# Tag Info

2

Here is my explanation. The results of the codes ∞ ∈ Reals False and ∞ ∈ Complexes False prove that ∞ is not a real/complex number, so its substitution in any function makes no sense. The result of Limit[HurwitzLerchPhi[-1, 1, x], x -> ∞] 0 is sometimes written as HurwitzLerchPhi[-1, 1, ∞]==0. It should be noticed that such notation may confuse in ...

0

Is this what you want: ClearAll[r, zR]; r: = Sqrt[x^2 + y^2]; zR: = z - f; u0[x_, y_] := PP[x, y] Exp[(-I k)/(2 f) (x^2 + y^2)] u1[x_, y_,z_] := ((-I)* Integrate[E^(((I/2)*k*((x - x0)^2 + (y - y0)^2))/z)*u0[x0, y0], y0, {x0, -Infinity, Infinity}])/(z*\[Lambda]) You should run: u1[r, zR, z0] /. z -> z0

2

I think that the functions you're trying to fit are inappropriate for the data (or at best no theoretical justification for the models is presented) and you only have 5 data points. (There's an old saying about restaurant review: The food was bad and the portions too small.) But you are right in that you need better starting values to get the "best&...

3

Consider your data: data = Sort@{{0, 54.61}, {100, 57.26243979492134}, {80, 53.839874154239816}, {50, 54.09456572258326}, {24, 56.15393883162748}}; ListLinePlot[data] Now consider your function: f = (x*(57.26 + 273.15) + (100 - x)*(54.61 + 273.15))/(x + (100 - x)) + q*(x*(100 - x)) // Simplify This is a quadratic function, but your ...

0

Updated code: (made mistake with formula) But it still oscillates wildly. f1[x_, y_, s_] := Sin[(Pi Gamma[x + I y])/(2 (x + I y))]^2/(x + I y)^s; f2[x_, y_, s_] := Sin[(Pi Gamma[x - I y])/(2 (x - I y))]^2/(x - I y)^s; Manipulate[ Plot[Im@(1/(Exp[2 Pi y] - 1) (f1[x, y, s] - f2[x, y, s])) /. {x -> myX, s -> 2}, {y, 0, 10}, PlotRange -> All], {{...

2

Are you sure 0 is not the right answer? The problem is not with the limit. Look at your function without limits. I changed your 0.5 to 1/2. ComplexExpand[Re[1/(a + b*I + (1/(a + b*I + (1/2/(a + b*I)))))]] // Simplify (*(a (4 a^4 + 8 a^2 (b^2 + 1) + 4 b^4 + 3))/((a^2 + b^2) (4 a^4 + 4 a^2 (2 b^2 + 3) + (3 - 2 b^2)^2))*) with a as a factor in the numerator ...

