9

To calculate Integrate[ArcTan[a*Sqrt[1 - k^2*Sin[z]^2]]/Sqrt[1 - k^2*Sin[z]^2], {z, 0,Pi/2}] do differentiation under the integral with respect to the parameter a D[ArcTan[a*Sqrt[1 - k^2*Sin[z]^2]]/Sqrt[1 - k^2*Sin[z]^2], a] and integrate over z : Integrate[1/(1 + a^2 - a^2*k^2*Sin[z]^2), z] (* ArcTanh[(Sqrt[-1 + a^2*(-1 + k^2)]*Tan[z])/Sqrt[1 + a^2]]/ (...


9

If Integrate returns unevaluated, expand the integrand (into a sum of simpler functions) and map Integrate onto the expanded expression. If the term-by-term integrations are successful, the evaluated expression will be the original integral. Clear["Global`*"] $Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) f[x_, ...


7

A way to get sometimes unexpected results : The following integral is ' easy' Integrate[Log[1 - x]/(1 - x), x] (* (-(1/2))*Log[1 - x]^2 *) Now write explicitly the geometric series Sum[x^k, {k, 0, Infinity}] (* 1/(1 - x) *) inside the integral (-(1/2))*Log[1 - x]^2 = Integrate[Log[1 - x]/(1 - x), x] = Integrate[Sum[x^k, {k, 0, Infinity}] Log[1 - x], x], ...


4

Your first integral might be solved in two steps! As far as I know, Finite Element Meshing is only provided up to dimension 3 in Mathematica. The error message "...method FiniteElement " led me to a stepwise approach, which works without error message: int[p_?NumericQ, q_?NumericQ, r_?NumericQ] := NIntegrate[1/Sqrt[(x - p)^2 + (y - q)^2 + (z - ...


4

Let us focus on the underlying indefinite integral, Integrate[((u + t)^m - u^m)^(n - 2)*(u + t)^(m - 1), u, Assumptions -> (n | m) ∈ Integers && n > 1] Unfortunately, it returns unevaluated. In contrast, Integrate returns a result for any n satisfying the assumptions. For instance, With[{n = 3}, Integrate[((u + t)^m - u^m)^(n - 2)*(u + ...


4

Symbolic functions like Reduce and Solve provide an exact solution of the given equation in terms of the Root objects. xs = x /. First @ Solve[ x^x^x == 36 && x > 0, x] Root[{-36 + #1^#1^#1 & , 2.10703646395674928520879246974369419433`20.601147030102787}] This is not an algebraic number and cannot be represented as a root of ...


4

$Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) Clear["Global`*"] B = Erfc[x] Exp[-x^2/2] + Sqrt[2] Erfc[x/Sqrt[2]] Exp[-x^2]; If you explicitly state that a is a real value, Mathematica will tell you that the result is conditional on a > 0 BI = Assuming[Element[a, Reals], Integrate[B, {x, a, ∞}]] (* ...


3

You can construct the Taylor series with SeriesData: SeriesData[x, 0, {1, 1, 1/2, 1/6}] (* 1 + x + x^2/2 + x^3/6 + O[x]^4 *) SeriesData is flexible enough to do Puiseux series, not just Taylor and Laurent series.


3

If you wrap the factors in front of the derivatives in HoldForm, you can get the output you want: RHS = ((h-y) (2 \[Mu]-x) D[p[x,t],t]+(x+y)D[l[x,t],{t,2}])/(2 h \[Mu] l[x,t]); T[x,t] == Collect[RHS, _Derivative[_][__], HoldForm] //TraditionalForm


3

You can add the option NonConstants -> {r} to tell D that r does, indeed, depend on non-explicit variables. This is the best way, I think—then you can use all of D's standard rules. Otherwise, you'd need to define pattern-matching to account for every single possible way r could occur in an expression, which is...a lot. (You could probably also replace r ...


3

Clear["Global`*"] expr = -(1/2) (g[1] - g[2]) (g[1] - g[3]) (g[2] - g[3]) (E^(-(1/2) g[4]^2) Sqrt[2 π] Erfc[g[4]/Sqrt[2]] - 2 E^-g[4]^2 g[4] + Sqrt[π] Erfc[g[4]] (-1 + 2 g[4]^2)); Collect terms with Erfc expr2 = Collect[expr, Cases[expr, _Erfc, Infinity]] (* -E^(-(1/2) g[4]^2) Sqrt[π/2] Erfc[g[4]/Sqrt[2]] (g[1] - g[2]) (g[1] - g[3]...


