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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

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The exponential function is E^(...) or Exp[...]. –  b.gatessucks Mar 12 '13 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)] –  Daniel Lichtblau Mar 13 '13 at 13:37
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3 Answers

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]
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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 '13 at 11:49
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You can combine both assumptions :

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

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The assumptions are slightly different (r>1) and allow Mathematica to find the analytic solution. Also addresses the likely mistyping of 'e' for 'E'... –  SEngstrom Mar 19 '13 at 16:37
    
Sorry, my mistake, you are right! –  István Zachar Mar 19 '13 at 17:26
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