permutations of solutions are solutions. p = 2; n = 10; hi = Table[i^-p, {i, 1, n}]; forward[s_] := Total@Table[Exp[-h*s] h, {h, hi}]; fi = Table[forward[s], {s,...

2

Okay so my answer is not a full answer but its a step in the right direction. We can't possibly solve the system and NSolve seems to hang around and not go towards a solution so we have to take the route of numerical optimizations. Initially, we set the equations to solve for: fs[s_]:=Sum[Exp[-h[i]*s]*h[i],{i,1,n}] equs=(N[fi[[#+1]]]-fs[#])&/@Range[0,n-1]...

4

The problem is not with NIntegrate -- the sol function gives Indeterminate for some points of the region reg1: SeedRandom; lsRPoints = RandomPoint[reg1, 200]; Select[Association[ Map[# -> sol[0.1, Sequence @@ #] &, lsRPoints]], ! NumberQ[#] &] During evaluation of In:= InterpolatingFunction::femdmval: Input value {-0.928406,-0.347651,0....

0

To long for a comment: MMA version 8.0 gives you an antiderivative that has no discontinuity (here called int1, your solution called int2) int1 = -(I (1 - 6 t^2 + t^4 + 8 t Z - 4 Z^2) (\[Pi]^2 - 4 I \[Pi] ArcTanh[(1 - t^2 + 2 I t u - 2 I u Z)/Sqrt[ 1 + t^4 + 4 t^3 y + 4 y Z + 4 Z^2 - 4 t (y + 2 Z) + t^2 (2 - 4 y Z)]] - 8 ArcTanh[(1 - t^2 +...

2

Because Solve produces the warning message, Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result. you might try rationalizing f1[s] (here f for simplicity): f = Simplify@Rationalize[-(0.000879451/((0.00024674 + s) (-s + 100. (1 - 1/z)))) + 0....

Top 50 recent answers are included