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 ...
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 ...
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 ...
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 ...
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&...
1
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 ...
1
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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