Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to finish some integration as following below:

$$\int_x^{\infty} \frac{\mathrm \beta^{\alpha+\gamma} X^{\alpha-1}(y-x)^{\gamma-1}\exp^{-\beta y}}{\Gamma(\alpha) \Gamma(\gamma)}dy\;$$

and

$$\int_0^{y} \frac{\mathrm \beta^{\alpha+\gamma} X^{\alpha-1}(y-x)^{\gamma-1}\exp^{-\beta y}}{\Gamma(\alpha) \Gamma(\gamma)}dx\;$$

For your step by step answer will really appreciated.

share|improve this question
1  
Thanks for your revised @Michael –  user91036 Sep 10 '13 at 16:44

1 Answer 1

For the first integral:

From the definition of the $\Gamma$-function and after substitutions you get

$$\int_x^\infty (y-x)^{\gamma-1} \cdot \exp(-\beta y) \, \mathrm{d}y= \exp(-x \cdot\beta)\cdot \beta^{-\gamma} \cdot \Gamma(\gamma)$$ when the real part of $\gamma$ and $\beta$ is strictly above $0$.

The substitutions should be $z=y-x$ from where you get the $\exp(-x\beta)$ part, and afterward s substitute to $u=\beta z$ from where you get $\beta^{-\gamma}$.

share|improve this answer
1  
thanks @dominic for the Hint, do you know how to integrate for the second integral? –  user91036 Sep 10 '13 at 18:59
1  
@lpchristine the second is just to integrate $x^\beta$ where is the problem ? –  Dominic Michaelis Sep 10 '13 at 20:19
1  
I am still confused to integrate the second part, what do you mean $x^\beta$? can you show it briefly Dominic? many thanks... –  user91036 Sep 11 '13 at 1:35
2  
After factoring out constants you have $$\int_0^y (y-x)^{\gamma-1} \, \mathrm{d}x$$ –  Dominic Michaelis Sep 11 '13 at 8:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.