8

Actually V 12.1 can do it directly, you just have to wait a little bit long: Clear["Global`*"]; int = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], {x, a, Infinity}, Assumptions -> 0 < d < c < b < a] May be OP used different Mathematica version? It will be good to post which version was used. Screen shot below: Old ...


8

This is a kind of elliptic integral, and it can be calculated if we prescribe appropriate assumptions. Integrate[ 1/Sqrt[(x - a) (x - b) (x - c) (x - d)], {x, -Infinity, 0}, Assumptions -> 0 < d < c < b < a] -((2(EllipticF[ArcSin[Sqrt[(a - d)/(b - d)]], ((a - c) (b - d))/((b - c) (a - d))] - EllipticF[ArcSin[Sqrt[(b (a - d))/(...


8

I presume that you wish to eliminate some variables in the last expression in terms of variables defined in the other expressions. Let me begin with two pieces of advice: Do not use subscripted variables. They may look nice, but they can cause problems. Simplification is in the eye of the beholder. Mathematica's idea of simplification, based on LeafCount,...


7

With the concept of greenfunction you might find a solution: update The homogenous solution of your ode is Sin[x] which fullfills the initial conditions! To calculate reenfunction first solve (homogenous initial conditions!) Y = DSolveValue[{y''[x] + y[x] == DiracDelta[x - ξ] , y[-Pi/2] == 0, y'[-Pi/2] == 0}, y[x], x] ; G = Function[{x, ξ}, Evaluate[Y] ...


7

Artes shows that the integral can be expressed in terms of the Legendre-Jacobi integrals. I will now present a solution that uses the Carlson elliptic integrals, for which I have written a package implementing them. In particular, our working formulae are a combination of formulae 19.29.4 and 19.29.6 in the DLMF: <<Carlson` (* load package after ...


5

This is what GenerateConditions$\to$All is supposedly supposed to do, but I find it doesn't usually work. From the documentation GenerateConditions$\to$All should return all possible answers using Piecewise But I find it doesn't work for your integral and practically no, what you seek is unfortunately not possible. Straight from the documentation on ...


5

Return to the original problem: s = DSolve[{y''[x] + y[x]==Sum[DiracDelta[x-2^n]/2^n,{n,0,Infinity}],y[-Pi/2]==-1,y'[-Pi/2]== 0}, y[x], x] According to the Mathematica documentation, this is a piecewise homogeneous differential equation with a special inhomogeneity. This is solved by a linear combination of the trigonometric function fitted to the ...


5

You are effectively calling Q with two different argument patterns (Q[foo, t] in the first call in the nested calls and Q[bar[t],t] in the outer call) but your Q is defined only for the first argument pattern. If you add the definition for the second signature, you get the desired result: ClearAll[Q] Q[f_, s0_] := Q[f, s0] = D[f[s0], s0] Q[f_[s0_], s0_] := ...


5

For the posted example, TensorReduce does the trick: TensorReduce[ Transpose[Transpose[A].Transpose[B]], Assumptions -> {A ∈ Matrices[{m, n}], B ∈ Matrices[{k, m}]} ] B.A


4

From a small note in the documentation for Arrays, The symmetry sym can be given in several forms. First, it can be given as expressions like Symmetric[{s_i, ..., s_k}] or Antisymmetric[{s_i,...,s_k}], with the slots s_i being different positive integers between 1 and the rank r. It can also be given as a list of generators of the form {perm,\[Phi]}, ...


4

A slick reformulation of bbgodfrey's answer is to recognize that the required computation is equivalent to evaluating an appropriate matrix power: With[{n = 5}, MatrixPower[{{-m, 1, 0}, {0, 0, 1}, {1, 0, 0}}, n - 2, {1, 0, 0}]] // Simplify {1 - m^3, -m, m^2} With[{n = 25}, MatrixPower[{{-m, 1, 0}, {0, 0, 1}, {1, 0, 0}}, n - 2, {1, 0, 0}]] // ...


4

Correct symbolic solution has already been given in the comment and answers, I'd like to show why your second attempt gives the incorrect result. What you've actually obtained is: Sin[1.1] (* 0.891207 *) In other words, the summation involving DiracDelta doesn't contribute to the numeric solution at all. So, why does this happen? Well, though there do ...


4

There used to be a nice FDFormula at WRI site, but it is gone now. But I used it before. Here is the result. I'll show some examples then the code at end getFormula[1, {-1, 0, 1}, "centered"] The first argument to getFormula is derivative order. So 1 for first order, 2 for second order. The second argument is list of points to generate the difference ...


4

In the newest version (i.e. 12.1) this integral evaluates a bit long, however changing the variable $x \mapsto t = x-a\;$ this can be evaluated a few times faster. int2 = Integrate[ 1/Sqrt[t (t + a - b) (t + a - c) (t + a - d)], {t, 0, ∞}, Assumptions -> 0 < d < c < b < a] 2 EllipticF[ ArcSin[ Sqrt[(b - d)/(a - d)]], ...


3

It looks like support for recognizing this property was added in M11.2 or M11.3. In M11.1 I get: Assuming[ (a|b|c) ∈ Reals && (X|Y) ∈ Matrices[{n,n}], TensorExpand[ KroneckerProduct[IdentityMatrix[n],a X].(b KroneckerProduct[c IdentityMatrix[n],Y]) ] ] KroneckerProduct[IdentityMatrix[n], a X].(b KroneckerProduct[c ...


