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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 ...


6

You have to tell Mathematica explicitly that n is supposed to be an integer: Assuming[n ∈ Integers, Limit[Sin[π Sqrt[4 n^2 + n]], n -> ∞]] (* ==> 1/Sqrt[2] *)


5

Here is a way to use Mathematica to find numerical limits for discrete variables: Needs["NumericalCalculus`"] u = IntegerPart; NLimit[Sin[Pi Sqrt[4 u[n]^2 + u@n]], n -> ∞] (* 0.707107 *)


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 ...


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

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

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

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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