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7

Chances are it will behave similarly to Integrate[Exp[-x t^3 ], {t, 0, Pi/2}]. I show a numerical verification below. We'll use the dominant series term, which we can compute, for the above integral, and compare to numerical integrations of the one we cannot compute analytically. ee1[x_] := Exp[-x*t^3] ee2[x_] := Exp[-x*t^3*Cos[t]]; Let's get the ...


1

There is nothing wrong with your mathematica code. The problem is with the volume element. The volume element in spherical coordinates is $$r^2\sin(\theta ) \;dr\,d\theta \,d\varphi $$ not $$r^2\sin(\varphi ) \;dr\,d\theta \,d\varphi $$ Simply replace $$Sin[phi]$$ with $$Sin[theta]$$ in your code, and you get the answer. Integrate[Integrate[Integrate[r ...


1

It is not a good idea to feed approximate numbers like 0.1 to symbolic methods. Have a look at the indefinite integral Integrate[g[Cos[\[Theta]], Sin[\[Theta]], x2, y2], \[Theta]] to see what's going on. Mathematica has to go very far into complex analysis to solve this integral symbolically. I suspect the several terms with branch points at \[Theta] == ...


1

Having a closed form for integral of an incomplete gamma function should be a big deal(!), but having a numerical approximation is simple. Define: f[k_,e_,limit_]:=NIntegrate[Gamma[k, 0, Exp[-2 e x^2]], {x, -limit, limit}]; For example: f[2,0.2,3] (* 0.591563 *) You can also see how f changes with k and e: Plot[{Legended[f[k, 0.2, 3.], "e=0.2"], ...


3

We can use any one of the line integrals that by Green's Theorem yield the area: dA = RandomChoice[{x Dt[y], -y Dt[x], (x Dt[y] - y Dt[x])/2}] param = {x -> r Cos[θ], y -> r Sin[θ]}; boundary1 = {r -> 6}; boundary2 = {r -> 4 + 4 Cos[θ]}; θ0 = θ /. Solve[{Equal @@ (r /. {boundary1, boundary2}), -Pi < θ < Pi}, {θ}]; Integrate[dA /. param /. ...


6

Integrate[Max[0, (4 + 4 Cos[t])^2/2 - 6^2/2 ], {t, -Pi, Pi}] or Integrate[(4 + 4 Cos[t])^2/2 - 6^2/2 , {t, -Pi/3, Pi/3}] 18 Sqrt[3] - 4 Pi Edit: another approach: Area[{r Sin[t], r Cos[t]}, {t, -Pi, Pi}, {r, 6, Max[6, 4 + 4 Cos[t]]}] or Area[CoordinateTransform[ "Polar" -> "Cartesian", {r, t}], {t, -Pi, Pi}, ...


3

As HyperGroups suggests, we can take advantage of the Area function new in v10. First, we'll represent the region implicitly. We want: $$ 6 < r < 4 + 4\cos\theta \\ 6 < \sqrt{x^2+y^2} < 4 + 4\cos\arctan\frac y x $$ First we'll plot this region to make sure we're correct: Show[ RegionPlot[6 < Sqrt[x^2 + y^2] < 4 + 4 Cos[ArcTan[x, y]], ...


4

Here's another approach just to illustrate the difficulty with multiple integrals that are not absolutely (i.e. $L^1$) convergent. The integral on any finite disk centered at the origin is zero. So it would be natural to conclude that the integral over the whole {r,z} plane is zero. In terms of polar coordinates: Integrate[ ((r BesselJ[0, r])/(2 (r^2 + ...


6

The inverse square root term is responsible for the errors as it diverges logarithmically at infinity. You could do one step at a time and see if you agree with the procedure of introducing a regularizing term, doing the integral and then removing it. int1 = Integrate[(r BesselJ[0, r])/(2 (r^2 + z^2)^(1/2)) Exp[-α z^2], {z, -Infinity, ...


2

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))


4

$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 ...


4

With N replaced by n and exact numbers, the function in the Question can be written as f[n_] = Sum[Binomial[n/2 - 1, a]*Binomial[n/2 - 1, a - 1]*(7/20)*(3/10)^(n - a - 1), {a, 2, n, 2}] Although Mathematica can perform the Sum, the result in terms of HypergeometricPFQ is not particularly enlightening. Instead, plot f[n]. ListLogPlot[Table[f[n], {n, ...


4

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 ...


0

You are using insufficient precision: theta2'[.8] (* -0.794774 + 0.280078 I *) Precision[.8] (* MachinePrecision *) Use exact arguments: theta2'[8/10] and get the exact result. Use N et. al. with desired precision to retrieve numeric values. N[theta2'[8/10], 10] (* -0.8874928427 + 4.596*10^-7 I *)


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 ...


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: ...


0

Version 10.1 gives integral result Integrate[(a + b*Cos[t1 - t2])/(c + d*Cos[t1 - t2]), {t1, 0, 2 \[Pi]},Assumptions -> {a \[Element] Reals, b \[Element] Reals, c >= 0, d <= 0, c + d >= 0, 0 <= t2 <= 2 \[Pi]}]


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 ...


0

I get the answer 0 using Integrate[x^(a - 1)/(1 - x) - c x^(b - 1)/(1 - x^c), {x, 0, 1}, Assumptions -> {a > 0, b > 0, c > 0}] v10.1


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


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 ...


0

There is another possibility, and that is to differentiate an InterpolatingFunction. Here is an example of a rather poorly behaving function whose derivative we can recover by this technique. The transfer function of a Butterworth filter looks like h[s_] := 1 / (1 + 2s + 2s^2 + s^3) for s complex. The Arg of this function, which, for purely imaginary ...


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 ...


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]



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