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5

You can gain insight by evaluating and then simplifying the indefinite integrals. Integrate[(1 + b x + c y)/(1 + e x + f y + I η), y, x, Assumptions -> {x ∈ Reals, y ∈ Reals, b ∈ Reals, c ∈ Reals, e ∈ Reals,f ∈ Reals, η ∈ Reals, η != 0}]; ans = Collect[%, {ArcTan[η/(1 + e x + f y)], Log[1 + e^2 x^2 + 2 f y + f^2 y^2 + 2 e (x + f x y) + η^2]}, ...


5

One way to approach this is to look for the source of the problem. In this case, the outer integral (on y) is irrelevant to the problem, because even the simpler integral int = Integrate[(1 + b x)/(1 + e x + I η), {x, -1/2, 1/2}, Assumptions -> {b ∈ Reals, c ∈ Reals, e ∈ Reals, η ∈ Reals, η != 0}] gives a conditional expression. But now the answer ...


4

The following approach transforms ans1 to ans2 and points the way to more general approaches. As noted in the Question, Simplify returns ans1 unchanged. Introducing a ComplexityFunction that defines fewer symbols as less complex helps some. cf[e_] := LeafCount[e] + 100 Count[e, _Symbol, Infinity] Simplify[ans1, ComplexityFunction -> cf] (* ...


4

Mathematica evidently won't simplify Integrate[f[t], {t, a, b}] + Integrate[f[t], {t, b, c}] to Integrate[f[t], {t, a, c}] on its own. However, you can easily write a function that does what you want, by using Mathematica's ability to have functions do pattern matching on their arguments. Here's a function that will do what we want: ...


3

Via TransformationFunctions: Clear[xf]; xf[e_] := e /. {Integrate[int_, {x_, a_, b_}] - Integrate[int_, {x_, a_, c_}] :> Integrate[int, {x, c, b}], coeff_ Integrate[int_, {x_, a_, b_}] :> Integrate[coeff int, {x, a, b}]}; sol = DSolve[{f'[t] + f[t]*g[t] == h[t], f[0] == f0}, {f[t]}, t]; Simplify[sol, ...


2

Maybe you can use the following two constructs to your advantage, which will keep the Conjugate, but evaluate and simplify the derivative inside. Using ReleaseHold, you can then evaluate even the Conjugate. Note that I left out the divisor in the Conjugate-case for clarity, but you can easily add that into the second function's definition. d[g_] := ...


2

You can split up the interval of integration: Integrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, 0, 1/2, 1}] (* 0 *) Still, it seems a bit inconvenient.


1

Updated answer On v10.0.2, this approach with Hold/ReleaseHold achieves the same result as v8.0.4, namely, it gives a limit of $\frac{2}{\pi}$. ReleaseHold@Limit[ Hold[ Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}] ], n -> Infinity] (* 2/Pi *) Although in a comment by Jinxed below, apparently this doesn't work in v10.1! Numerically, one can ...


1

Easiest and straightforward may be to define a unit conversion function like so: uc[expr_, targetunit___]:=UnitConvert[#,targetunit]&/@expr/.UnitConvert[a_,b___]->a Usage: uc[x Quantity[y, ("Milliwatts"*"Seconds"^2)/"Joules"]] (* x (Quantity[y/1000, "Seconds"]) *) or giving a desired target unit: uc[x Quantity[y, ...


1

Indeed, the integral should give zero even with finite bounds. This workaround seems to give the desired result: Limit[ Integrate[ DiracDelta[1 - x] DiracDelta[x] f[x], {x, ϵ, 1}], ϵ -> 0] (* ==> 0 *) But it works only because it effectively cuts off the lower bound and thus the lower delta function.


1

With v10.0.2 it appears to work only for infinite bounds or using NIntegrate $Version "10.0 for Mac OS X x86 (64-bit) (December 4, 2014)" Integrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, -Infinity, Infinity}] 0 Integrate[DiracDelta[x, 1 - x] f[x], {x, -Infinity, Infinity}] 0 NIntegrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, 0, ...



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