3

Using some assumptions and an immediate assignment gives an analytic integral, which is easy to plot: F[x_, a_] = Assuming[-1 <= x <= 1 && a > 0, Integrate[Abs[x - z]^(-1 - 2 a), {z, -Infinity, -1}] + Integrate[Abs[x - z]^(-1 - 2 a), {z, 1, Infinity}]] (* (1 - x)^(-2 a)/(2 a) + (1 + x)^(-2 a)/(2 a) *) The plot looks exactly like ...


3

For a real function we can transform RealAbs[Sin[x - y]]^(-2/3) to 1/(1 - Cos[x - y]^2)^(1/3), then we have Integrate[1/(1 - Cos[x - y]^2)^(1/3), {x, 0, Pi}, {y, 0, Pi}] Out[]= (\[Pi]^(3/2) Gamma[1/6])/Gamma[2/3] Numerical result is 22.88949310061915. It can be compare to NIntegrate[1/(1 - Cos[x - y]^2)^(1/3), {x, 0, Pi}, {y, 0, Pi}, Exclusions -> {x ...


3

What you wrote is almost correct. You just need to changef' := 1/g[x] to f'[x_] := 1/g[x]. Using L[x], you will obtain the desired result


3

Sorry, not allowed to comment on glS' answer. I found the following useful, which builds a rule to change variables in (all!) integrals, building on changeVariables above: changeVariableRule[changeFunc_, newVar_] := (Integrate[expr_, {x_, start_, stop_}] :> Integrate @@ changeVariables[expr, x, changeFunc, newVar, {start, stop}])


3

Try this: https://reference.wolframcloud.com/language/ref/FunctionSingularities.html For instance: FunctionSingularities[Tan[x], x] or FunctionSingularities[ArcTan[x^y], {x, y}, Complexes]


2

Here is my explanation. The results of the codes ∞ ∈ Reals False and ∞ ∈ Complexes False prove that ∞ is not a real/complex number, so its substitution in any function makes no sense. The result of Limit[HurwitzLerchPhi[-1, 1, x], x -> ∞] 0 is sometimes written as HurwitzLerchPhi[-1, 1, ∞]==0. It should be noticed that such notation may confuse in ...


2

This should be able to be solved with the DifferentialForms.m package https://library.wolfram.com/infocenter/MathSource/482/. Here is a sample script using one of the above examples: Clear["Global`*"] << DifferentialForms` p1 = Exp[x] Cos[y] + Exp[y] + Sin[x] + y + 2 x q = d[p1] (*Take exterior derivative*) p2 = Simplify[HomotopyOperator[q]] (...


2

u[x_,y_]= x^2 + x y - y^2 v[x_,y_]:= 2 x y + y^2 Now, δx/δu|v = (δu/δx)^(-1)|v. Therefore, we first calculate (δu/δx)|v. For v constant we have: dv|v== δv/δx dx + δv/δy dy == 0 From this, dy|v: dy|v== -(δv/δy)^(-1) δv/δx dx with this in du|v: du|v== δu/δx dx -δu/δy (δv/δy)^(-1) δv/δx dx and therefore: (δu/δx)|v== δu/δx -δu/δy (δv/δy)^(-1) δv/δx δx/δu|v == (...


2

Try the following code fun = Sum[I^(-n) BesselJ[n, r] Exp[I n phi], {n, -k, k}]; pl = Position[fun, DifferenceRoot[_]]; fun1 = Table[{Subscript[f, i] @@ fun[[pl[[i, 1]], 0, 1, 1]] == fun[[pl[[i, 1]], 0, 1, 2]]}, {i, Length@pl}] /. {\[FormalY] -> y, \[FormalN] -> n}; ReplacePart[fun, Table[pl[[i]] -> Subscript[f, i], {i, Length@pl}]] // ...