3

Does this work for you? I only tested it on your example ClearAll[f, x]; f[x_] := Exp[x^2] (f'[x] /. f[x] -> HoldForm[f[x]]) // Simplify This will work with different variable name (f'[y] /. f[y] -> HoldForm[f[y]]) (1 + D[f[x], x]) /. f[x] -> HoldForm[f[x]] To do f'[x]^2, make sure to do the derivative first then square the result, like this ...


3

RiccatiSolve and DiscreteRiccatiSolve can handle symbolic matrices as long as the eigenvalues of the Hamiltonian are numeric so that the solver can determine those on the left- and right-half planes. In this case aa = {{1, 0}, {0, 1}}; bb = {{1, 0}, {0, a}}; qq = {{1, 0}, {0, 1}}; rr = {{1, 0}, {0, 1}}; and the eigenvalues of the Hamiltonian matrix are ...


2

I don't think this would be simple to achieve automatically (i.e. if you do not know which variables should be "separated", and the corresponding ranges), but you can help MMA along: integrand = f[a*x] g[c1*z1 + c2*z2 + c3*z3] h[e*x] k[d1*z1 + d2*z2 + d3*z3]; split = Times @@@ GatherBy[List @@ integrand, FreeQ[x]] (* {f[a x] h[e x], g[c1 z1 + c2 z2 + c3 z3]...


2

Actually, I do believe that this question is of pure mathematics character and does not belong here. Nonetheless, I provide a short answer. A lot of information on special functions can be found in the so called Bateman Manuscript Project. There are download links at the bottom of the page. Specifically, you need the volume I, Sec. 6.5 As I mentioned in the ...


2

Version 12.1 solves the first integral without difficulty. With[{z = 1/2 + I t}, Integrate[z^(-q - 1) (1 - z)^(-λ - 1), {t, -Infinity, Infinity}, Assumptions -> {q >= 0, λ > 0}]] (* (2 π Gamma[1 + q + λ])/(Gamma[1 + q] Gamma[1 + λ]) *) I shall give the second integral some thought tomorrow.


2

I suggest you using Derivative: Q[f_] := Derivative[1][f] Q[f_, s0_] := Derivative[1][f][s0] Q[f, s0] f'[s0] Q[Q[f], s0] f''[s0] BTW, the number 1 in the first [...] after Derivative can be other integers, so that the nested application can actually be lifted.


2

On Mathematica 12.1 I can get the Eigenvectors and Eigenvalues. I noticed each eigenvalue was 1/3 times some Root expression so I created some replacement rules to express the eigenvectors in terms of the $\mu_i$ m = {{p3 + p8/Sqrt[3], p1 - I p2, p4 - I p5}, {p1 + I p2, -p3 + p8/Sqrt[3], p6 - I p7}, {p4 + I p5, p6 + I p7, -((2 p8)/Sqrt[3])}}; eval ...


2

Mathematica's antiderivative has poles and or not continuous, so FTOC can not be used. Either that (I have not figured which it is), or the anti-derivative is simply wrong. But using Rubi, it produces anti-derivative which allows using FTOC on it and does not produce complex values. And now the result of integration is real. << Rubi` q = 14; Arubi =...


2

This is your function: f[g_,t_]:=-4Sin[g t/2]/g^2/t-4Sin[g t/4]^2/g-2Cos[g t/2]/g+Pi t/2 With transformed variable $z=g*t$ you get a new function: h[g_,z_]=f[g,t]/.t->z/g $$\frac{\pi z}{2 g}-\frac{4 \sin ^2\left(\frac{z}{4}\right)}{g}-\frac{4 \sin \left(\frac{z}{2}\right)}{g z}-\frac{2 \cos \left(\frac{z}{2}\right)}{g}$$ which you plot (with, for ...


1

Since you end up substituting numerical values, I recommend doing that earlier rather than later. This should greatly simplify the complex task you set for Integrate and is likely to lead to better results. You also evaluate the integral for real values of Δp, so it would make sense to make Integrate aware of that possible simplification as well, through ...


1

This stuff doesn't want to simplify. My best effort so far: fz[g_] := (Dt[Re[g], z] - I Dt[Im[g], z])/2 (*Wirt df/dz*) fw[g_] := (Dt[Re[g], z] + I Dt[Im[g], z])/2 (*Wirt df/dOverscript[z, _]*) cz[g_] := fz[Conjugate[g]] (*dOverscript[f, _]/dz*) cw[g_] := fw[Conjugate[g]] (*dOverscript[f, _]/dOverscript[z, _]*) fzz[g_] := (fz@*fz)[g] fzw[g_] := (fw@*fz)[...


1

Here's my work-around. Not sure if it can be applied to your real problem, but it can at least help you to construct a sample closer to your real problem, I think. We define the following 2 functions: ClearAll[unitStep, v52] SetAttributes[v52, HoldAll] v52[expr_Integrate] := Check[expr /. HeavisideTheta[0] -> 1/2, ∞ NIntegrate @@ Unevaluated@expr // ...


1

f[x_, y_] := (1 - Exp[-y x] - y x); Use TagSet or TagSetDelayed If the variables u and v are not literal ClearAll[f] f /: (f[s[v_], u_] /; u =!= v) = f[s[v], v]; f[s[v], u] - f[s[v], v] (* 0 *) f[s[x], y] - f[s[x], x] (* 0 *) If the variables u and v are literal ClearAll[f] f /: f[s[v], u] = f[s[v], v]; f[s[v], u] - f[s[v], v] (* 0 *) f[s[x], y]...


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