1 | initial version |
After executing the start of your code :
sage: h
(p1, alpha, beta) |--> 1/(p1^(2/3)*(-p1 + 1)^(2/3))
sage: k`k`
(p1, xi) |--> -(e^(-p1*xi) - 1)*e^(-p1*xi)
sage: hk
(p1, alpha, beta, xi) |--> -(e^(-p1*xi) - 1)*e^(-p1*xi)/(p1^(2/3)*(-p1 + 1)^(2/3))
Both h
and k
are symbolic functions (more precisely callable symbolic expressions needing respectively three and four arguments. But so is hk which needs four arguments.
iuntegrate
works on symbolic expressions, not on such functions. What you intended is probably :
sage: definite_integral(hk(p1, alpha, beta, xi), p1, 0, 1)
integrate((e^(p1*xi) - 1)*e^(-2*p1*xi)/(p1^(2/3)*(-p1 + 1)^(2/3)), p1, 0, 1)
This integral turns out to be undoable by the four built-in integrators ; mathematica_free
fails to return anything. However, Mathematica
boasts :
sage: mathematica.Integrate(hk(p1, alpha, beta, xi), (p1, 0, 1)).sage()
sqrt(pi)*(bessel_I(-1/6, 1/2*xi)*e^(1/2*xi) - 2^(1/6)*bessel_I(-1/6, xi))*xi^(1/6)*e^(-xi)*gamma(1/3)
Note, however, that Mathematica
fails to integrate the indefinite integral :
sage: mathematica.Integrate(hk(p1, alpha, beta, xi), p1)
-Integrate[(-1 + E^(-(p1*xi)))/(E^(p1*xi)*(1 - p1)^(2/3)*p1^(2/3)), p1]
which renders impossible the verification by re-differentiation. The only possible "raincheck" would be to plot the numerical value of the Mathematica ex^ression againts the numerical integration for various value of xi
, and to check that these value do not depend on alpha
nor beta
...
Summary : Here may be tygers...
HTH,