5

$Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) int[c_] = Integrate[Exp[-c*x + x^2], {x, 0, c}] (* E^(-(c^2/4)) Sqrt[π] Erfi[c/2] *) Erfi is real for real c FunctionDomain[int[c], c] (* True *) Plot[int[c], {c, -15, 15}] For an alternate representation using DawsonF (EDIT: Eliminated unnecessary assumption) int[c] // ... 3 modified answer (11.04.2021) (Thanks @DanielLichtblau for his helpful comment) Here my ideas to make it work: Mathematica evaluates the first integral to Integrate[Exp[a u^2 + b v^2 +c u v], {v, -∞,∞}, {u, -∞, ∞},Assumptions -> {Element[{a, b, c}, Reals] ] (*ConditionalExpression[(2 \[Pi])/(Sqrt[-b]Sqrt[-4 a + c^2/b]), 4 a b^2 < b c^2] *) result is ... 2 The function (Re[z] - I*Im[z])/Abs[z]^2 is $$f(z) = {\bar z \over |z|^2} = {\bar z \over z\, \bar z}={1\over z}$$ which is meromorphic (holomorphic over$\mathbb C\backslash \{0\}$). Actually (Re[z] - I*Im[z])/Abs[z]^2 // FullSimplify returns 1/z and FunctionMeromorphic[1/z, z] returns True. As a side note one would be tempted to apply FullSimplify to ... 2 Clear["Global*"] What you wrote is (E^-2 x + 3 x)/x // Simplify (* 3 + 1/E^2 *) Presumably you meant to write expr = (E^(-2 x) + 3 x)/x; First, check that l'Hopital's rule is applicable Limit[#, x -> Infinity] & /@ {Numerator[expr], Denominator[expr]} (* {∞, ∞} *) The derivatives are D[{Numerator[expr], Denominator[expr]}, x] (* {3 - ... 7 The accepted answer shows you what is happening, but it doesn't really illustrate why it is happening. If you look at the documentation for Apply[f,expr], you will note that what it does is it replaces the head of expr with f. But for the given expression, the head of expr is not the full expression itself, but Plus. You can see this by typing Head[f[y] + ... 8 Use Trace Apply[D[#, y] &, f[y] + g[y]] // Trace // Column What you want is Map Map[D[#, y] &, f[y] + g[y]] Derivative[1][f][y] + Derivative[1][g][y] 2 e /. f -> Sin Cos[x] + Sin[x] 2 In V12.2, we get the expected answer: Integrate[ Power[u, sigma - 1]*Exp[-c*Power[u, sigma]], {u, 0, Infinity}, Assumptions -> {c > 0 && 0 < sigma && Element[c, Reals] && Element[sigma, Reals]}] (* 1/(c sigma) *) I assume they fixed the bug. 2 Mathematica doesn't yield results using generalized functions unless Fourier/Laplace transforms are involved. So, differentiate using a transform. LaplaceTransform[-I (Log[-x] - Log[x]), x, s] (* -I ((EulerGamma + Log[s])/s - (EulerGamma - I \[Pi] + Log[s])/s) *) InverseLaplaceTransform[s %, s, x] (* \[Pi] DiracDelta[x] *) 2 As mentioned by flinty in the comment, v12.2 can solve the problem directly, with a Solve::ifun warning generated. If you need the implicit solution as given by Maple, you can: Trace[ DSolve[1/x (-1 - (2 a/Sqrt[a^2 - 3 x L f^2 Exp[3 y[x]]])) == y'[x], y[x], x], Solve[_, y[x]], TraceInternal -> True] // Flatten (* {HoldForm[ Solve[(3/8)*(3*Log[... 4 Mathematica seems to have problems with Sqrt. "Squaring" gives ode = ( -(2 a/Sqrt[a^2 - 3 x L f^2 Exp[3 y[x]]]))^2 == (x y'[x] + 1)^2 (*(4 a^2)/(a^2 - 3 E^(3 y[x]) f^2 L x) == (1 + x Derivative[1][y][x])^2*) DSolveis now able to solve the ode DSolve[ode, y[x], x][[1]] (*Solve[(3 (6 Sqrt[a^2 x^2 (a^2 - 3 E^(3 y[x]) f^2 L x)] ArcTan[Sqrt[-... 3$$\, _3F_1\left(n+1,n+1,n+\frac{3}{2};2 n+3;-1\right)=\\\sum _{k=0}^{\infty } \frac{\left((n+1)_k\right){}^2 \left(n+\frac{3}{2}\right)_k (-1)^k}{(2 n+3)_k k!}=\\\sum _{k=0}^{\infty } \frac{(-1)^k 2^{1-2 k} (1+n) \Gamma (1+k+n) \Gamma (2+2 k+2 n)}{\Gamma (1+k) \Gamma (1+n) \Gamma (3+k+2 n)}=\\\sum _{k=0}^{\infty } \frac{(-1)^k 2^{1-2 k} (1+n) \Gamma (1+... 8 Yes this must be a bug. Splitting the integral "by hand" gives the correct result: Integrate[Integrate[(x - y)^n, {y, 0, x} ], {x, 0, 1}] +Integrate[Integrate[( y - x)^n, {x, 0, y} ], {y, 0, 1}] (*ConditionalExpression[2/(2 + 3 n + n^2), Re[n] > -1]*) addendum Mathematica v12.2 evaluates the integral OP asked for only correct if ... 0 Right bracket in the wrong place? Perhaps Plot[DH1[x, c, theta1]], {x, 0, 100} should be Plot[DH1[x, c, theta1], {x, 0, 100}] 1 First: Your integral equals 0 ! Integrate[II, {\[Phi], 0, 2 \[Pi]} ] (*0*) Second: Mathematica evaluates in finite time (~84s) the integral if you provide additional condition R>0 Integrate[II, {\[Phi], 0, 2 \[Pi]} , {\[Theta], 0,\[Pi]}, Assumptions -> R > 0] (*0*) 4 We can make the functional transformation via substitutions, without destroying the symbolic T[t] by giving it a definition, and the calculus will be done automatically: x[t_] = Q/T[t]; Simplify[Derivative[1][x][t] /. Q -> X T[t]] % /. T -> (1/a[#1] &) One can change the X to x if desired. It seems undesirable from a programming point of view to ... 0 It would appear that this simple staircase has "jaggies" where no slope can be defined. I think your function lacks defined slopes, too. Plot[CantorStaircase[x], {x, 0, 1}] 5 You are making a mistake. If T[t] = 1/a[t] , then T'[t] = D[(1/a[t]), t] = -a'[t] / a[t]^2 Further, replacement is a wholly structural operation, it can not do derivatives. To achieve what you want, you would have to set T[t]= 1/a[t] Clear[a, T] T[t_] = 1/a[t]; x T'[t]/T[t] 2$Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) Clear["Global`*"] expr = Cosh[x + s]^-2*Cosh[x]^-2; An indefinite integral (anti-derivative) of expr is ad = Integrate[expr, x] // Simplify (* -Csch[s]^2 (2 Coth[s] (Log[Cosh[x]] - Log[Cosh[s + x]]) + Sech[s] Sech[s + x] Sinh[x] + Tanh[x]) *) Verifying that ad ...

