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8

This will work with any number of independent integrants. Define: repl[l_] := # /. Thread[#[[All, 1]] -> Table[x, {Length[#]}]] &[l] inTfaC[int_] := Times @@ MapThread[Integrate[#1, #2] &, repl /@ {First[#], Rest[#]} &[ int /. {Integrate -> List, Times -> List}]] Now verify: test = Integrate[p[x] p[y] q[z] r[s] r[u], {x, ...

7

The "trick" which works frequently when Mathematica refuses to calculate a definite integral is to calculate first the indefinite integral, then take the limits at the ends of the integration interval and subtract the results. Here we go. Let the integrand be f = -((cA Log[1 + mgl^2/sa1])/(-1 + x)) + (cA Log[-(sa1/(-mgl^2 - sa1))])/(1 - x) + (cA cF Log[-1 ...

5

Given expression = (20*t^2*Erf[b + t])/(E^t^2*Sqrt[a + t^2]) can be achieved via replacement of the Head List @@ expression To find constant factors: GatherBy[List @@ expression, NumberQ] It is so because Times here is single Head for all first-level factors: TreeForm[expression] This also can prove to be useful: FactorList[expression] ...

4

OK, with your new formula I'm able to give an incomplete answer now. The difficulty in implementing the forumla $$-\sum _{n=1}^{\infty } \frac{B_n(1) f^{(n-1)}(0)}{n!}$$ is how to symbolically compute the n-th derivative, which is discussed here. Use the solution in that post, we can easily obtain this: ramanujanSum[f_] := Block[{x, n}, FullSimplify[ ...

3

As stated in the OP Mathematica finds this value of the integral within seconds f0 = Integrate[x^2 Log[1 - E^-x], {x, 0, \[Infinity]}] (* Out[21]= -(\[Pi]^4/45) *) Most probably Mathematica employs this standard procedure: 1) solve the indefinte integral, i.e. find an antiderivative 2) check continuity of the antiderivative 3) use the fundamental ...

3

You could try something like the following: Table[dS[i] = -β[i]*S[i], {i, 4}]; To see the definitions for dS use the function Definition. Definition[dS] (* dS[1]=-S[1] β[1] dS[2]=-S[2] β[2] dS[3]=-S[3] β[3] dS[4]=-S[4] β[4] *) See that you can perform operations on dS. dS[1] + dS[2] (* -S[1] β[1] - S[2] β[2] *)

3

seems to work: expr /. Times[xx_?(! FreeQ[#, x] &) , p_plusd, rest___] :> Times[xx /. x -> 1, p, rest] and terms like: 2 Log[(sa1 (-2 + x))/(-mgl^2 + sa1 (-2 + x))] plusd[1/(1 - x)] and Log[2 + mgl^2/sa1 - x] plusd[-(2/(-1 + x))] were reduced to 2 Log[-(sa1/(-mgl^2 - sa1))] plusd[1/(1 - x)] and Log[1 + mgl^2/sa1] plusd[-(2/(-1 + x))] ...

2

Adapting linearExpand from my answer to How to do algebra on unsolved integrals?, we can come up with some transformations to factor separable multiple integrals. The function someFunction internally deals with and returns Inactive integrals, which can be evaluated with Activate, if appropriate or desired. Examples someFunction[Integrate[p[x] p[y], {x, ...

1

booleanTable := Module[ {var, iter}, f = Input["Enter Boolean function"]; var = BooleanVariables@f; iter = Sequence @@ ({#, {True, False}} & /@ var); Column[{Table[ Evaluate[{var, f} // Flatten], Evaluate[iter]] // Flatten[#, Length[var] - 1] & // Prepend[#, {var, f} // Flatten] & // ...

1

Not an answer but too long for comment: If I modify slightly your input and choose 'n=2` n = 2; Solve[Table[{Subscript[β, 1] (1 - Subscript[a, i] - Subscript[b, i] - Subscript[c, i]) Sum[ Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 1, n}] - Subscript[δ, 1] Subscript[a, i] + Subscript[δ, 2] ...

1

Here is a simple solution. Think of everything as functions. Then define: A = Function[T , Function[{i,j,k}, u[i+1,j,k]*(T[i+1,j,k]+T[i,j,k]) - u[i-1,j,k]*(T[i-1,j,k]+T[i,j,k]) + v[i,j+1,k]*(T[i,j+1,k]+T[i,j,k]) - v[i,j-1,k]*(T[i,j-1,k]+T[i,j,k]) + ...

1

Mathematica has logical symbols such as $\exists$, $\forall$, $\in$, $\wedge$, and so forth and can be used in statements such as ForAll[{a, b}, a > 0 && b > 0, (a + b)/2 >= Sqrt[a b]]. One performs simple logical resolution by Resolve[] applied to such an expression. I suspect Mathematica can simplify or resolve such logical and ...

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