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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