Functions that integrate to a gamma function?

[daviddoria]

maple syntax:

int(theta^y * exp(-theta*(1-alpha) ) , theta)

I have a distribution that I need to integrate, and I know the result should have a gamma function in it.

The only thing I have found helpful is:
http://en.wikipedia.org/wiki/Gamma_function

My function is kind of in that form (theta^something * exp(something) ), but the "somethings" don't seem to be able to be manipulated into that form. Any hints on how to go about this?

Thanks!

David

 

[Crosson]

You could match it with this generalized form of the gamma function (sometimes called the Plica Function):

$$\Gamma (a,z)=\int_z^{\infty } t^{a-1} e^{-t} dt$$

Or you could also match it with the exponential integral function (n is an integer):

$$E_n (z)=\int_1^{\infty } \frac{e^{-z t}}{t^n} dt$$

If you must match it to a non-generalized gamma expression, show us the final form you are aiming for.

 

[daviddoria]

I don't understand how I would match it with those? The problem is I have

t^alpha

and I need
t^(alpha - 1)

The only way I would know to "mold" mine into the correct one is to multiply by t^(-1) and t, then take t outside the integral and combine t^alpha with t^(-1) to get t^(alpha-1). However, t is the integration variable, so I can't do that!

any other thoughts?

Thanks,

David

 

[Mute]

Use the recursive property of the gamma function:

$$\Gamma(\alpha+1) = \alpha \Gamma(\alpha)$$

This does require, however, that your limits of integration are 0 to infinity.

 

[Crosson]

I don't understand how I would match it with those? The problem is I have

t^alpha

and I need
t^(alpha - 1)

However, t is the integration variable

Use integration by parts, differentiate t^alpha.

 

[coomast]

If the integral you are looking for is the following (assuming the limits are correct):

$$\int_{0}^{\infty}\theta^y \cdot e^{-\theta(1-\alpha)} d\theta$$

You can transform it using:

$$\theta(1-\alpha)=u$$

This gives you something related to the Gamma function.