12
Actually, the contours $C$ in the two integrals are different.
By definition, $C$ goes from $+\infty$ through an clockwise path return to $+\infty$, encircling all poles of $\prod_{j=1}^m\Gamma(b_j-s)$ (each pole exactly once) and none pole of $\prod_{j=1}^n\Gamma(1-a_j-s)$. In this case, the integral-contours of $\textrm{G}^{4, 0}_{2, 4}\Big({1,c+\frac12\...
10
This is an interesting example where new improvements in version Mathematica 12.2 of elliptic functions handling appear important to get a correct result.
TraditionalForm[
intd[x_] = FullSimplify[
Integrate[ Sqrt[(37 - (45 37 x^2)/(74 150))^2 Sin[t]^2
- (40 - (27 37 x^2)/(16 150))^2 Cos[t]^2],
...
9
Have a look at the definition of the hypergeometric function: you can see that $_2F_1(a,b;2;1)=\frac{\Gamma(2-a-b)}{\Gamma(2-a)\Gamma(2-b)}$:
h[a_,b_] = Sum[Pochhammer[a, k]*Pochhammer[b, k]/Pochhammer[2, k] * 1/k!, {k, 0, ∞}]
(* Gamma[2 - a - b]/(Gamma[2 - a] Gamma[2 - b]) *)
The general form of this hypergeometric function at $z=1$ is described on ...
9
If I use the integral representation of the Gaussian hypergeometric function, and then differentiate that before plugging in the arguments, I get the following expression:
$$\frac{2\pi^4}{5}+8 \int_0^1 \left(2(\operatorname{artanh}(1-2t))^2-\pi ^2\right)(\operatorname{artanh}(1-2t)\log(1-t))^2 \, \mathrm dt$$
which does not symbolically evaluate in ...
8
Increasing MaxExtraPrecision to the value 1000 helps.
$MaxExtraPrecision = 1000;
N[Derivative[2, 4, 0, 0][Hypergeometric2F1][0, 1, 2, 1],20]
(*291.98909605411660258*)
Try this, you will see what is going on:
$MaxExtraPrecision = 100;
N[Derivative[2, 4, 0, 0][Hypergeometric2F1][0, 1, 2, 1],20]
(*N::meprec: Internal precision limit $MaxExtraPrecision = ...
7
For non-integer $M$ we can get an explicit expression for this derivative, as given on the Wolfram Functions site:
f[M_] = Derivative[1, 0][BesselK][-M, 2] // FunctionExpand
(* huge result *)
f[0.37]
(* -0.0179285 *)
Plot[f[M], {M, -1.5, 1.5}]
For specific values of $M$ we can get explicit expressions for this derivative directly:
Table[...
6
Here is an indirect way to prove this. Applying DifferentialRootReduce[] to your res gives a tenth-order linear ODE:
Short[ode = First[Head[DifferentialRootReduce[res, x]]][y, x][[1]]]
(-8741760 - 1918960 x^3 + 143800 x^6 - 2560 x^9) y[x] +
(-7160448 x + 2599248 x^4 - 337056 x^7 + 6144 x^10) y'[x] +
(11531328 x^2 - 1639768 x^5 + 265156 x^8 - 4864 x^...
5
When generalized, polynomials turn into functions. See Hermite Function
HermiteH[ν, z] ==
2^ν*Sqrt[Pi]*
((1/Gamma[(1 - ν)/2])*
Hypergeometric1F1[
-(ν/2), 1/2, z^2] -
((2*z)/Gamma[-(ν/2)])*
Hypergeometric1F1[
(1 - ν)/2, 3/2, z^2]) // FullSimplify
(* True *)
5
Maybe worth showing this:
Verification of log scaling, although with a significant but small error in the first point:
plot = Plot[Zeta[x], {x, 2, 20}, ScalingFunctions -> "Log"];
{xvals, logzvals} =
Transpose@First@Cases[plot, Line[p_] :> p, Infinity];
Rest@Log@Zeta[xvals] == Rest@logzvals
(* True *)
First@Log@Zeta[xvals] - First@...
5
This might get you most of the way there. First break up the integrand into 6 pieces:
integrand = Sqrt[θ]/(Sqrt[p] BesselK[1, Sqrt[p θ]]) x BesselK[0, x Sqrt[p θ]] (Sqrt[θ]/
Sqrt[s] (rTD BesselK[1, rTD Sqrt[s θ]] (BesselI[0, x Sqrt[s θ]] -
BesselK[0, x Sqrt[s θ]] BesselI[0, Sqrt[s θ]]/BesselK[0, Sqrt[s θ]]) +
BesselI[0, x Sqrt[s θ]] (x BesselK[1, x ...
5
This is an example of the general problem of analytically-continuing a multi-valued function along a continuous path.
In the case of an algebraic function such as $w=\sqrt{z^8}$, we can write it as $f(z,w)=w^2-z^8=0$ and in your case, letting $z(t)=1+it$, write:
$$
\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}=\frac{4i(1+it)^7}{w}
$$
We next solve the (multi-...
4
The message General::munfl only occurs for machine numbers. So, you can fix the issue by using arbitrary precision numbers internally, and then converting to a machine number:
function[t_, V_, l_] := N @ ReleaseHold[
SetPrecision[Hold[1/Sin[t]*LegendreP[-1/2+V*I,l+1,Cos[t]]], 30]
]
Then:
function[.00001, 10, 50]
1.34228*10^-189 + 0. I
3
Use Cartesion coordinate and define a region to cut the surface and cut the domain of integrate.
f[x_, y_] := BesselJ[0, r]^2 /. r -> Sqrt[x^2 + y^2];
plot = Plot3D[f[x, y], {x, y} ∈ Disk[{0, 0}, 12],
PlotPoints -> 50, PlotRange -> All];
reg = ImplicitRegion[x <= -5 && Sqrt[x^2 + y^2] <= 12, {x, y}];
reg // Region
NIntegrate[f[x, ...
3
An argument could be made that the result being returned by ClebschGordan[] is generically correct; that is, the expression that comes from the hypergeometric representation of the Clebsch-Gordan coefficient is correct except at a countable number of values.
In particular,
ClebschGordan[{2, 0}, {l2, 0}, {2, 0}] // FullSimplify
Piecewise[{{(Sqrt[5] (-1 + ...
3
This is not a bug. Just the scale is small enough to be noticed by the eye!
You can try this code, which is just your code added a more length of the y-axis :
Plot[Zeta[x], {x, 2, 20}, ScalingFunctions -> "Log",
PlotRange -> {{0, 20}, {0, 2}}]
Now the result is :
in which the logarithmic scale is more emphasized.
1
A related, but potentially more flexible approach than Daniel Huber's. First do the most minimal transformations necessary to allow Mathematica to interpret this as an expression
hexpr = "H(2,2,x),H(2,1,x),H(2,1,2,x),H(1,2,3,4,x),H(1,2,3,4,5,x)" //
StringReplace[{"(" -> "[", ")" -> "]"}] // ...
1
12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)
gives these results using antiderivate with integration limits 0 and Pi:
Plot[{Re[NIntegrate[Sqrt[(37 - 45*37*(x^2/(74*150)))^2*Sin[t]^2 -
(40 - 27*37*(x^2/(16*150)))^2*Cos[t]^2], {t, 0, Pi},
WorkingPrecision -> 50]],
Im[(1/400)*(32000 - 333*x^2)*
EllipticE[(1900160000 - 28416000*x^2 + ...
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