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7

There is also a newer package, HolonomicFunctions, that has an implementation of Chyzak's generalization of Zeilberger's algorithm. To perform the desired task, use the following commands: smnd = Simplify[ G /. HoldPattern[HypergeometricPFQ[pl_List, ql_List, x_]] :> (Times @@ (Pochhammer[#, k] & /@ pl)) / (Times @@ (Pochhammer[#, k] & /@ ql)...


7

First of all you can simplify the integral by the substitution $t\mapsto t/\omega$, which just changes the value by a known factor. After this, the integrand is: (Cos[t]^2 Cos[α + t])/(-1 + A Cos[α + t])^2 Which is integrated to obtain int = Integrate[(Cos[t]^2 Cos[α + t])/(-1 + A Cos[α + t])^2, t]; Plotting the int for some values of A, omega, and ...


6

One problem is that Exp[y] evaluates to Power[E, y], so that the integral does not match the (held) pattern with Exp. Another is that other functions sometimes evaluate to other forms, such as Sin: Sin[Sin[x] + Cos[x] - x] (* -Sin[x - Cos[x] - Sin[x]] *) Here is a fix that works on the example. I added a constant factor c_ to take care of the -1 ...


6

Everything you want in the question can be done by defining the derivative of the Norm: Derivative[1][Norm][z_] := z/Norm[z] D[Norm[x - y], {x}] (* ==> (x - y)/Norm[x - y] *) Simplify[D[Norm[x - y], {y}]] (* ==> (-x + y)/Norm[x - y] *) Here, the syntax I used for the derivatives is such that it would remain valid if x or y were replaced by vectors ...


4

Rather than define a function, you can define a replacement rule using Hold and RuleDelayed rule = Hold[Integrate[ Exp[p_. Cos[x_] + q_. Sin[x_]]* Sin[a_. Cos[x_] + b_. Sin[x_] - m_. x_], {x_, 0, 2 Pi}]] :> I*Pi*((b - p)^2 + (a + q)^2)^(-m/ 2)*(((p^2 - q^2 + a^2 - b^2) + I*(2*(p*q + a*b)))^(m/2)* BesselI[m, Sqrt[(p^2 +...


3

Michael E2 and Bob have solved the integral. Here is an alternative method which might be of interest as well. I have used the similar method already in How to solve this integration?. We solve the integral transforming it into a complex contour integral which, after a simple binomial expansion, can easily be soved by the Cauchy theorem. The remaining ...


3

How about adding some assumptions (I think the following is reasonable): res = Integrate[(x1 + x2 - 1)*(Boole[ x1 + x2 >= s && x1 >= t1 && x2 >= t2]), {x1, 0, 1}, {x2, 0, 1}]; FullSimplify[res, 0 < s < 1 && 0 < t1 <= t2 < 1] $$\begin{cases} \frac{1}{2} (\text{t1}-1) (\text{t2}-1) (\text{t1}+\text{...


3

We can simplify the derivation of coolwater appreciably by getting rid of the mixture of t and alpha. It turns out that in the present aproach the only jumps of the antiderivative occur at t = Pi which can be easily taken into account. Here we go: $Version (* Out[176]= "10.1.0 for Microsoft Windows (64-bit) (March 24, 2015)" *) In the first step we "...


2

To some extent (and with some care) this can be done with FeynCalc. At least I used it several times when I needed to compute gradients and divergences of Cartesian vectors. The trick is to work with D-dimensional 4-vectors and take the limit $D \to 3$ at the end. Since FeynCalc doesn't distinguish between upper and lower indices, the results are the same as ...


1

The function cannot be integrated with unknown n, because its form changes with different values for n: Table[Integrate[Sqrt[2 - Cos[x]] Exp[I n x], x], {n, 0, 3}] 1: 2 EllipticE[x/2, -2], 1/3 (-4 EllipticE[x/2, -2] + 6 EllipticF[x/2, -2] - I Sqrt[2 - Cos[x]] (-5 + 2 Cos[x/2]^2 + Cos[x] + 2 I Sin[x])) 2: 2/15 (I + Cot[x]) (13 ...


1

Here is how to do the contour integral. Shown for some specific parameters. A = 1; B = -1/2; a = 1; f[z_] := (E^(-a z))/(A + B (Cosh[z])) Integrate[f[rz - I Pi], {rz, Infinity, 0}] + Integrate[f[z], {z, -I Pi, I Pi}] + Integrate[f[rz + I Pi], {rz, 0, Infinity}] // Simplify 4/3 I (-3 + 2 Sqrt[3]) Pi You will need to add appropriate assumptions to ...



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