I get the following when using auxiliary variables to make true your
assumptions.
>>> eq
(0.2*exp(t*(4*d1 + 7*d2 + 7*d3)) + 0.2*exp(t*(7*d1 + 4*d2 + 7*d3)) +
0.2*exp(t*(
7*d1 + 7*d2 + 4*d3)) + 0.4*exp(t*(6*d1 + 6*d2 + 6*d3 + 2*sqrt(d1**2 - d1*d2
- d1
*d3 + d2**2 - d2*d3 + d3**2))))*exp(-8*t*(d1 + d2 + d3))
>>> eq = nsimplify(eq)
>>> p,q=var('p,q',positive=True)
>>> factor_terms(nsimplify(eq.subs(d1,d3+p+q).subs(d2,d3+p).expand()))
(1 + 2*exp(-2*p*t)*exp(-q*t)*exp(2*t*sqrt(p**2 + p*q + q**2)) + exp(-3*p*t)
+ ex
p(-3*p*t)*exp(-3*q*t))*exp(-6*d3*t)*exp(-2*p*t)*exp(-q*t)/5
>>> _.integrate((t,0,oo))
Piecewise((2/(5*(1 + q/(d3*(3 + 2*p/d3)))*(1 - sqrt(p**2 + p*q +
q**2)/(d3*(1 +
q/(d3*(3 + 2*p/d3)))*(3 + 2*p/d3)))*(6*d3 + 4*p)) + 1/(5*d3*(1 + 4*q/(d3*(6
+ 5*
p/d3)))*(6 + 5*p/d3)) + 1/(5*d3*(1 + q/(d3*(6 + 5*p/d3)))*(6 + 5*p/d3)) +
1/(10*
d3*(1 + q/(2*d3*(3 + p/d3)))*(3 + p/d3)),
(Abs(periodic_argument(polar_lift(d3)*
polar_lift(3 + p/d3), oo)) <= pi/2) &
(Abs(periodic_argument(polar_lift(d3)*pola
r_lift(6 + 5*p/d3), oo)) <= pi/2) &
(Abs(periodic_argument(polar_lift(d3)*polar_
lift(1 + q/(d3*polar_lift(3 + 2*p/d3)))*polar_lift(3 + 2*p/d3), oo)) <
pi/2) & N
e(sqrt(p**2 + p*q + q**2)/(d3*(1 + q/(d3*(3 + 2*p/d3)))*(3 + 2*p/d3)), 1) &
(-2*
sqrt(p**2 + p*q + q**2) +
2*cos(Abs(periodic_argument(polar_lift(d3)*polar_lift(
1 + q/(d3*polar_lift(3 + 2*p/d3)))*polar_lift(3 + 2*p/d3), oo)))*Abs(d3*(1
+ q/(
d3*(3 + 2*p/d3)))*(3 + 2*p/d3)) > 0)), (Integral((1 +
2*exp(-2*p*t)*exp(-q*t)*ex
p(2*t*sqrt(p**2 + p*q + q**2)) + exp(-3*p*t) +
exp(-3*p*t)*exp(-3*q*t))*exp(-6*d
3*t)*exp(-2*p*t)*exp(-q*t)/5, (t, 0, oo)), True))
On Thursday, October 26, 2017 at 11:13:34 AM UTC-5, Max Linke wrote:
>
>
> <https://screenshots.firefoxusercontent.com/images/c1f24747-118b-41ba-9785-012212161d83.png>
>
>
> Hi
>
> I'm trying to solve a integral with sympy. To get meaningful results I
> need to tell sympy that the assumptions d1>d2>d3 hold for the variables
> in the integral. I tried this unsuccessfully:
>
> ```python
> with assume.assuming(Q.is_true(d1 > d2), Q.is_true(d2 > d3)):
> integrate(C2, (t, 0, oo)
> ```
>
> I've attached a image of the function C2 and a here is a notebook
> <https://gist.github.com/kain88-de/3f019fff2ab8b5299edc9b0fb19caf93> with
> my attempt at solving the integral.
>
> I appreciate help how to define these assumptions in sympy.
>
> With google I haven't found anything helpful on this topic. Only a 4 year
> old SO post <https://stackoverflow.com/a/16435975/2207958> that this is
> not possible.
>
> A colleague already checked with mathematica for me that given the
> assumption d1>d2>d3 the integral
> converges. So a solution does exist.
>
> best Max
>
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