3

Both answers are wrong. Here are the correct answers: The first integral temp1 is off by Log[z]/Pi, the second one by 2/(Pi z). But, you know, integration is hard. Have pity. But it should be reported to WRI. Plot[D[-Log[z]/Pi + temp1, z] - f[z] // Evaluate, {z, 1/10, 10}, WorkingPrecision -> 16] Plot[D[2/(Pi z) + temp2, z] - g[z] // Evaluate, {z, 1/...


3

To add to what Kuba said: you can avoid the infinite loop by creating a button that starts the evaluation: Button["Evaluate notebook", NotebookEvaluate[EvaluationNotebook[], InsertResults -> True], Method -> "Queued" ] Note the Method option. Without it, the FE will freeze up completely.


2

Do integration first and then take the limit k-> inf. Use indefinite integration. Integrate only finds a solution for m==0. g[x_, m_, k_] = 1/((2*(x - m))^(2*k) + 1) mint[x_, k_] = Integrate[g[x, 0, k], x] (* x Hypergeometric2F1[1, 1/(2 k), 1 + 1/(2 k), -4^k x^(2 k)] *) Use a trick. Tell Limit that x^(2 k) is always positive. (Valid for Integer k)...


2

Does this help? f2[x_?NumericQ, m_] := Limit[1/((2*(x - m))^(2*k) + 1), k -> Infinity]; Module[{m = 0}, NIntegrate[f2[x, m], {x, m - 1/2, m + 1/2}]] (* 1. *)


1

Clear["Global`*"] nmax = 7; Table[{n, Assuming[n/2 > a, Hypergeometric2F1[-a, n/2 - a, n/2, z] // FunctionExpand // FullSimplify]}, {n, 1, nmax}] // Prepend[#, {"n", Hypergeometric2F1[-a, n/2 - a, n/2, z]}] & // Grid[#, Frame -> All] & // TraditionalForm Table[{n, Assuming[n/2 > a, ...


1

For general $n$ and $a$ no, but for many values it can simplify. For instance: a = 3/2; Table[Hypergeometric2F1[-a, n/2 - a, n/2, z] // FunctionExpand // FullSimplify, {n, 1, 5}] { 3 z + 1, (8 (z - 1) EllipticK[z] + 2 (z + 7) EllipticE[z])/(3 π), 1, (4 (z - 1) (z + 3) EllipticK[z] + 4 ((7 - 2 z) z + 3) EllipticE[z])/(15 π z), (Sqrt[z] ((8 - 3 z) z + 3) ...


1

using my myRootSearch and myRootSearchShow ClearAll[f]; f[x_, a_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin@Sin[a*x] Manipulate[ Column[{Quiet@myRootSearchShow[f[#, a] &, 0, 50, ImageSize -> 250], Quiet@myRootSearch[f[#, a] &, 0, 50]}], {a, 0, 4}]


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