# Tag Info

1

Some progress can be made by integrating over y only (with the correction that {…} be replaced by (…)). Integrate[x DiracDelta[r x - y] Exp[1/g^2 (Cos[x2 - x] + Cos[x2] + Cos[y + x2])], {y, 0, 2 Pi}, Assumptions -> r > 0 && 0 < x < 2 Pi] (* E^((Cos[x - x2] + Cos[x2] + Cos[r x + x2])/g^2) x HeavisideTheta[2 Pi - r x] *) In contrast, ...

2

It looks like you've already got several good answers, but I thought I would mention Apply. f2[x_, y_] := x^2 + y^3 g[t_] := {t^2, 3 t + 1} f2@@g[1] 65 The head of the result of evaluating g[1] is List, and Apply replaces this with f2. Essentially, List[1, 4] --> f2[1, 4].

2

Does this work for you? ClearAll[f2, g] f2[{x_, y_}] := f2[x, y] f2[x_, y_] := x^2 + y^3; g[t_] := {t^2, 3*t + 1} f2[g[1]] 65

3

Φ[r_, θ_, ϕ_] := {r Sin[θ] Cos[ϕ], r Sin[θ] Sin[ϕ], r Cos[θ]} grad = Grad[f@Φ[r, θ, ϕ], {r, θ, ϕ}] TeXForm @ Style[grad, TextAlignment -> Left] $\scriptsize\left\{\cos (\theta ) f^{(\{0,0,1\})}(\{r \sin (\theta ) \cos (\phi ),r \sin (\theta ) \sin (\phi ),r \cos (\theta )\})+\\ \ \ \ \sin (\theta ) \sin (\phi ) f^{(\{0,1,0\})}(\{r \sin (\theta ) ... 5 We couldn't be well-satisfied taking the black box system results when having in mind that no computer system is free of bugs and that its various aspects related to symbolic integration may seem not quite perfect (see e.g. this remarks). And so our purpose is to calculate this integral with more laborious approach, at least with a few steps which could be ... 4 Consider the calculation that Mathematica does internally, which is a chain rule: D[foo[12 bar], bar] // Trace // TableForm You can see that it needs to compute foo'[12 bar], and the way it does this is to create a special internal indexed variable SystemPrivateDerivativeX[1] and computes the derivative with respect to that. Because this doesn't have ... 2 One can either use SumConvergence or Sum with GenerateConditions -> True (thanks to a comment by Artes). As an example: Sum[Cos[n] x^n, {n, 1, Infinity}, VerifyConvergence -> True, GenerateConditions -> True] ConditionalExpression[-((x (1 + E^(2 I) - 2 E^I x))/(2 (E^I - x) (-1 + E^I x))), x != E^I && E^I x != 1 && Abs[x] < 1] ... 4 I will strictly focus on the question of equivalence of the two integrals. They are not equivalent. One hint is the behavior at$q\rightarrow0$. However, even more spectacularly it can be seen in the asymptotic limit$q\rightarrow\infty$. The first integral asymptotically tends to zero as: $$I_1\simeq cq^{-3/2}\exp(-q/4),$$ where$c$is a constant. ... 1 You could do it numerically? f[z_, s_] := Sin[Gamma[z]/z]^2/z^s; II[x_, s_] := NIntegrate[(f[x + I y, s] - f[x - I y, s])/(Exp[2 Pi y] - 1), {y, 0, Infinity}] Then ParallelTable[{x, II[x, s] // Im}, {s, 1, 2, 1/2}, {x, 1, 5, 0.05}] // ListLinePlot[#, PlotRange -> All] & 1 This can be solved easily using DSolve: (note that this is also easy to do manually) DSolve[{ Norm@D[{x, f[x]}, x] == Norm@D[{Cos[ϕ[x]] f[x], Sin[ϕ[x]] f[x]}, x] // ComplexExpand, ϕ[x0] == ϕ0 }, ϕ[x], x ] (* Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.*)... 