maths question about an integral

Originally posted by humy
Oh damn! You are right. And that was a misprint in my OP and which I propagated throughout my earlier workings causing me (and you ) confusion. Terribly sorry about that. My fault. That should have been;

∫[0, ∞] h (θ^(m-1)) / ( e^( θ*(h + s) )) dθ

I will now systematically iterate through all my files to correct that edit error wherever I have propagated it -that could take me hours.

Of course the easiest way of doing this is to make a change of variables to a standard integral. Up to a factor of h your integral is I(m, h + s), where:

I(m, a) = ∫[0, ∞] θ^(m-1) exp( -aθ ) dθ

Change variables so that x = aθ, then we have:

I(m, a) = ∫[0, ∞] (x/a)^(m-1) exp(-x) dx/a
= (1/a)^m ∫[0, ∞] dx x^(m-1) exp(-x)

The definition of the gamma function is:

Γ(t) = ∫[0, ∞] dx x^(t-1) exp(-x)

So:

I(m, a) = Γ(m)/a^m ∀a, m

So your integral is hΓ(m)/(h + s)^m. Γ(m) = (m - 1)! for positive integer m (which we can look up on Wikipedia or prove using one of the methods above). This gives your result of h (m - 1)! / (h + s)^m.

I am now stuck on a closely related integration problem:

How to algebraically show that:

∫[θ=0, ∞] ( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) ) dθ
= m * S^m / ((S+x)^(m + 1))

m ∈ ℕ,
m>0,
x, S, θ ∈ ℝ,
x, S, θ > 0,

( so all positive and none zero although it doesn't matter that θ starts at zero for this integral )

I have checked that the above equation is correct using several reliable numerical methods with my thoroughly tested computer programs I made for that so absolutely sure that is correct but I seem to get nonsense when I try and show it algebraically.

Here is my workings below that must be erroneous somewhere so at what point do I go wrong below? :

∫[0, ∞] ( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) ) dθ
= ∫[0, ∞] (θ^(m – 1)) (S^m) (e^(θx)) / ( θ * (m-1)! * e^(θS) ) dθ
= ∫[0, ∞] (θ^(m – 1)) (S^m) e^(θ(x - S)) / ( θ * (m-1)! ) dθ
= ∫[0, ∞] (θ^(m – 2)) (S^m) e^(θ(x - S)) / (m-1)! dθ
= ( (S^m) / ((m-1)!) ) * ∫[0, ∞] (θ^(m – 2)) e^(θ(x - S)) dθ ( by applying the multiplication by constant rule )
= ( (S^m) / ((m-1)!) ) * (m - 2)! / (S - x)^(m - 1)
= ( (S^m) / (m - 1)) / (S - x)^(m - 1)
= (S^m) / ( (m - 1) * (S - x)^(m - 1) )
= (S^m) * ((S - x)^(1 - m) ) / (m - 1)
-which is total nonsense because this gives me negative values if I now integrate this in respect to x ( not in respect to θ like the integral above ) and that is not allowed here because such an integral in respect to x is supposed to be a probability!
In contrast; ∫[x=0, ∞] m * S^m / ((S+x)^(m + 1)) dx = 1 as required here for it to be a probability distribution.

Originally posted by humy
I am now stuck on a closely related integration problem:

How to algebraically show that:

∫[θ=0, ∞] ( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) ) dθ
= m * S^m / ((S+x)^(m + 1))

m ∈ ℕ,
m>0,
x, S, θ ∈ ℝ,
x, S, θ > 0,

( so all positive and none zero although it doesn't matter that θ starts at zero for this integral ) ...[text shortened]... , ∞] m * S^m / ((S+x)^(m + 1)) dx = 1 as required here for it to be a probability distribution.

Your integral is:
I = ∫[θ=0, ∞] ( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) ) dθ

Getting the constants out of the integral gives:

I = (S^m)/(m - 1)! ∫[θ=0, ∞] (θ^m) exp(-θ(S + x)) dθ

which is the integral from above, but with m instead of m - 1 so

I = S^m Γ(m+1)/(m - 1)!(S + x)^(m + 1)
= m S^m/(S + x)^(m + 1)

Originally posted by DeepThought
[b]Your integral is:
I = ∫[θ=0, ∞] ( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) ) dθ

Getting the constants out of the integral gives:

I = (S^m)/(m - 1)! ∫[θ=0, ∞] (θ^m) exp(-θ(S + x)) dθ

I am sure you are right but where did you get that;

"(θ^m) exp(-θ(S + x))"

from?

( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) )
= (θ^(m – 2)) (S^m) e^(θ(x - S)) / (m-1)!

-correct?

and if you now take out the constant
(S^m) / (m-1)!
out of that RHS;
(θ^(m – 2)) (S^m) e^(θ(x - S)) / (m-1)!
you are left with
(θ^(m – 2)) e^(θ(x - S))
i.e.
(θ^(m – 2)) (S^m) e^(θ(x - S)) / (m-1)!
= ( (S^m) / ((m-1)!) ) * (θ^(m – 2)) e^(θ(x - S))

and that above " (θ^(m – 2)) e^(θ(x - S))" of mine
doesn't equal your
"(θ^m) exp(-θ(S + x))"
-somebody tell me where I am going wrong here.

Just notice it would have made much more practical sense for me to take out the constants right at the start before starting to rearrange the expression but never mind.

Originally posted by humy
it would have made much more practical sense for me to take out the constants right at the start before starting to rearrange the expression but never mind.[/b]

I just tried doing just that and I immediately got it!

∫[0, ∞] ( (θ^(m – 1)) (S^m) / ( (m-1)! * e^(θS) ) ) * ( θ/(e^(θx)) ) dθ

= ( (S^m) / (m-1)! ) * ∫[0, ∞] ( (θ^(m – 1)) / e^(θS) ) * ( θ/(e^(θx)) ) dθ

let c = ( (S^m) / (m-1)! )
= c∫[0, ∞] ( (θ^(m – 1)) / e^(θS) ) * ( θ/(e^(θx)) ) dθ
= c∫[0, ∞] ( (θ^m) / e^(θS) ) / e^(θx) ) dθ
= c∫[0, ∞] (θ^m) / ( e^(θS) * e^(θx) ) dθ
= c∫[0, ∞] (θ^m) / (e^(θ(S + x)) dθ
(equivalent to what DeepThought said )

and now at last I am well and truly on the right track.

Thanks DeepThought.