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11

region = ImplicitRegion[ 0 < z < 4 - x*y && 0 <= x <= 2 && 0 <= y <= 1, {x, y, z}]; RegionPlot3D[region, BoxRatios -> {1, 1, 1}, Axes -> True] Volume[region] (* 7 *) RegionMeasure[region] (* 7 *) Integrate[1, {x, 0, 2}, {y, 0, 1}, {z, 0, 4 - x*y}] (* 7 *) Integrate[1, Element[{x, y, z}, ...


9

I believe that s = ArcTan[Sqrt[-4 E^(I a)]] N[Limit[s, a -> 0]] (* 4.71239 + 0.549306 I *) is a bug in Limit. Plotting the function s Plot[Evaluate[ReIm[s]], {a, -1, 1}] indicates that s assumes the value above nowhere in the vicinity of a == 0. (The same is true in the complex plane.) Furthermore, Limit[s, a -> 0] (* π + I ArcTanh[2] *) ...


6

Some insight can be gained by plotting Sqrt[Exp[I*t]^2 - 1] in the complex plane. Plot3D[Evaluate[ReIm[Sqrt[Exp[I*t]^2 - 1] /. t -> tr + I ti]], {tr, 0, Pi}, {ti, -1, 1}, AxesLabel -> {tr, ti, f}] Branch points occur at t == n Pi, n an integer, with branch cuts extending from the branch points to t == n Pi + I ∞. Visibly, there also are ...


4

Since Mathematica 10 there's been ImplicitRegion. We define that region this way: IR = ImplicitRegion[-9 <= 6 x + 5 y <= 9 && -7 <= 3 x + 6 y <= 7, {x, y}]; then we can calculate this integral directly: Integrate[(18 x^2 + 51 x*y + 30 y^2)^2, {x, y} ∈ IR] 5292 This integral can be calculated without the newest version of the ...


3

Here is a simple answer Define k[n] as a listable function : SetAttributes[k, Listable] k[n_] := \!\( \*SubsuperscriptBox[\(\[Integral]\), \(3\), \(10\)]\( \*FractionBox[\(1 + Cos[f\ \((1 + 2\ n\ )\)\ \[Pi]]\), \(1 + Cos[f\ \[Pi]]\)] \[DifferentialD]f\)\) then define a list compose of integer of the desired Length --- say 10 nn = Range[10] ask ...


3

It is definitely a bug. And it can be formuated even more sharply. $Version (* Out[156]= "10.1.0 for Microsoft Windows (64-bit) (March 24, 2015)" *) Define f[n_, a_] := (1 + (-1)^n/n^a)^n If a>0 the limit exists and is equal to unity. If a == 0 we have the problem of the OP, where one might say that "two alternative limits exist". This is ...


1

This post is cosmetic. It adds nothing to Artes excellent and instructive answer (which I have voted for). I post it to illustrate visualization of the region (a parallelogram in this instance) and use same change of variables as Artes. I think SliceContourPlot3D is also a a nice way to visualize. m = {{6, 5}, {3, 6}}; mi = Inverse[m]; mi3 = ...


1

By using SetDelayed (short form :=) to define the function g1 or g2 you prevent the evaluation of b1 or b2 respectively into the assigned expression, so e.g. g1 has for its complete definition: ?? g1 Global`g1 g1[s_]:=Piecewise[b1,0] Parameter substitutions into the right-hand-side are made before further evaluation, and since there is no literal s ...


1

Simplify[(1/T)* Integrate[2*t/T*Exp[-I*2*Pi/T*k*t], {t, 0, T/2}, Assumptions -> T > 0] + (1/T)* Integrate[2*(T - t)/T*Exp[-I*2*Pi/T*k*t], {t, T/2, T}, Assumptions -> 0 < T/2 < T], k \[Element] Integers] (* (-1 + (-1)^k)/(k^2 Pi^2) *) check your solution Simplify[(Exp[Pi (-I) k] - 1)/(Pi^2 k^2), k \[Element] Integers] (* ...


1

Use higher precision f[x_] = -(10^-20 x)/(0.99005 - E^(10^-12 x)) // Rationalize // Simplify; int = Integrate[f[x], {x, 0, 10^9}] // FullSimplify; int // N[#, 20] & // Chop[#, 10^-20] & (* 0.47118211649097404645 *)


1

First, if you use Integrate, you should define your function exactly. Don't use approximate numbers like 0.99005. f[x_] = -(x/(10^20*(99005/100000 - Exp[x/10^12]))) Integrate[f[x], {x, 0, 10^9}] (* Complicated exact expression involving Log and PolyLog *) Evaluating this approximately indeed yields a small imaginary part. %//N (* 0.471182 - ...



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