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6

The expression in question is (we have replaced sigma by s) g = Integrate[ Exp[-(x - y)^2/(2 y s^2)] f[y] 1/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}] $\int_0^{\infty } \frac{e^{-\frac{(x-y)^2}{2 s^2 y}} f[y]}{\sqrt{2 \pi } \sqrt{s^2 y}} \, dy$ First of all we notice that for y>0 "Limit[Exp[-(x - y)^2/(2 y s^2)] 1/Sqrt[2 \[Pi] y s^2], s -> 0] = ...


5

The antiderivative is correct, in the sense its derivative gives back the original integrand Clear[x] integrand = 4279/Sqrt[6817/10000 + 3183/10000*(1 + x)^3]; mmaResult = Integrate[integrand, x]; integrandBack = Simplify[D[mmaResult, x]]; Plot[{integrandBack, integrand}, {x, 0, 20}, PlotTheme -> "Detailed"] But the definite integral does not give ...


3

Perhaps better: NMinimize[{m/(m (1 - c) + 1), 2^(m*(1 - c)) - 2.718/((m^.5)*2*3.14*c^(c*m + .5)*(1 - c)^((1 - c)*m + .5)) <= 0 && 0.006 <= c <= 1 && 4 <= m <= 2500}, {m, c}, MaxIterations -> 300] (* {1.11184, {m -> 4., c -> 0.350592}} *)


3

As a general rule, avoid using l' and m', which Mathematica interprets as derivatives. Mathematica has difficulty performing the integration for symbolic constants l, m, ll and mm, but can verify orthonormality for a finite range of those values: Table[Assuming[{l, ll, m, mm} \[Element] Integers, Integrate[ Conjugate[ ...


2

There are a very large number of syntax errors. The most serious relate to use of protected symbols: D, I. This is a correction that should work. Please adjust plot range to your needs: sirds[α_, β_, δ_, μ_] := {S[t], SS[t], i[t], R[t], d[t]} /. First@NDSolve[{S'[t] == -α*S[t]*i[t] - δ*SS[t], i'[t] == α*S[t]*i[t] - β*i[t] - μ*i[t], R'[t] ...


2

First I'll provide a symbolic workaround, and then I'll explain why your attempt doesn't work. The integral can be symbolically evaluated, like this: ModelInternalEnergy[Td_, T_] := Evaluate[Simplify[ Alpha*T^4* Integrate[x^3/(Exp[x] - 1), {x, 0, Td/T}, Assumptions -> Td/T > 0] + ground]] which for reference gives ground - 1/15 ...


2

The output of the Mathematica code is ConditionalExpression[ 1/2 (2 a - a Sqrt[1 - a^2] + Sqrt[3 + 2 a - a^2] - a Sqrt[3 + 2 a - a^2] - 2 b + b Sqrt[1 - b^2] - Sqrt[3 + 2 b - b^2]+ b Sqrt[3 + 2 b - b^2] + 4 ArcSin[(1 - a)/2] *Routine clean-up*-ArcSin[a] - 4 ArcSin[(1 - b)/2] + ArcSin[b]), -1 < b < 1 && ( -1 < a < b || b ...


1

My earlier answer resulting from misreading the question. Taken at face value, you are asking for sum=Sum[3 - 4/(1 - (-3)^(n + 1)), {n, 1436, 2015}] which produces a rationale number with an enormous number of digits. It is, however, almost precisely equal to 3*(2015 - 1436 + 1), or 1740, as one would expect. N[sum,1000] (* ...


1

Did you try using assumptions? Using assumptions with version 10.0.0, I get the same results as with version 10.0.2 $Version "10.0 for Mac OS X x86 (64-bit) (June 29, 2014)" Clear[f] f[n_Integer] = Integrate[HermiteH[n, x]*Exp[-x^2], {x, 0, Infinity}, Assumptions -> {Element[n, Integers]}] (2^(-1 + n)*Sqrt[Pi])/Gamma[1 - n/2] ...


1

It might be Errr... a bug or some singular points(not on the real axis) might be involved and Integrate chose a strange Integrate route. define f[x_] := Integrate[4279/Sqrt[0.6817 + 0.3183*(1 + x)^3], x] then fdat = Transpose@{Range[0, 20, 0.1], f[x] /. x -> Range[0, 20, 0.1]} plot the f generated by Integrate: ListLinePlot[{{#[[1]], ...


1

but why For the why part, you need little bit of math. You can always start by looking at series expansion of all terms expr = (Exp[x] - Exp[Sin[x]])/(x - Sin[x]); Limit[expr, x -> 0] (*1*) Now expand all in taylor (used 4 terms to make it easy to see) expr = (Normal@Series[Exp[x], {x, 0, 4}] - Normal@Series[Exp[Sin[x]], {x, 0, 4}])/(x - ...


1

First of all correcting the sign in the exponential we get f2 = Integrate[ Exp[r^2/(2 M^2)] (b^2 - 3 r^2)^2 (Abs[b^2 + r^2] - (b^2 - r^2))/( r (r^2 - b^2)), {r, 0, I \[Infinity]}, Assumptions -> {M > 0, b > 0] (* Out[278]= 18 b^2 M^2 + 4 b^4 E^(b^2/(2 M^2)) ExpIntegralEi[-(b^2/(2 M^2))] *) Taking the pricipal value has no influence as ...


1

Outline As you didn't provide boundary and initial conditions and the function pa'[t] this solution must be generic. Your equation for p[u,t] is linear (I guess pa'(t) means D[p[u,t],u]/.u->a) and can therefore be solved by standard mathematical methods once you provide the boundary and initial conditions. Physically it describes diffusion in a cylinder. A ...


1

In V10,D can symbolically differentiate NIntegrate if it is Inactive. In the case that the integral can only be evaluated with NIntegrate the following defines an arbitrary Derivative of the OP's function numericalModelInternalEnergy. ClearAll[numericalModelInternalEnergy, numericalModelHeatCapacity]; Block[{NIntegrate, Alpha, ground, T, Td}, integrand = ...



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