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$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] // ...


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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 ...


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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 ...


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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 ...


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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 ...


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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&...


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Using some assumptions and an immediate assignment gives an analytic integral, which is easy to plot: F[x_, a_] = Assuming[-1 <= x <= 1 && a > 0, Integrate[Abs[x - z]^(-1 - 2 a), {z, -Infinity, -1}] + Integrate[Abs[x - z]^(-1 - 2 a), {z, 1, Infinity}]] (* (1 - x)^(-2 a)/(2 a) + (1 + x)^(-2 a)/(2 a) *) The plot looks exactly like ...


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You can do numerical integration and a 3D plot like: Plot3D[NIntegrate[Abs[x - z]^(-1 - 2 a), {z, -Infinity, -1}] + NIntegrate[Abs[x - z]^(-1 - 2 a), {z, 1, Infinity}], {x, -1, 1}, {a,0,1}] a 1-d plot could look like Plot[Evaluate[ Table[((1 - x)^(-2*a) + (1 + x)^(-2*a))/(2*a), {a, 0.1, 1, 0.2}]], {x, -1, 1}]


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