Integration involving Gamma Distribution

[cse63146]

Homework Statement

Evaluate $$\int^{\infty}_0 \frac{1}{B^{\alpha} \Gamma(\alpha)} \theta^{\alpha + x -1}e^{- \theta(1+B)/B} d \theta$$

Homework Equations

The Attempt at a Solution

$$\int^{\infty}_0 \frac{1}{B^{\alpha} \Gamma(\alpha)} \theta^{\alpha + x -1}e^{- \theta(1+B)/B}d \theta$$

$$= \frac{1}{B^{\alpha} \Gamma(\alpha)} \int^{\infty}_0 \theta^{\alpha + x -1}e^{- \theta(1+B)/B}d \theta$$

Let $$u = \theta(1+B)/B$$ and $$du = \frac{1 + B}{B} d\theta$$

$$= \frac{1}{B^{\alpha} \Gamma(\alpha)} \int^{\infty}_0 \frac{1 + B}{B} \frac{Bu}{1+B}^{\alpha + x -1}e^{- u} du$$

I'm stuck here. Not sure what to do next.

 

[mighty2000]

Can you please provide some guidance?

Hi there, it looks like you're on the right track with your integration. Let's continue from where you left off:

= \frac{1}{B^{\alpha} \Gamma(\alpha)} \int^{\infty}_0 \frac{1}{B} u^{\alpha + x -1}e^{- u} du


= \frac{1}{B^{\alpha + 1} \Gamma(\alpha)} \int^{\infty}_0 u^{\alpha + x -1}e^{- u} du


Now, we can use the definition of the Gamma function to simplify the expression:

= \frac{\Gamma(\alpha + x)}{B^{\alpha + 1} \Gamma(\alpha)}


= \frac{\Gamma(\alpha + x)}{B \Gamma(\alpha + 1)}


= \frac{\Gamma(\alpha + x)}{B \alpha!}


Therefore, our final solution is:

\int^{\infty}_0 \frac{1}{B^{\alpha} \Gamma(\alpha)} \theta^{\alpha + x -1}e^{- \theta(1+B)/B} d \theta = \frac{\Gamma(\alpha + x)}{B \alpha!}

I hope this helps! Let me know if you have any further questions.