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Another method is to use PowerExpand: PowerExpand[Log[E^a], Assumptions -> True] a + 2 I π Floor[1/2 - Im[a]/(2 π)] Addendum As an aside, Log behave exactly like ArcSin here: ArcSin[Sin[x]] … Pi/2 < x < Pi/2] x For larger domains, Simplify doesn't work: Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi] ArcSin[Sin[x]] Again, using PowerExpand is useful: p = PowerExpand[ArcSin[Sin[x]], Assumptions …

Per the documentation you can use Esc ints Esc to enter the StandardForm symbol for Integers

New in Mathematica 12 is PositiveReals (and others like NonNegativeIntegers, etc): Solve[x^2 == 1, x, PositiveReals] {{x -> 1}}

One possibility is to just replace your variable with one whose powers are all explicit integers, and then use Series: e = PowerExpand[ a^(-c2)/((a*c1)^(c2) + c3) /. a -> z^(1/c2), Assumptions …

Note what happens when you add them: $Assumptions = Element[d, Matrices[{3, 3}, Reals, Symmetric[{1,2}]]]; d + IdentityMatrix[3] {{1 + d, d, d}, {d, 1 + d, d}, {d, d, 1 + d}} The Listable attribute …

One possibility is to mimic the behavior of symbolic constants like Pi, E, etc: N[x, _] ^= 4; NumericQ[x] ^= True; Then: Sqrt[x^2] x without even using Simplify.

TrueQ@$IntFlag := Block[{$Foo=True}, MakeBoxes[Integrate[a],TraditionalForm] ] Protect[Integrate]; Then: Integrate[f[x]Sin[x],{x,0,L0},Assumptions->True] //TeXForm $\int_0^{\operatorname{L0 …

ToRadicals supports the undocumented option Assumptions. … Making use of this option gives: sol = Solve[x^3==a^2,x,Reals] ToRadicals[sol, Assumptions -> a>0] {{x -> Root[-a^2 + #1^3 &, 1]}} {{x -> a^(2/3)}} …

You could use SeriesCoefficient instead: Assuming[n ∈ Integers && n>1, Simplify @ SeriesCoefficient[ Exp[x] x^(-1-n), {x, 0, -1} ] ] 1/n!

For example: TensorReduce[ w + TensorTranspose[w, {1, 3, 2}], Assumptions -> w ∈ Arrays[{n, n, n}, Complexes, Antisymmetric[{1,2,3}]] ] 0 …

I think using Reduce would be a better approach, although it's not completely straightforward: Reduce[x^a < 1 && a > 0 && x > 1, Reals] False (updated with another approach using Resolve) Ano …

You could try using DSolveValue. First, define your sum using inactive integrals: h[α_] := Inactive[Integrate][x g'[x],{x,0,α}] + Inactive[Integrate][α g'[x],{x,α,∞}] Even though the integrals are …

The region of integration is: reg = ImplicitRegion[0 < y < a && 0 < x < z && 0 < z < y, {x, y, z}]; Using this region in Integrate: sol = Integrate[Exp[(a-x)^3], {x, y, z} ∈ reg, Assumptions -> a > 0] …

Another possibility is to use the new in M12.1 function Asymptotic: asym = Assuming[ r0 > 3 M > 0 && θ > Pi/2, Simplify @ Asymptotic[expr, θ -> Pi/2] ] Coefficient[asym, Log[θ - Pi/2]] //Simpl …

For this example, you could use FourierCoefficient instead of Integrate: FourierCoefficient[1, t, k, FourierParameters->{-1,1}] 2 π DiscreteDelta[k]