Multiple assumptions in integral

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

• The exponential function is E^(...) or Exp[...]. Mar 12, 2013 at 12:12
• Can only give one Assumptions-> option but it can have a List or conjunction (And). So Assumptions->{r>0,L>0} would be a correct syntax. I get Integrate[(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)] Mar 13, 2013 at 13:37

You can make multiple assumptions by passing a list:

Integrate[(L r x^(r - 2))/(r - 1)! e^(-L x), {x, 0, Infinity},
Assumptions -> {r > 0, L > 0}]

Or, if you prefer, a boolean expression:

Integrate[(L r x^(r - 2))/(r - 1)! e^(-L x), {x, 0, Infinity},
Assumptions -> r > 0 && L > 0]
• Hmm ok i tried to get a solution, but as far as i can see, it doesnt not give you 1. Maybe someone have a solution?
– Anru Narenthirarajah
Mar 13, 2013 at 11:49

You can combine both assumptions :

Integrate[(L r x^(r - 2))/(r - 1)! e^(-L x), {x, 0, Infinity},
Assumptions -> {r > 0, L > 0}]

Your integral is available analytically with:

Integrate[(L*r *x^(r - 2))/(r - 1)! Exp[-L*x], {x, 0, Infinity}, Assumptions -> {r > 1, L > 0}]

... with the result

(L^(2 - r) r)/(-1 + r)

This is unity only for careful choices of L and r so I wonder if your question isn't about a normalization rather than an absolute check on a probability density function.

• The assumptions are slightly different (r>1) and allow Mathematica to find the analytic solution. Also addresses the likely mistyping of 'e' for 'E'... Mar 19, 2013 at 16:37
• Sorry, my mistake, you are right! Mar 19, 2013 at 17:26