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1

If you do l = 0.; k = 0; m = 1; Plot[MeijerG[{{2 - l + s}, {}}, {{0, 2 + k - l + s, 3 + k - l + s, 2 + k - l + m + s}, {}}, 30], {s, -0.1, 0.1}] you see that at l=0 && s=0 the function is not defined. There may be other places in parameter space where this is also the case (see Wikipedia entry for MeijerG).


1

$Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global`*"] freq = 5*^9; Omg = 2*Pi*freq; rho = 15; vco = 5907.5 (*core velocity*); vcl = 5944 (*clad velocity*); m = 0; U = Sqrt[Omg^2/vco^2 - Omg^2/v]; W = Sqrt[Omg^2/vcl^2 - Omg^2/v]; EDIT 2: Replacing Rationalize with Rationalize[#, 0]& to ensure ...


1

This is not a answer how to find, only a solution. From this site closed form expression is: $\sum _{n=1}^{\infty } \frac{P_n(x)}{n^2}=\frac{1}{24} \left(4 \pi ^2+12 \tanh ^{-1}(x)^2-3 \log ^2\left(\frac{1}{4} \left(1-x^2\right)\right)-48 \text{Li}_2\left(\frac{\sqrt{1-x}}{\sqrt{2}}\right)+12 \text{Li}_2\left(\frac{1-x}{2}\right)\right)$ f[x_, m_] := Sum[...


2

The integral Integrate[Log[2/(1 - q z + Sqrt[1 + q^2 - 2 q z])]/q, {q, 0, 1}] may be found in the following way : Calculate the derivative of the integrand with respect to z and determine the antiderivative with respect to q. Then find its limits at q -> 1 (easy) as well as at q -> 0 (harder, use Series[_, {q, 0, 0}]) and integrate their difference ...


0

I didn't succeed in making Mathematica "understand" your integral. The wellknown simpler case Integrate[Exp[I \[Omega] t], {t, -Infinity, Infinity}] == 2 Pi DiracDelta[\[Omega]] isn't understood by Mathematica too: Integrate[Exp[I \[Omega] t], {t, -Infinity, Infinity}] (*Integrate::idiv: Integral of E^(I t \[Omega]) does not converge on {-\[...


2

Look up BesselJZero. BesselJZero[2, 2]/3. (* 2.80575 *)


3

Based on @BobHanlon's very interesting answer one can proceed a little bit forward to get an approximated limit( !not a proof, only applied numerics!) . As Bob mentioned there is a dominant harmonic in the solution sum[m]~a+b Sin[2Pi (t-c)/8], which might be detected by Fouriertransformation. Alternatively I try NonlinearModelFit to get the harmonic data = ...


10

The result is 1/4 (ArcCosh[3] ArcSinh[1] - ArcSinh[1 - Sqrt[2]] Log[7 - 4 Sqrt[2] - 2 Sqrt[2 (10 - 7 Sqrt[2])]]) You can get to it using the integral representation of LegendreQ and then pulling the integral before the sum. With this integral repesentation of LegendreQ (omitting a purely imaginary part) I1=Integrate[(1/Sqrt[2] + I Cosh[t]/Sqrt[2])^(-1 - n), ...


9

Clear["Global`*"] Defining the sum recursively: LegendreQ[n, Sqrt[2]/2]/(n + 1) /. n -> 0 // Simplify (* 1/2 Log[3 + 2 Sqrt[2]] *) sum[0] = Log[3 + 2 Sqrt[2]]/2.0`20; sum[m_Integer?Positive] := sum[m] = sum[m - 1] + LegendreQ[m, Sqrt[2]/2]/(m + 1) Calculating the points for a plot (this is slow) sum[1000] (* 0.30718615098022934340 *) ...


7

Update One of the (brute force) estimates I posted earlier using the Method option: Method -> {"WynnEpsilon", "ExtraTerms" -> 200, "Degree" -> 2} is in agreement with the estimates from Bob Hanlon's and Ulrih Neumann's answers: 0.3071246932 First comment/answer (Not an answer, extended comment -- I have to ...


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