# Tag Info

9

I have no idea why Mathematica complains about the convergence. When you expand the integrand, then integrate each of the terms and add, you will find the exact result: E^(-((201 x1^2)/101)) x1^2 (15 - 20 x1^2 + 4 x1^4) (5913508078417951503 + 1124782662003060300000 x1^2 - 2983951574394000000000 x1^4 + 2591462040000000000000 x1^6 - 797900000000000000000 ...

7

One way to achieve this is: expr = {-5 E^(-2 t) Cos[3 t] + 12 E^(-2 t) Sin[3 t], -12 E^(-2 t) Cos[3 t] - 5 E^(-2 t) Sin[3 t], 4 E^(-2 t)} Then: Collect[expr, E^(-2 t)] /. {x_ u_, x_ v_, x_ w_} :> x[u, v, w]

6

int = Integrate[1 + Log[Cos[x]^4], {x, -2/3, 2/3}] // Simplify 1/9 I (3 [Pi]^2 + 4 ((-4 - 3 I) + 12 I Log[2] + 9 PolyLog[2, -E^(((4 I)/3))])) int // N 0.91905 - 3.94746*10^-16 I The imaginary part results from numerical noise. You can remove it with Chop int // N // Chop 0.91905 To see that it is just noise increase the precision ...

5

A good comprehensive answer should explain why InverseFunction "didn't work", however there's been no explanation so far. A unique inverse function can be found in a region if there its jacobian is nondegenerate, i.e. its determinant doesn't vanish (Inverse function theorem) . For one - variable function it means that the derivative doesn't vanish. ...

4

In order to get the expression yo simplify to the fullest extent, one could make the additional assumption that the label $\alpha$ always falls within the range of the summation whose index is $\beta$. To do this, we have to add some rules which are easier to write if the starting expression is brought into a slightly different form: \frac{\partial ...

4

To begin with, in the definition of fx include the parameter "a" as an argument and write x_ instead of x : fx[x_, a_] := Piecewise[{{(1/2) a*E^(-a*x), x >= 0}, {(1/2) a*E^(a*x), x < 0}}] If you now you tell Mathematica that "t" is real and that a>0 there is no problem integrating fx: Integrate[fx[x, a], {x, -Infinity, t}, Assumptions -> {a ...

3

Starting from Harry's approach v = {-5 E^(-2 t) Cos[3 t] + 12 E^(-2 t) Sin[3 t], -12 E^(-2 t) Cos[3 t] - 5 E^(-2 t) Sin[3 t], 4 E^(-2 t)}; factor = Intersection @@ (Factor /@ v); expr = Inactive[Times][factor, Simplify[v/factor]] v == (expr // Activate) // Simplify True

3

maybe we can use Intersection to find the factor: v = {-5 E^(-2 t) Cos[3 t] + 12 E^(-2 t) Sin[3 t], -12 E^(-2 t) Cos[3 t] - 5 E^(-2 t) Sin[3 t], 4 E^(-2 t)}; factor = Intersection @@ (Factor /@ v); factor[Simplify[v/factor]] (*(E^(-2 t))[{-5 Cos[3 t] + 12 Sin[3 t], -12 Cos[3 t] - 5 Sin[3 t], 4}]*) I have to say the output is a little bit ugly. Maybe ...

3

Integrate[E^(-((201 x1^2)/101)) x1^2 (15 - 20 x1^2 + 4 x1^4) (5913508078417951503 + 1124782662003060300000 x1^2 - 2983951574394000000000 x1^4 + 2591462040000000000000 x1^6 - 797900000000000000000 x1^8 + 80000000000000000000 x1^10), {x1, -Infinity, Infinity}, PrincipalValue -> True] (*(368559503694222488015947879582225000 Sqrt[( 101 ...

3

To add to Bob Hanlon's answer: int = Integrate[1 + Log[Cos[x]^4], {x, -2/3, 2/3}] // Simplify; int // Im // PossibleZeroQ PossibleZeroQ::ztest: Unable to decide whether numeric quantities {-(16/9)+[Pi]^2/3+4/9 Re[9 PolyLog[2,-Power[<<2>>]]],-16+3 [Pi]^2+4 Re[9 PolyLog[2,-Power[<<2>>]]]} are equal to zero. Assuming they are. >> (* True *) ...

2

Expectation, interestingly, does not have the same problem as Integrate, and it is much faster. If you "have to do this with hundreds of similar polynomials," then I recommend using Expectation. First the exponential factor of the integrand corresponds to the pdf of a normal distribution, scaled by a constant: 1 / Sqrt[(201/(101 π))] * ...

2

Here's a slight rewording of Nasser's answer in the proposed duplicate. Of course, you've got to specify the contour along which to integrate somehow; the standard way to do this is in terms of a parametrization of the contour. Note that this contour gamma[t] is defined in terms of a parameter r, which is not present in the result. f[z_] = 1/z; gamma[t_] ...

1

There are a number of errors in the code, e.g. the unit tangent vector should be r'[t]/Norm[r'[t]]. There are sometimes difficulties with Norm and simplification. To illustrate some points I will consider curvature of plane curve and then 3 dimensional curve. Plane curvature This uses $\kappa =\frac{x'y''-x''y'}{(x'^2+y'^2)^{3/2}}$ pcv[exp_, t_] := ...

1

Try this and see if it works. int[x_] = Integrate[Log[a Cos[x]^2 + b Sin[x]^2], x]; int[2 Pi] - int[0] I got the result very fast.

1

Your syntax is wrong for Assuming Clear[fx]; fx[x_] := Piecewise[{ {(1/2) a*E^(-a*x), x >= 0}, {(1/2) a*E^(a*x), x < 0}}] Assuming[ {x \[Element] Reals && a \[Element] Reals && a > 0 && t \[Element] Reals}, Integrate[fx[x], {x, -Infinity, t}]] Piecewise[{{E^(a*t)/2, t <= 0}}, ((1/2)*(-1 + ...

1

D[f[g[t]*h[t]], t] /. t -> 0 Derivative[1][f][ g[0] h[0]] (h[0] Derivative[1][g][0] + g[0] Derivative[1][h][0]) EDIT: I misread your post. Same approach D[f[g[t], h[t]], t] /. t -> 0 Derivative[1][h][0]* Derivative[0, 1][f][g[0], h[0]] + Derivative[1][g][0]* Derivative[1, 0][f][g[0], h[0]]

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