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1

According to the documentation, Format[] seems to be done for that : Unprotect[LogIntegral] Format[LogIntegral[z_]] := li[z] Protect[LogIntegral]


5

I'll make this brief: it's a job for a MakeBoxes rule. In this case a particularly simple one: MakeBoxes[li : LogIntegral, StandardForm] := InterpretationBox["li", li] Now LogIntegral prints as li.


1

If you trying to compare the built in Gamma function with the integral definition of $\Gamma(z)$ then first define it properly myGamma[z_] := Simplify[Integrate[(t^(z - 1))*Exp[-t], {t, 0, Infinity}], Assumptions -> Re[z] > 0]; Then we can check & see that they agree. Clear[z]; Gamma[z] == myGamma[z] (*True*) If you want to see it ...


1

Introduction For many transcendental functions, NSolve can solve for the roots, but not in this case. Since the roots are real we can apply the "Chebyshev-proxy rootfinder" method (CPR) which is based on the "colleague matrix" of a truncated Chebyshev series approximation to the function (see this answer by J.M. and the book by Boyd (2014)). The first ...


1

This is a good example of the "dynamic range" problem in Boyd's CPR method (see also this answer by J.M.). As α moves toward -100, the OP's oscillatory Whittaker function decays exponentially. According to Boyd, where $ε_{mach}$ is the machine epsilon and $f_{max}$ is the maximum of $f(x)$ over an interval $[a,b]$, "If a function $f(x)$ has a magnitude as ...


2

the culprit here is AccuracyGoal: a = 3; c = 6; d = 0.00033; b = 2; NIntegrate[ x^3 (SphericalBesselJ[0, b x] + SphericalBesselJ[2, b x])/(4 d^2 x^2 + 9)^6 1/ 2 E^(-(1/2) x^2 - I x c - 1/2 a^2) (1 + E^(2 I x c) - I E^(2 I x c) Erfi[(x - I c)/Sqrt[2]] - I Erfi[(x + I c)/Sqrt[2]]), {x, 0, 8}, MaxRecursion -> 22, AccuracyGoal ...


3

Analysis of the error (bug?) We can see from the trace below that the second limit, which carries out a ratio test for the product, mistakenly yields -17 (which would indicate divergence, if correct). Trace[ NProduct[(n^2)!/stirling[n^2], {n, 1, Infinity}], _Limit, TraceInternal -> True, TraceForward -> True] There might have been some ...


3

n Binomial[n - 1, k - 1] == k Binomial[n, k] // FullSimplify True


1

Normal@Series[BesselK[1, r Λ]/BesselK[1, Λ], {r, 0, 1}, Assumptions -> Λ > 0] gives You can also use Simplify@Normal@Series[BesselK[1, r Λ]/BesselK[1, Λ], {r, 0, 1}, Assumptions -> Λ > 0] or Simplify[Normal@ Series[BesselK[1, r Λ]/ BesselK[1, Λ], {r, 0, 1}], Assumptions -> Λ > 0] or Assuming[{ Λ > 0}, ...


4

In Mathematica notation $Assumptions = ϕ ∈ Reals F[ϕ_, m_] := Integrate[1/Sqrt[1 - m Sin[θ]^2], {θ, 0, ϕ}] Plot[F[ϕ, -1], {ϕ, - Pi, π}, PlotStyle -> Red] In Maple notation F[\[Phi]_, k_] := Integrate[1/Sqrt[1 - k^2 Sin[\[Theta]]^2], {\[Theta], 0, \[Phi]}] Plot[F[\[Phi], I], {\[Phi], -2 Pi, 2 \[Pi]}] In maple


0

Perhaps this suffices: f[x_, y_] := NIntegrate[ UnitBox[s, t] BesselK[0, Sqrt[(x - s)^2 + (y - s)^2]], {s, -Infinity, Infinity}, {t, -Infinity, Infinity}] Visualizing: Plot3D[f[x, y], {x, -2, 2}, {y, -2, 2}, PlotRange -> Full, Mesh -> False, PlotPoints -> 25] Plot could be improved but I do not have time at present.


0

Today I believe I encountered the same problem when trying to reproduce the result of this paper about Lamb's problem and solutions mentioned above doesn't help in my specific case. After struggling for a while I managed to find a work-around and I think it's worth sharing. In short, if the integrate contains singular point(s) in addition, you may need to ...


14

This isn't necessarily how these functions are implemented, but MathematicalFunctionData gives a way to access definitions that are equivalent to the ones Mathematica uses. (* There are a total of 348 functions to choose from *) Length[functions = MathematicalFunctionData[]] 348 functions[[1]]["Definition"] {Function[{\[FormalX]}, Inactivate[ ...


2

You're not going to be able to find all the roots, because there are an infinite number of them. But you can use FindRoot directly to find any subset within a range. x = 3; eqn = BesselY[1, b] BesselJ[1, b x] - BesselJ[1, b] BesselY[1, b x] == 0; sol = FindRoot[eqn, {b, #}] & /@ Range[20] Here are the first few: Sort[DeleteDuplicates[sol[[All, 1, ...


2

x = 3; f[b_] = BesselY[1, b] BesselJ[1, b x] - BesselJ[1, b] BesselY[1, b x] ; FindInstance[{f[b]==0, 0 <= b <= 10}, b, Reals, 7] sol = b /. % // N (* {1.63562, 3.17884, 4.73809, 6.30272, 7.86971, 9.43793} *) There are only 6 real roots. You also can do it with Solve or Reduce: Solve[{f[b]==0, 0 <= b <= 10}, b, Reals] Plot[f[b], {b, 0, ...



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