2

Extended comment, not an answer: Perhaps StreamPlot shows you the reason: StreamPlot[{1, -2 E^-x ^2 x - Which[x >= 0, x /(0.1` + (0.9`/(1 + 5 t)) - x ),x < 0, x /(0.1` + (0.6`/(1 + 15 t)) + x ), True, 0]}, {t, 0,5}, {x, -.5, .5}] The trajectory x->0 seems to be unstable!


2

Clear["Global`*"] $Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) inttau[r_] = -(1/ Sqrt[0.0345106153943703 - ((37.3042 - r) (-25.578 + r) (62.8822 + r))/(3000 r)]); tauanalytical[r_] = Integrate[inttau[r], r]; tauanalytical /@ Range[5, 30, 5] (* {-1.80511 + 99.3459 I, -5.63104 + 99.3459 I, -...


2

Clear["Global`*"] f[x_, y_, t_] := (t x^(-3) + (1/ 50000000) x^(-4/3) + (80000^2 E^(-200 π y) π^2 Sqrt[y])/ 3 x - (4000 π y E^(-100 π y))/(x^2))/(800000000 π) FunctionDomain[f[x, y, t], {x, y}] (* x > 0 && y >= 0 *) Since you "know approximately the values on x and y that minimizes the function", add ...


1

With a command like N[TableForm[Table[{n, x, Hypergeometric2F1[1, (1/2)*(13 - 9*n), (3/2)*(5 - 3*n), x], Hypergeometric2F1[1, 6 - (9*n)/2, 7 - (9*n)/2, x]}, {n, 1, 5}, {x, 0, 1.3, 0.13}]]] you see that starting from n (integer) > 1 the Hypergeo functions show infinity the one or the other. For argument (sqrt[r/b0]) > 1 the numerical values get ...


1

If a and J are real and positive, then this gives a partial result: Integrate[√(a*Sin[θ]^2 - J^2*Sin[θ]^4), {θ, 0, π}, {ϕ, 0, 2*π}, Assumptions -> a > 0 && J > 0] (* ConditionalExpression[ 2*Pi*(Sqrt[a] + (a/J - J)* ArcCosh[Sqrt[a/(a - J^2)]]), a > J^2] *)


1

Fixing obvious mistakes and slightly redesinging the code in order to avoid NIntegrate error messages, we get: nOfControlPoints = 5; controlPoints = Subdivide[1, 10, nOfControlPoints - 1]; Clear[fctn, model]; model[y__Real] := Interpolation[Transpose[{controlPoints, {y}}]] fctn[{y__Real}][T_Real] := NIntegrate[model[y][lam]*intensity[lam, T], {lam, 1, 10}...


1

ClearAll[vv] vv[r_List] := Total[1/Abs[Subtract @@@ Subsets[r, {2}]]] Manipulate[Quiet @ Plot[vv[Array[r, n] /. r[1] -> r1], {r1, 0, 10}, AxesLabel -> {"r[1]", "vv[r]"}], {{n, 3}, 1, 20, 1}, Delimiter, Dynamic[Grid[Table[With[{i = i}, {HoldForm@r[i], Slider[Dynamic[r[i]], {0, 10}], Dynamic[r[i]]}], {i, 2, n}]]], ...


1

A mathematical approach: $\sum_{k=1}^n (-1)^{k+1} s_k$ translates as follows s = {a, b, c, d, e, f, g, h}; Sum[(-1)^(k+1) s[[k]], {k, 1, Length[s]}] (* a - b + c - d + e - f + g - h *)


1

Some more options Clear[a, b, c, d, e, f, g, h]; s = {a, b, c, d, e, f, g, h}; Total[Table[s[[n]] - s[[n + 1]], {n, 1, Length[s] - 1, 2}]] Sum[s[[n]] - s[[n + 1]], {n, 1, Length[s] - 1, 2}] Total[SequenceCases[s, {j_, k_} :> j - k]] Total[Subtract @@@ Partition[s, 2]] All give a - b + c - d + e - f + g - h


1

I'm left unsure what is the correct answer from the above analyses. Personally, I believe the integral should be evaluated as an improper or principal-valued integral. In this case, the singularity is the line y=x. Below I create a principal-valued integral function f1[x], then integrate this function over the requisite domain: f1[x_?NumericQ] := ...


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