8 Another option is to use the package MoreCalculus by Kuba << MoreCalculus diff[x_] := y''[x] + (epsilon - x^2) y[x] == 0; DChange[diff[x], {x^2 == s}, {x}, {s}, y[x]] $$\epsilon y(s)+4 s y''(s)+2 y'(s)=s y(s)$$ 9 deq = y''[x] + (epsilon - x^2) y[x]; deq /. {y -> (y[#^2] &)} /. x -> Sqrt[s] (*(epsilon - s) y[s] + 4 s y''[s] + 2 y'[s]*) 1 As per @MelaGo... Normal@LinearModelFit[data,{x,x^2,x^3,x^4,x^5,x^6},x] 6.67 - 42.6435 x + 16.1427 x^2 - 2.09464 x^3 + 0.126902 x^4 - 0.00367168 x^5 + 0.0000409458 x^6 3 Reduce also seems to work on the original f, though the answer it gives is somewhat long and involves a collection of Tan and ArcTan. f[x_, y_] := (4 y^2 - 6) (Cos[y \[Pi]] Cos[x] - Cos[2 y \[Pi]] Sin[x]); Reduce[f[x, y] == 0, {x, y}] 15 There are infinitely many (continuum) solutions and basically one cannot list them all e.g. because of the Cantor theorem. This is the reason why numerical approach is unsatisfactory. However we can make use of FindRoot to find a finite numerical subset of the solution space. Numerical solutions First we set up a net of values of x and starting point of y ... 6 A fix is to give SystemPrivateDerivativeX the NHoldAll attribute (which it probably should have, since it seems to be used as a dummy indexed variable): SetAttributes[SystemPrivateDerivativeX, NHoldAll] f[x_] = BesselI[0, 1.0 x]; f'[x] (* 1. BesselI[1, 1. x] *) 3 I would do it this way. Show[ Plot[{1/x}, {x, 0, 26}, PlotRange -> {Automatic, {0, .6}}, PlotStyle -> AbsoluteThickness[3]], Plot[{1/x}, {x, 2, 6}, PlotRange -> {Automatic, {0, .6}}, PlotStyle -> Transparent, Filling -> Axis, FillingStyle -> Red], Plot[1/x, {x, 8, 24}, PlotRange -> {Automatic, {0, .6}},... 0 A simple answer: Show[Plot[{1/x, If[2 <= x <= 6, 0, 1/x]}, {x, 0, 26}, Filling -> {1 -> {2}}, AxesOrigin -> {0, 0}, FillingStyle -> {Red}, PlotRange -> {0, 0.6}, PlotStyle -> Black], Plot[{1/x, If[8 <= x <= 24, 0, 1/x]}, {x, 0, 26}, Filling -> {1 -> {2}}, AxesOrigin -> {0, 0}, FillingStyle -> {Blue}, ... 2 Integrate[f[x], x] gives an antiderivative whose derivative is generically equal to f[x]. From the docs: For indefinite integrals, Integrate tries to find results that are correct for almost all values of parameters. A primitive function of a continuous function$f$such as the OP seeks is given by $$F(x) = \int_a^x f(t) \; dt \,.$$ prim = Integrate[3/(... 4 Yes, it is possible. First compute in a standard way Integrate[3/(5-4 Cos[x]),x] (*2 ArcTan[3 Tan[x/2]]*) Plot[2 ArcTan[3 Tan[x/2]],{x,-10π,10π}] Now try the Rubi package Get["Rubi"] Int[3/(5-4 Cos[x]),x] (*x+2 ArcTan[Sin[x]/(2-Cos[x])]*) Plot[x+2 ArcTan[Sin[x]/(2-Cos[x])],{x,-10π,10π}] 4 I have MA 11 and the problem still persist. Consider the following simplification that works for unsubscripted variables f1=(-p (a-x) ((a-b) P (b-x)+Q (b+x) (a-b+2 x)) +q (a+x) (-(a-b) Q (b+x)+P (b-x) (-a+b+2 x))) /((b-x) (-a+x) (a+x) (b+x)); rule1={(p+q)->1,(P+Q)->1}; FullSimplify[f1]/.rule1 $$\frac{Q}{b-x}+\frac{q}{x-a}-\frac{p}{a+x}+\frac{P}... 1 I got an answer to my question which I have asked, to connect with this I am writing my reply through this since I don't have enough point to ask in the comment section. E2 = 0.3; L = 2.5; M = 1; Chandra2 =NDSolve[{u'[\[Phi]]^2 - (2 M u[\[Phi]]^3 + u[\[Phi]]^2 - (2 M)/L^2 u[\[Phi]] + (1 - E2^2)/L^2) == 0, u[0] == 4/3}, u, {\[Phi], 0, 1.1}] PolarPlot[... 1 Version (* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *) Clear["Global`*"] p[V_, T_] = (R*T/(V - b)) - (a/(V^2)); Include an assumption Assuming[V2 > V1, Integrate[p[V, T], {V, V1, V2}]] (* ConditionalExpression[ a (-(1/V1) + 1/V2) + R T (-Log[-b + V1] + Log[-b + V2]), (V1 > 0 && (V1 > Re[b] || V2 < Re[b] || Re[b] ... 2 And there is no output sometimes when Integrate seems to hang or take long time, use the option GenerateConditions -> False ClearAll[p, V, T, V1, V2, R, a, b]; p[V_, T_] := (R*T/(V - b)) - (a/(V^2)); Integrate[p[V, T], {V, V1, V2}, GenerateConditions -> False] I still got no answer I do not know why on your system you get no answer. For me, it ... 