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5

Confirmed for 10.4.1 as Alexei. However, WolframAlpha still gives the c/s result. See result from Wolfram|Alpha And using the properties of Dirac delta function, I would say that c/s is the correct one. May be a bug of the new 10.4.1? You can test with numeric constants, and Mathematica gives the correct answer. But symbolic result is wrong, ...


4

MapAt[D[#, x] &, Piecewise[{{x^2, 0 <= x <= 1}, {x, x <= 4}}], {{1, ;; , 1}}] Note: this function works for expressions with Head Piecewise. For general expressions that contain Piecewise subexpressions, it can be used with ReplaceAll as follows: Y[x_] = Piecewise[{{x^2, 0 <= x <= 1}, {x, x <= 4}}]*F + Piecewise[{{x^4, 0 <= ...


3

The problem is that you are not using large enough precision goal for an oscillatory integrand with very small absolute values over the integration range. Look at the log plot -- there are more oscillations than the ones visible with Plot: The remedy is to use higher precision goal. Try this: NIntegrate[ Re[(x^4 Sin[x b])/(9 + 4 d^2 x^2)^6 1/ 2 ...


3

Analysis of the error (bug?) We can see from the trace below that the second limit, which carries out a ratio test for the product, mistakenly yields -17 (which would indicate divergence, if correct). Trace[ NProduct[(n^2)!/stirling[n^2], {n, 1, Infinity}], _Limit, TraceInternal -> True, TraceForward -> True] There might have been some ...


2

If you just want to make a plot of the region in question, this is good RevolutionPlot3D[{{x, (x + 8)^4}, {x, 8 x + 64}}, {x, -8, -6}, {th, 0, 2 π}, Mesh -> None, Axes -> False, Boxed -> False, BoxRatios -> {1, 1, .3}] But try as I might, I can't seem to find a way to get the volume of the region using Volume - this would involve ...


2

Here's a symbolic-algebraic way, based on the cylindrical algebraic decomposition (CAD) that Reduce computes. (* Disk/Washer method *) Clear[x, y, z]; eqns = {y == (x - 2)^4, 8 x - y == 16}; axis = x == 10; {depV} = Variables[Subtract @@ axis]; {indepV} = Complement[{x, y}, {depV}]; vars = {indepV, depV}; components = Map[ Reduce[#, vars] &, (* ...


2

Lets start with a simpler case Integrate[q0/(-1 + a q^2), q] $\frac{\log \left(1-a \text{q}^2\right)}{2 a}$ When you put limit [0,A], it has no problem with q=0. But it is not defined when $aA^2>1$. So you always have to obey that condition. You can check that by Integrate[q0/(-1 + a q0^2), {q0, 0, A}] In your second case Integrate[q0/(-1 + 12. ...


2

Is this what you mean? Solve[Integrate[1/x, {x, v0, vt}, Assumptions -> 0 < v0 <= vt] == g/v0, v0] (* Out: {{v0 -> -(g/ProductLog[-(g/vt)])}} *) Or perhaps with a few more assumptions, and heeding Solve's suggestion to use Reduce, which would get us a more complete complete answer including "special cases" (i.e. solutions that are valid only ...


2

happy fish ,he said "nth derivative is not natively supported" Yes it's true,but from here. Method1: For simple functions you can use InverseFourierTransform. f[s_] := Sin[s]; nthDeriv1[f_, s_, n_] := FullSimplify[InverseFourierTransform[(-I k)^n FourierTransform[f, s, k], k, s], {n \[Element] Integers, n > 0}] nthDeriv1[f[s], s, n] $$\sin ...


2

I'd suggest a simpler approach that seems to work just fine: Use arbitrary precision input if you can. In your case, instead of d -> 0.001/5, use d -> 5/1000 integrand = Re[(x^4 Sin[x b])/(9 + 4 d^2 x^2)^6 1/2 E^(-(1/2) x^2 - I x c - 1/2 a^2) (Erfc[(I x - c)/Sqrt[2]] + E^(2 I x c) Erfc[(I x + c)/Sqrt[2]]) /. {b -> 5, ...


1

Let, $$ f=\frac{z-e^{i \pi/2 n}}{z^{2 n}+1} $$ Then, in Mathematica code, the following seems to work, Limit[f, z -> Exp[(I Pi)/(2 n)], Assumptions -> n >= 1/2] which gives the result, $$ -\frac{e^{i \pi/2 n}}{2n} $$ The problem isn't that Mathematica is handling the $0/0$ limit incorrectly. It's that Mathematica, without additional information, ...


1

Try Integrate[ Log[Abs[2 (0.27059805007309845` - 0.7071067811865475` s)]], {s, .3826, .3829}] (*-0.00280689*) returns no errors, since it is the same with your integral.


1

As suggested by the Documentation Center, you can remove excess (last 3) conditions. s = NDSolve[{D[u[x, t], t] == -D[u[x, t], x] - u[x, t], D[v[x, t], t] == -D[v[x, t], x] - v[x, t], D[w[x, t], t] == -D[w[x, t], x] - w[x, t], D[z[x, t], t] == -D[z[x, t], x] - z[x, t], u[x, 0] == 1, v[x, 0] == 1, w[x, 0] == 1, z[x, 0] == 1, u[0, t] ...


1

If you need the output to be a function of v0 and vt, you do not need necessarily to do explicit symbolic integration. I hope this might help: Integration[v0_?NumberQ, vt_?NumberQ] := NIntegrate[1/x, {x, v0, vt}] Integration[5, 3] My output is -0.510826



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