computation of an integral

edit

FWIW :

sage: var("a")
a
sage: foo=1/((x-a)^(1/2)*x^(3/4))
sage: foo.integrate((x, 0, oo))
---------------------------------------------------------------------------
RuntimeError                              Traceback (most recent call last)

[ Snip... ]

ValueError: Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a>0)', see `assume?` for more details)
Is a positive, negative or zero?` arguments fails

Well...

sage: with assuming(real(a)<0): gee=foo.integrate((x, 0, oo)) ; gee
beta(1/4, 1/4)/(-a)^(1/4)
sage: gee.simplify_factorial()
gamma(1/4)^2/(sqrt(pi)*(-a)^(1/4))

Following Sage's suggestion gives us an usable result for a<0 (and fails for a>=0, BTW).

Let's try someting else. Adding algorithm="sympy" to `integrate arguments fails, because the result cannot (yet) be translated back to Sage. Try it manually :

sage: import sympy
sage: sympy.integrate(*map(sympy.sympify, (foo, (x, 0, oo))))
Piecewise((-a*exp(3*I*pi/4)*gamma(1/4)**2/(sqrt(pi)*polar_lift(a)**(5/4)), (Abs(arg(a) + pi) < pi) | (Ne(1/a, 0) & (Abs(arg(a) + pi) < pi) & Ne(Abs(arg(a) + pi), pi))), (Integral(1/(x**(3/4)*sqrt(-a + x)), (x, 0, oo)), True))

which ultimately means $ \frac{a e^{\frac{3 i \pi}{4}} \Gamma^{2}\left(\frac{1}{4}\right)}{\sqrt{\pi} \operatorname{polar_lift}^{\frac{5}{4}}{\left(a \right)}} $ if abs(arg+pi)<pi, the latter reducing to real(a)<0 in I'm not confused.

Sympy fails to simplify the complicated expressions it uses (and cannot be translated to sage). But, FWIW :

sage: foo._mathematica_().Integrate((x, 0, oo))
ConditionalExpression[Gamma[1/4]^2/((-a)^(1/4)*Sqrt[Pi]), Re[a] < 0]

The real situation is therefore a tad moree complex than described.

• Sage's can find an explicit form of the integral under a condition which it cannot find.

• Sympy cannot find simplified expressions of its solution.

• Mathematica agrees with Sage's integratin and seems to agree to Sympy's condition.

HTH,