AngouriMath
SolveAPowerTimesAnExponentialOfAQuadraticInALogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A power of x times an exponential of a quadratic in a logarithm, x^p G^(Q(L)) with
L = ln(c x^r) and Q a quadratic with a square term, onto the
Gaussian. Witht = L and L' = s/x , dx = x dt/s , and
x^(p + 1) e^(-(p + 1) L/s) is a constant, so the integral is that constant over
s times the integral of e^(Q(t) ln G + (p + 1) t/s) : the Gaussian with a
linear term, which the table answers. A polynomial beside it is summed a power at a time,
and the logarithm of a power of a linear is the same question underu = d + e x .
Rubi's 2.3,F^(f (a + b ln(c (d + e x)^n))^2) (g + h x)^m and
F^(f (a + b ln(c (d + e x)^n)^2)) (g + h x)^m .
https://github.com/asc-community/AngouriMath/issues/1501
Gaussian. With
linear term, which the table answers. A polynomial beside it is summed a power at a time,
and the logarithm of a power of a linear is the same question under
Rubi's 2.3,
https://github.com/asc-community/AngouriMath/issues/1501
Remarks
The constant is written x^(p + 1) e^(-(p + 1) L/s) , as
SolveAPowerTimesAHalfOddPowerOfTheLogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable) writes it and as Rubi does,
rather than withln(c x^r) split into ln c + r ln x , which holds only
wherec is positive. A quadratic without its square term is a power of
c x^r , which the rules for powers answer.
SolveAPowerTimesAHalfOddPowerOfTheLogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable) writes it and as Rubi does,
rather than with
where
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