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9

Here is a visualization using MeshFunctions (as mentioned by Daniel in the comment): f = x y z; g = x^2 + 10 y^2 + z^2; gp = With[{r = 3}, RegionPlot3D[g < 5, {x, -r, r}, {y, -r, r}, {z, -r, r}, PlotStyle -> Orange, PlotPoints -> 40, Mesh -> None, ViewPoint -> Front, PlotTheme -> "Classic"] ]; With[{r = 3}, Manipulate[ ...


5

Let us define the equation: Clear[eq]; eq[m_, f_] := 1/(x - 1) - (m + 1)/(x^(m + 1) - 1) == f; where x stays for Exp[f/(k t)]. If one applies the function Solve to it, Mma clearly answers that it cannot provide exact solution of this equation. It should not be expected, therefore, that one can find any other analytical solution. Numerically, it is ...


5

You can gain insight by evaluating and then simplifying the indefinite integrals. Integrate[(1 + b x + c y)/(1 + e x + f y + I η), y, x, Assumptions -> {x ∈ Reals, y ∈ Reals, b ∈ Reals, c ∈ Reals, e ∈ Reals,f ∈ Reals, η ∈ Reals, η != 0}]; ans = Collect[%, {ArcTan[η/(1 + e x + f y)], Log[1 + e^2 x^2 + 2 f y + f^2 y^2 + 2 e (x + f x y) + η^2]}, ...


5

One way to approach this is to look for the source of the problem. In this case, the outer integral (on y) is irrelevant to the problem, because even the simpler integral int = Integrate[(1 + b x)/(1 + e x + I η), {x, -1/2, 1/2}, Assumptions -> {b ∈ Reals, c ∈ Reals, e ∈ Reals, η ∈ Reals, η != 0}] gives a conditional expression. But now the answer ...


4

Here is another way: multi[f_, x_, a_, b_, n_] := Inactive@Integrate @@ {f @@ Array[x[#] &, {n}]}~Join~Array[{x[#], a[#], b[#]} &, {n}] I used Inactive because I don't know if the purpose is purely typesetting, or if you want it to try to evaluate (if so remove it, or use Activate later). multi[g, y, ymin, ymax, 10] (* Inactive[Integrate][ ...


4

With using number of assumptions, and breaking thing step by step: (I do things step by step, just to see where the problem is when it shows up, much easier to debug this way) integrand = 1/(r g) Exp[-p (1 + r a) - b q ((1 + r a)/(1 + g x))] Exp[-a/r] Exp[-b] Exp[-x/g]; z0 = Assuming[Re[(1 + q + a q r + g x)/(1 + g x)] > 0, ...


3

Because NDSolve cannot accommodate the x=0 boundary condition, it is necessary to perform this computation by discretizing the PDE in x. The resulting do-it-yourself procedure is discussed in Introduction to Method of Lines. For illustrative purposes, assume that x is divided into five equal segments. n = 5; h = 1/n; with a variable defined at each node, ...


2

Using Sequence and Transpose to splice in the integration parameters: mi[f_, x_, a_, b_]:=Integrate[f @@ x, Sequence @@ Transpose@{x, a, b}]; Usage (exemplary): f[v_,w_,y_]:=v^2+w^2+y^2 x={x1,x2,x3}; a={a1,a2,a3}; b={b1,b2,b3}; mi[f,x,a,b] (* -(1/3)(a1-b1)(a2-b2)(a3-b3)(a1^2+a2^2+a3^2+a1 b1+b1^2+a2 b2+b2^2+a3 b3+b3^2) *)


1

Something like this: f[x_, m_] := 1/(x - 1) - (1 + m)/(x^(1 + m) - 1) g[y_, m_] := x /. NSolve[f[x, m] == y, x]


1

I'm not sure if this will be complete enough for you, but here's a snippet of code that worked for me: vars = Table[x[i], {i, 1, 7}] Do[x[i, 0] = RandomInteger[{0, 5}], {i, 1, 7}] Do[x[i, 1] = RandomInteger[{6, 10}], {i, 1, 7}] Integrate[Plus @@ vars, Sequence @@ Table[{x[i], x[i, 0], x[i, 1]}, {i, 1, 7}]] In this case I create a list of variables x[i] ...



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