# Problem with Integrate [closed]

 Integrate[(c*al*be)/Beta[a, b]*x^ -
(be + 1)*Exp[t*x]*Exp[-al*a*c*x^-be]*(1 - Exp[-al*c*x^-be])^(b - 1),
{x, 0, ∞},
Assumptions -> {a > 0 ,  b > 0, c > 0, al > 0, be > 0, t > 0}]


When I evaluate the above expression in Mathematica, I do not get a solution.
What's the problem?

For t = 0, the above expression is a proper "pdf". With a simple transformation we can prove the value of the integral is 1. But Mathematica not get solution. For manual solution, I get the following:

$$\frac{c^* al^* be}{B(a,b)} \int_0^\infty x^{-be-1} e^{-al^* a^* c^* x^{-be}} (1-e^{-al^* c^* x^{-be}})^{b-1}\,\mathrm{d}x$$ Write $z=e^{-al^* c^* x^{-be}}$, whence $$\mathrm{d}z = e^{-al^* c^* x^{-be}} (al^* be^* c^* x^{-be-1})\,\mathrm{d}x$$ and $$\frac{\mathrm{d}z}{z} = (al^* be^* c^* x^{-be-1})\,\mathrm{d}x$$ By substitution, $$\frac{1}{B(a,b)} \int_0^1 z^{a-1} \left(1-z\right)^{b-1} \,\mathrm{d}z = 1.$$

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## closed as off-topic by Louis, m_goldberg, MarcoB, happy fish, Mr.Wizard♦Jul 15 at 6:42

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Perhaps you correct Exp[tx] to Exp[t*x]. Otherwise your assumption with t>0 does not make sense. If you correct it in this way, Mathematica can not solve this Integral anymore. Perhaps you check carefully again how your integrand is defined. – partial81 Jun 28 '13 at 13:29
Integrand is defined properly, I also do not find solution with Mathematica, also when I put t=1 and remove assumption t>0, Whats the problem? – SAAN Jun 28 '13 at 13:40
As little simplification you could use one parameter for c*al. I do not see a reason why to use c and al when both are used all the time as product. – partial81 Jun 28 '13 at 13:41
Look e.g. at this: With[{c = 1, a = 1, b = 3, al = 2, be = 1, t = 2}, Integrate[(c*al*be)/Beta[a, b]*x^-(be + 1)*Exp[t*x]*Exp[-al*a*c*x^-be]*(1 - Exp[-al*c*x^-be])^(b - 1), {x, 0, Infinity}]]. You really needn't all those exponential functions inside Integrate. If you get rid of them you can go forward. Try e.g. FullSimplify[ Integrate[z^(a - 1) (1 - z)^(b - 1), {z, 0, 1}] == Beta[a, b], Assumptions -> (a > 1 && b > 1)] – Artes Jun 28 '13 at 16:40
I'm voting to close this question as off-topic because the issue it raises is not a Mathematica issue but a mathematics issue. That it is formulated in terms of Mathematica is not sufficient to make it an appropriate question for Mathematica.SE. – m_goldberg Jul 13 at 19:01