6

Try the definite integral Integrate[Cosh[x + s]^-2*Cosh[x]^-2, {x, -Infinity,Infinity}] (*2 Csch[s]^2 (-2 + Coth[s] Log[E^(2 s)]) if ...*) which Mathematica conditional solves. Numerical "confirmation" int[s_?NumericQ] := NIntegrate[ Cosh[x + s]^-2*Cosh[x]^-2, {x, -Infinity, Infinity}] Plot[{int[s], 2 Csch[s]^2 (-2 + Coth[s] Log[E^(...

7

Edition r[u_] = {Sin[u], Cos[u], u/10}; curve = ParametricPlot3D[{Sin[u], Cos[u], u/10}, {u, 0, 20}]; Show[VectorPlot3D[{x, y, z}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, VectorScaling -> True, VectorMarkers -> "Arrow", VectorPoints -> Table[r[u], {u, 0, 20, .1}]] /. Arrow[{p1_, p2_}] :> Arrow[{(p1 + p2)/2, p2}], curve] Or use ...

6

A 2D example: field[{x_, y_}] = {-y^2, x y}; curve[t_] = {t, t^3} + 0.3; Show[ ParametricPlot[curve[t], {t, -1 , 1}, PlotRange -> {{-1, 1}, {-1, 1}}], Graphics[{Arrowheads[{{Automatic, 1}}], Table[Arrow[{curve[t], curve[t] + field[curve[t]]}], {t, -1, 1, .1}]} ] ] And a 3D example: field[{x_, y_, z_}] = 0.5 {-y^2, x y, Sqrt[z y]}; curve[t_] = {t, ...

10

vf[{x_, y_}] = {-1 - x^2 + y, 1 + x - y^2}; curve[t_] := Sqrt[t] {Cos[t], Sin[t]}; Using ParametricPlot: scale = .2; ParametricPlot[curve[t], {t, 0, 3 Pi}, PlotStyle -> Thick, Epilog -> {Red, Thick, Arrow[{curve @ #, curve @ # + (scale Norm[#] Normalize[#] & @ vf[curve@#])}] & /@ Rest[Subdivide[0, 3 Pi, 50]]}, PlotRange -&...

