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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[ ...


7

Here's how to make Mathematica integrate this, but a lot is done by hand. Cos[β] Exp[I z Cos[β - α]] == Cos[β] Cos[z Cos[α - β]] + I Cos[β] Sin[z Cos[α - β]] The integral of the first summand is zero. Make the substitution γ == α - β and expand the Cos[β] to Cos[α] Cos[γ] + Sin[α] Sin[γ]. Also note that the function is Pi-periodic. The integral ...


6

In this case you can find minimum by comparing values at zero derivates: TakeSmallestBy[{A /. #, #} & /@ Solve[D[A, x] == 0, x, Reals], First, 1] {{Sqrt[3], {x -> 0}}} TakeSmallestBy is a v10.1 function similar to MinimalBy, but performs numerical comparisons.


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 ...


5

This question is being automatically bumped as unanswered. However, we have an authoritative answer in comments: Investigating as a regression. You can put a "bugs" tag on it if you like. --Daniel Lichtblau


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 ...


4

Although Daniel pointed correctly out that problems related to Integrate have been the subject of many discussion here, I found it worthwhile to study this case in detail, because a condition including Mod was not discussed up to now, as far as I know. The aim is to find out if there is a bug, and if so, where exactly it is sitting, and/or, if possible, to ...


4

I believe this equation are quiet unstable for the initial values, so there is a two solutions. You can either specify AccuracyGoal: ListPlot@NDSolveValue[{-w''[x] + 2/x w'[x] + w[x] == 0, w[1/10^6] == 10^-2, w[5] == 1}, w, {x, 1/10^6, 5}, AccuracyGoal -> 10] Or use the DSolveValue, the equation are solvable analytically: ...


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

It seems to me that Integrate can do some strange things with your function g. From plotting g, we can see the integral should clearly be zero. g[x1_, y1_, x2_, y2_] = -Log[Sqrt[(x2 - x1)^2 + (y2 - y1)^2]] Plot[g[Cos[θ], Sin[θ], 1/10, 1/10], {θ, 0, 2 π}] Further, Integrate[g[Cos[θ], Sin[θ], 1/10, 1/10], {θ, 0, 2 π}] gives zero as expected, but ...


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

Perform the three "easy" integrals first: gg = Assuming[T > 0 && H > 0 && G > 0 && F > 0 && p > 0 && q > 0, 1/(F G H T) Integrate[ Exp[-y/T] Exp[-a/F] Exp[-b/H] Exp[-x/G], {a, 0, \[Infinity]}, {x, 0, \[Infinity]}, {b, 0, \[Infinity]}] ] (* ...


2

I believe the main problem with original integration is due to Mathematica try to integrate with a and b being complex numbers. I have some doubts that it's even possible to analytically integrate with complex constants. Integrate[Log[a Cos[x]^2 + b Sin[x]^2], {x, 0, 2 Pi}, Assumptions -> a > 0 && b > 0] (* π (Log[(a b)/16] + 2 Log[(1 + ...


2

fixed in 10.1 (windows): code: Clear[x] Integrate[(1 - x)*(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}]/Pi


2

fixed in 10.1 windows code N[Integrate[Sqrt[1 + x^3], {x, -1, 3}]]


2

With version 9.0.1, f[x_] := (p^2 + k^2 - 2 p k x)/(x - (p^2 + k^2 + 1 - ((p^2 - k^2)^2)/4)/(2 p k)); ans9 = Integrate[f[x], {x, -1, 1}, PrincipalValue -> True] (* ConditionalExpression[1/2 k p (-8 + ((-4 + (k^2 - p^2)^2) ArcCoth[(8 k p)/(-4 + k^4 - 4 p^2 + p^4 - 2 k^2 (2 + p^2))])/(k p)), k^4 + p^4 < 4 + 4 p^2 + 2 k^2 (2 + p^2) && (k ...


2

I would try to see if you can use Distribute for this: Distribute@ Integrate[f[x] + DiracDelta[x - y] g[x], {x, -Infinity, Infinity}] Unlike Map, Distribute is especially (though not exclusively) intended for use with sums.


2

One approach, admittedly not elegant, is Map[Integrate[#, {x, -∞, ∞}] &, f[x] + DiracDelta[x - y] g[x]] (* ConditionalExpression[g[y] + Integrate[f[x], {x, -∞, ∞}], Element[y, Reals]] *) Incidentally, the code in the Question can be rewritten as Integrate[#, {x, -∞, ∞}] & @ (f[x] + DiracDelta[x - y] g[x]) and the code at the beginning of this ...


2

Trace[Integrate[ Integrate[ Integrate[ E^(-v1 - v2 - v3) (v1 + v2 + v3)/3, {v3, 2 v1 - 5/2 v2, v2}], {v2, 4 v1/7, 5 v1/7}], {v1, 0, Infinity}]] Check one line before the result. I think this is what you want.


2

Well, n[1, 1, 1] (* {1.38996, 1.85383, 1.37325} *) n[5, 5, 5] (* {1.38996, 1.85383, 1.37325} *) When you first define your variables and then assign n[...] to this expression, the expression will evaluate and then be stored as that value. Regardless of what values you pass to n it will always return the same thing. You can read more about how Mathematica ...


2

$Version "10.0 for Mac OS X x86 (64-bit) (September 10, 2014)" As entered Mathematica returns the wrong result. Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, Pi}, Assumptions -> {x > 0, y > 0, x > y}] Pi*(1/(8*y) - (3*y)/(8*x^2)) However, a workaround is to convert the trig functions to exponentials ...


1

This appears to be a bug in V10.0.x which was fixed in V10.1.0. $Version "10.1.0 for Mac OS X x86 (64-bit) (March 24, 2015)" Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, Pi}, Assumptions -> {x > 0, y > 0, x > y}] -((π y)/(4 x^2))


1

Something like this?: rect4[f_, a_, b_, n_] := With[{ex = Integrate[f[y], {y, a, b}], r = Range[0, n]}, With[{h = (b - a)/2.^r}, {r, #, {"/"}~Join~Ratios@#}\[Transpose] &@ Abs[ex - (b - a) Mean /@ f /@ Range[a, b - h, h]]]] MatrixForm@rect4[#^2 &, -1, 1, 6]


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

fixed in 10.1 (windows). Now integral remains unevaluated. code: $Assumptions = t \[Element] Reals && t > 0 && t < 1 f[x_] = Abs[Re[Exp[I*x]/(1 - t*Exp[I*x])]] Integrate[f[x], {x, 0, 2 \[Pi]}]


1

bug fixed in 10.1 (windows) code ArcLength[Line[{{0, 0}, {1, 0}, {2, 0}}]] ArcLength[Line[{{0}, {1}, {2}}]] ArcLength[Line[{{0, 0}, {1, 0}, {2.0, 0}}]] ArcLength[Line[{{0}, {1}, {2.0}}]]



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