5
Let me first emphasize that the requested procedure cannot be unique and is best performed by hands. Nonetheless, a completely automatic approach can be suggested:
MeijerGIntegral[expr_, var_, dum_] :=
Module[{s, sA, sB, tvar, si, fA, fB, xx},
s = MellinTransform[expr, var, tvar];
si = First[
Cases[FactorList[
s], {Gamma[ tvar] | Gamma[ a_ ...
5
This is just an additional comment to @Artes answer. (All the real work was provided in that answer.)
The term
(Sqrt[z (y + w)/a])/n
can be simplified without loss of generality with
c Sqrt[y + w]
So instead of 3 parameters ($z$, $a$, and $n$), you only need 1. Doing so gets you explicit solutions for all positive integer values of $t$. For example:
t = ...
4
The bug is fixed at least since v12.2:
Series[Pochhammer[1 + n, n], {n, Infinity, 1}]
(* E^SeriesData[n, Infinity, {-1 + 2*Log[2] + Log[n]}, -1, 3, 1]*
SeriesData[n, Infinity, {Sqrt[2], 0, -1/12*1/Sqrt[2]}, 0, 4, 2] *)
Plot[{(2^(2 n) Gamma[1/2 + n])/Sqrt[π], % // Normal} // Evaluate, {n, 0, 2},
PlotStyle -> {Automatic, Dashed}]
4
Too many symbolic constants usually turn out to be obstructive to calculate integrals symbolically. Another problem comes up when no restriction for parameters is given. We can set specific values to a few constants using With and (or) restrict constants with an option Assumptions in Integrate e.g.
With[{n = 1, a = 1, z = 1},
Table[{t, Integrate[ Exp[-...
3
Clear["Global`*"]
Assuming that the x in your code is actually l to agree with the equation image
eqn = ((2*κ)/(k + κ)^2) - 2*l*I*Exp[-2*I*l*k] == 0;
sol = Reduce[eqn, k]
Assuming that the conditions are met to produce the ProductLog results
sol2 = {sol[[-1, -1]] // ToRules}
Verifying the solutions,
FullSimplify[eqn /. sol2]
(* {True, True} *)
...
3
Since 11.2 (2017) Mathematica does have the built-in tools to deal with the Eisenstein series:
$$G_{2k}(\tau)=\sum_{(m,n)}{'}\frac{1}{(m+n \tau)^{2k}}$$
being the functions:
WeierstrassInvariantG2
WeierstrassInvariantG3
Calculating $G_{2k}(\tau)$ can be implemented as follows:
g2[2, t_] = WeierstrassInvariantG2[{1/2, t/2}]/60;
g2[3, t_] = ...
3
As @yarchik points out in the comments, your Meijer G-functions can be expressed as Mellin integrals. The integrands are found with
f1[t_] = MellinTransform[MeijerG[{{}, {}}, {{0, 1/2, 1}, {}}, x], x, t]
(* 2^(1 - 2 t) Sqrt[π] Gamma[2 t] Gamma[1 + t] *)
f2[t_] = MellinTransform[MeijerG[{{}, {}}, {{0, 1/2, 1/2, 3/2}, {}}, x], x, t]
(* 2^(1 - 2 t) ...
2
The following is true
LaplaceTransform[Exp[-(x/t)^2], t, s]
(* MeijerG[{{}, {}}, {{0, 1/2, 1}, {}}, (s^2 x^2)/4]/(Sqrt[π] s) *)
Note that your integral is given by the value obtained when s=1.
If Mathematica could evaluate
InverseLaplaceTransform[
MeijerG[{{}, {}}, {{0, 1/2, 1}, {}}, (s^2 x^2)/4]/(Sqrt[π] s), s, t]
you would get something close to your ...
1
This works:
f1 = 9*q^2*r0 - q^2*Sqrt[81*r0^2 - 12*q^2]
f2 = 9*q^2*r0 + q^2*Sqrt[81*r0^2 - 12*q^2]
r2 = (1/144^(1/3))*((f1^(1/3) + f2^(1/3)) +
I*Sqrt[3]*(f1^(1/3) - f2^(1/3)))
r3 = (1/144^(1/3))*((f1^(1/3) + f2^(1/3)) -
I*Sqrt[3]*(f1^(1/3) - f2^(1/3)))
a = (2*r2 + r3)/(r2 + 2*r3)
b = r2/(2*r2 + r3)
\[Phi][r_, r0_, q_] = (1/Sqrt[r2*(r2 + 2*r3)])*2*
...
1
Without arrow in the first line it can be preproduced with this code
LG[r_, ϕ_, p_, l_, w_,
z_] := (Sqrt[(2 p!)/(π (p + Abs[l])!)] 1/
w E^(-r^2/w^2) ((r Sqrt[2])/w)^Abs[l] LaguerreL[p, Abs[l],
2 r^2/w^2] E^(I l ϕ + I z));
LGA[r_, p_, l_, w_, z_] :=
Sqrt[(2 p!)/(π (p + Abs[l])!)] 1/w E^(-r^2/w^2) ((r Sqrt[2])/w)^
Abs[l] LaguerreL[p, Abs[...
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