1

As much as I know, Mma does not simplify any integrals. However, one can help it. Try the following. Let us first introduce a rule: intRule = Inactivate[Integrate[exprA___ exprB__ , {var_, lim__}], Integrate] /; FreeQ[{exprB}, var] :> exprB Inactivate[Integrate[exprA , {var, lim}], Integrate]; Here is your integral (inactivated): expr1 = ...

0

( Not an answer, a crash proclaim and verification request. ) The code below crashes the Mathematica kernel on my computer, Version 12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020). Please, run it on your machines/versions and proclaim the results. E.g. on Windows 10, Mathematica Version 12.1, etc. (It takes 20-35 min to get the crash.) AbsoluteTiming[ ...

1

We set t=1/N and use NDSolve. f[x_] := (x^2 +4)^4/2^x; NDSolve[{F'[t] == (f[1/t + 1] - f[1/t]) (-1/t^2), F[1] == Integrate[f[x], {x, 1, 2}], WhenEvent[F[t] <= 1/10, Print[Ceiling[1/t]]]}, F, {t, 1/1000, 1}] 48 i[n_] = Integrate[f[x], {x, n, n + 1}]; {i[47], i[48], i[49]} // N {0.132621, 0.0783269, 0.0461038}

2

Let f[n_] := NIntegrate[(x^2 + 4)^4/2^x, {x, n, n + 1}] though the above integral can be expressed in a closed form. First, the result of NMinimize[{f[n], n <= 10}, n, Method -> "DifferentialEvolution"] {244.597, {n -> -0.120219}} kicks out n<=10. Second, we find NMinimize[{n, f[n] <= 0.1 && n >= 10}, n, Method -> &...

2

If the integral converges it is allowed to change the coordinates {x,y}->{r,\[CurlyPhi]} to polar and to consider the eqivalent integral Integrate[r "integrand",{r,0,Infinity},{\[CurlyPhi],Pi/2,Pi}]! integrand= ((8 - 8 I) x y (1 + x^2 + y^2) ((3 + 4 I) + (2 + 4 I) x^2 -x^4 - 2 ((1 + 2 I) + x^2) y^2 +3 y^4))/(\[Pi] (-((-2 - I) + x)^2 + y^2) (-((...

1

expr1 = Simplify[ RotationMatrix[angle/n, {a, b, c}], {angle \[Element] Reals, n \[Element] Integers, a \[Element] Reals, b \[Element] Reals, c \[Element] Reals}]; expr2 = Simplify[ MatrixPower[expr1, n], {angle \[Element] Reals, n \[Element] Integers, a \[Element] Reals, b \[Element] Reals, c \[Element] Reals}]; expr3 = Limit[expr2,...

2

See update below as per user64494 Also, see update to integration method suggested by Daniele below Do you know beforehand if it converges? I can create an integral function innerI(x) of the inner integral from say y from 0 to 2000 as a function of x, then integrate that integral from say x from 0 to -400 and the absolute value doesn't appear to be ...

7

$d^3p$ is the volume element of momentum space, I think. So, $p$ is a point in 3 dimensional space; therefore an infinitesimal volume element in this space will be sort of...thrice $d$'d, one time for each axis. In other words, $d^3p = dp_1\ dp_2\ dp_3$, where $p=(p_1, p_2, p_3)$. Note that $\int d^3p$ is notationally equivalent to $\iiint dp_1\ dp_2\ dp_3$, ...

2

Another way is as follows. Assuming[{a*b < 0, \[CapitalOmega] \[Element] Reals}, Integrate[ 1/(I*\[Omega] + I*\[CapitalOmega] - a)*1/(I*\[Omega] - b), {\[Omega], -Infinity, Infinity}]] $\fbox{$-\frac{2 \pi }{a-b-i \Omega }\text{ if }a>0$}$

3

Integrate[1/(I*w+I*Ω-a)*1/(I*w-b), {w,-Infinity,Infinity}, Assumptions->(a∈Reals&&b∈Reals&&Ω∈Reals), GenerateConditions->False ] (π (Sign[a]-Sign[b]))/((a-b-I Ω) Sign[a] Sign[b])

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