I want to add two assumptions, so I can get this probability density function to equal 1, though I can't get a solution.
Integrate[(L r x^(r - 2))/(r - 1)! e^(-L x), {x, 0, Infinity},
Assumptions -> r > 0, Assumptions -> L > 0]
My equation. With the assumptions r > 0
and L > 0
E^(...)
orExp[...]
. $\endgroup$Assumptions->{r>0,L>0}
would be a correct syntax. I getIntegrate[(L r x^(r - 2))/(r - 1)! e^(-L x), {x, 0, Infinity}, Assumptions ->{r>0,L>0}] // InputForm Out[2]//InputForm= ConditionalExpression[(L*r*Gamma[-1 + r]*(L*Log[e])^(1 - r))/(-1 + r)!, r > 1 && ((Re[Log[e]] == 0 && r < 2) || Re[Log[e]] > 0)]
$\endgroup$