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0

The work-around posted by george2079 in this answer will work for you. LogLogPlot[(-1 + E^#)/#&[N[Rationalize[x, 0], 30]], {x, 10^-15., 10^-5}]


0

This is not straightforward. You have some hope of stumbling into approximately correct results because of the aggressively decreasing exponentials in your sum, but I make no promises that you can get any particular degree of precision. You should probably construct little machines for each $n$ that sequentially produce values of $\beta_{n,k}$ for ...


1

Use higher precision f[x_] = -(10^-20 x)/(0.99005 - E^(10^-12 x)) // Rationalize // Simplify; int = Integrate[f[x], {x, 0, 10^9}] // FullSimplify; int // N[#, 20] & // Chop[#, 10^-20] & (* 0.47118211649097404645 *)


1

First, if you use Integrate, you should define your function exactly. Don't use approximate numbers like 0.99005. f[x_] = -(x/(10^20*(99005/100000 - Exp[x/10^12]))) Integrate[f[x], {x, 0, 10^9}] (* Complicated exact expression involving Log and PolyLog *) Evaluating this approximately indeed yields a small imaginary part. %//N (* 0.471182 - ...


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 - ...


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 ...


3

From the LegendreP help page:


0

One more possibility would be to directly manipulate lists of coefficients of polynomials. To do that, we need to define a few required operations: SetAttributes[{add, mult}, Orderless]; add[c1_?VectorQ, c2_?VectorQ] := Total[PadRight[{c1, c2}]]; mult[{0}, c2_?VectorQ] := {0}; mult[c1_?VectorQ, c2_?VectorQ] := ListConvolve[c1, c2, {1, -1}, 0]; diff[{_}] := ...


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] == ...


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 + ...


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 ...


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: ...


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 ...


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

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

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}}


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 *) ...


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 ...


2

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


8

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.


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 ...



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