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11

There appears to be a bug, not in Integrate or in BesselK, but in the vertical-axis Ticks of LogLogPlot. Consider the simple case, LogLogPlot[Exp[x], {x, 10^-10, 1}, PlotRange -> All] as it should be. However, LogLogPlot[Exp[x], {x, 10^-10, 1}, WorkingPrecision -> 50, PlotRange -> All] In fact, any value of WorkingPrecision except ...


9

It seems that the analytic result is correct, but the precision is lost when converting it to a number. For example, if we use a higher precision, we get consistent results between numerical and analytical integration: f[a_, b_] = Integrate[x^2 Exp[-a x^2 - b x^4], {x, -∞, ∞}, Assumptions -> {a > 0, b > 0}] g[a_, b_] := NIntegrate[x^2 Exp[-a ...


8

The culprit, as suspected by xslittlegrass, is indeed numerical instability; in particular, this is because of the perverse combination of modified Bessel functions exhibited in the result returned by Mathematica. Using a recurrence identity satisfied by the modified Bessel function of the first kind, we can simplify the expression returned, like so: ...


7

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.


7

It will work perfectly once you write it correctly x /. NSolve[{3 BesselJ[1, x] + x D[BesselJ[1, x], x] == 0, 16 > x >= 0}, x] {2.9496, 5.84113, 8.87273, 11.9561, 15.0624} You made a mistake when you write D[BesselJ[1, x]] and also you are missing 3 in first term.


6

Just an update. This is now built in in WL: GeoProjectionData["PeirceQuincuncial"] {"PeirceQuincuncial", {"Centering" -> {90, -90}, "GridOrigin" -> {0, 0}, "ReferenceModel" -> 1}} GeoGraphics["World", GeoProjection -> "PeirceQuincuncial", GeoBackground -> "ReliefMap"]


5

Identities from MathematicalFunctionData can be useful when Mathematica can't seem to rewrite things in a form we want. whit[k_, m_, z_] = Activate[MathematicalFunctionData[ WhittakerW, "AlternativeRepresentations"][[4]][k, m, z][[1, 2]]] int = FullSimplify[whit[0, a, x] whit[1, b, x]] Integrate[int/x, x] (* large output of HypergeometricPFQs *) ...


4

As it turns out, although Mathematica is unable to deal with the integral as it stands, using the Kelvin functions yields an answer equivalent to the one returned by Maple. In particular, Integrate[(y/x) (KelvinBer[1, x]^2 + KelvinBei[1, x]^2), {y, 0, r}, {x, 0, y}] 1/32 r^4 HypergeometricPFQ[{1/2}, {3/2, 3/2, 2, 2}, r^4/64] where I used one of the ...


3

From the LegendreP help page:


3

Mathematica can't this integral to solve,but we can convert WhittakerW function to BesselI function.I'm use Maple to convert. $$W_{0,a}(x)=\frac{1}{2} \sqrt{\pi } \sqrt{x} \csc (\pi (a+1)) \left(I_a\left(\frac{x}{2}\right)-I_{-a}\left(\frac{x}{2}\right)\right)$$ $$W_{1,b}(x)=\frac{1}{2} \sqrt{\pi } \sqrt{x} \csc (\pi b) \left(-\frac{1}{2} x ...


3

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

Just for reference, the sum depicted in the OP has a representation in terms of the $q$-hypergeometric function. In particular, $$\sum_{n=0}^{\infty}\frac{(-1)^nq^{6n^2}}{(q^3;q^3)_n(-q;q)_{3n}}={}_1\phi_4\left({{0}\atop{-q,-q^2,-q^3,0}}\mid q^3,-q^6\right)$$ Compare: Series[QHypergeometricPFQ[{0}, {-q, -q^2, -q^3, 0}, q^3, -q^6], {q, 0, 15}] 1 - q^6 + ...


2

Difficulties encountered in solving the dispersion relation in the Question are due not so much to convergence of the integral as to the branch point in complex γ- space, which occurs where the argument of ArcTanh[] is equal to 1. Based on the related article cited in a comment above, the integration contour {γ, 1, Infinity} must pass below all non-analytic ...


2

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


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 ...


2

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 ...


2

You will probably only be able to do this numerically. The function being integrated is expr = Exp[x]/x*(ExpIntegralE[1, x])^2; LogPlot[expr, {x, 10^-5, 1}, AxesLabel -> {"x", "expr"}] Using numeric integration data = Table[{a, NIntegrate[expr, {x, a, Infinity}]}, {a, 10.^Range[-5, 0, .2]}]; ListLogPlot[data, Joined -> True, PlotRange ...


1

a bit of an extended comment, but i case anyone doesnt see the issue clearly, LegendreP[46, x] is a 46 order polynomial, with all even powers of x and alternating signs on the coefficients. We can separate out the positive and negative terms: {neg, pos} = { Total@MapIndexed[# x^(4 First@#2 - 4) &, #[[1 ;; ;; 2]]], Total@MapIndexed[# x^(4 First@#2 - ...


1

Your equation is too complicated for Solve. You can however try NSolve with a range for x. I choose here 0<x<1. NSolve[{C1*BesselK[0, 3.7268*10^-4*x] == 1.3*10^-6, x == 53.66*C1*BesselK[1, 3.7268*10^-4*x], 0 < x < 1}, {C1, x}] {{C1 -> 1.30002*10^-7, x -> 0.136815}}


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

On the other hand, we can use DifferentialRootReduce[] on LegendreP[1/2 (-1 + Sqrt[17]), x] to see what linear differential equation is satisfied by it: Operate[FullSimplify, DifferentialRootReduce[LegendreP[1/2 (-1 + Sqrt[17]), x], x]] DifferentialRoot[Function[{y, x}, {-4 y[x] + 2 x y'[x] + (-1 + x^2) y''[x] == 0, y[0] == ...



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