3 I like the analytical solution @Artes. Nevertheless, if we need to find a numerical solution using NDSolve[], then we can differentiate the equation and use the first-order equation at one point as a boundary condition, for example, E2 = 3/10; L = 5/2; M = 1; eq = {u''[x] == 3 M u[x]^2 - u[x] + M/L^2, u[0] == 4/3, u'[0] == -(Sqrt[(54743/3)]/75)}; U = ... 6 First of all, I observed that your matrix can be represented as KroneckerProduct of smaller matrixes which are themselfes KroneckerProduct (or TensorProducts) of vectors: Ux = Table[m^2 Sin[m x], {m, 1, k}]; Vx = Table[p^2 Sin[p x], {p, 1, k}]; Uy = Table[Sin[n y], {n, 1, k}]; Vy = Table[Sin[q y], {q, 1, k}]; Ax = KroneckerProduct[Ux, Vx]; Ay = ... 1 The references from wuyudi seem meaningful but not very targeted. The attempt by Cesareo is probably a good ansatz, but not a solution. The sum of 2 + Sqrt1 is bigger than 1 and the series in the argument fairly divergent. A proper plot will give a proximate solution in the limits of the numerical accuracy of the local Mathematica installation. You are in ... 14 Studying basic solutions at the theoretical physics it is advantageous when one can get an exact solution. At the first sight one can see that the solution can be given in terms of elliptic functions (and elliptic integrals), though there are some tricks to remember in order to play with them seamlessly. Instead of dealing with approximate numbers we are ... 0 I cannot try this without the definition of the function \alpha(a,b;c), but here's a general solution: Define the integral A(c)=\int_{-\infty}^{\infty}\alpha(a,b;c)\,da\,db: A[c_?NumericQ] := NIntegrate[α[a, b, c], {a, -∞, ∞}, {b, -∞, ∞}] You can plot A(c) to get an idea of the desired solution c: Plot[A[c], {c, 0, 1}] You can find a numerical ... 5 Try the replacement u=x^3. Then dx=du/3u^2/3and Integrate[1/3 u^(-2/3)/Sqrt[1 - 2 u], u] /. u -> x^3 yields (* (x^3)^(1/3) Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3] *) Have fun! 0 A simple answer: (*ref/Grad Maclaurin*) orderedForm[poly_, var_List] := HoldForm[+##] & @@ MonomialList[poly, var][[Ordering[-Total[#] & @@@ CoefficientRules[poly, var], All, GreaterEqual]]]; (*f[x_,y_,z_] :=g[x,y,z]*) grads = NestList[Grad[#, {x, y, z}] &, f[x, y, z], 6] /. {x -> 0, y -> 0, z -> 0}; \[... 3 Depending on your needs consider PrePrint and MakeBoxes. Something like this: MakeBoxes[ Integrate[a__, Assumptions -> _], fmt : TraditionalForm ] := MakeBoxes[Integrate[a], fmt] Now when you view the integral in TraditionalForm: Integrate[f[x] Sin[(n π x)/L], {x, 0, L}, Assumptions -> k > 0 && L > 0] // TraditionalForm$$\... 7 While the function x Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3] can be obtained by integration of the integrand$\frac{1}{\sqrt{1-2x^3}}$in Mathematica 12 as observed by Bob Hanlon, it cannot be obtained in Mathematica 11.2 and earlier versions, even though D[x Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3],x] yields$\frac{1}{\sqrt{1-2x^3}}$also in version 11.2.... 5 Works well in version 12$Version (* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *) int = Integrate[1/Sqrt[1 - 2 x^3], x] (* x Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3] *) The derivative returns the original expression D[int, x] (* 1/Sqrt[1 - 2 x^3] *)

7

Slightly awkward, but certainly possible: Assuming[v ∈ Vectors[n, Reals], Integrate[TensorExpand[a Sin[Norm[a v]]/Norm[a v]], {a, 0, 1}]] (1 - Cos[Norm[v]])/Norm[v]^2

Top 50 recent answers are included