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I want to calculate the integration of, $$ \frac{1}{y! \Gamma(\alpha) \beta^\alpha}\int_{0}^{\infty} \lambda^{y+\alpha-1} \exp\left\{ \frac{-\lambda}{\frac{\beta}{\beta+1}} \right\} d\lambda \ \ \ \ \ \ \ \ \ \ (1) $$ The answer says, $$ \begin{pmatrix}y + \alpha -1 \\ y \end{pmatrix} \left( \frac{\beta}{\beta+1}\right)^y \left( \frac{1}{\beta+1} \right)^\alpha$$

I am not sure where to start with. When we encounter this type of integration, what is the first thing we should consider? And how can I calculate Equation (1)?

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    Take a look at the definition of the Gammafunction $\Gamma$ (in google/wikipedia). I think this immediately will help you. – 2017-02-01

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By definition of Gamma Function,

$$\Gamma (n)=\int\limits_{0}^{\infty} t^{n-1} e^{-t}dt$$

$$ I=\frac{1}{y! \Gamma(\alpha) \beta^\alpha}\int_{0}^{\infty} \lambda^{y+\alpha-1} \exp\left\{ \frac{-\lambda}{\frac{\beta}{\beta+1}} \right\} d\lambda \ \ \ \ \ \ \ \ \ \ $$ Take subsitution $s=\frac{\beta+1}{\beta}\lambda$,

$ I=\frac{1}{y! \Gamma(\alpha) \beta^\alpha}\int_{0}^{\infty} (\frac{s \beta}{\beta+1})^{y+\alpha-1} \exp{(-s)} (\frac{\beta ds}{\beta+1}) $

$ =\frac{1}{y! \Gamma(\alpha) \beta^\alpha}\bigg(\frac{\beta}{\beta+1}\bigg)^{y+\alpha}\int_{0}^{\infty} s^{y+\alpha-1} \exp{(-s)} ds $

$ =\frac{1}{y! \Gamma(\alpha) \beta^\alpha}\bigg(\frac{\beta}{\beta+1}\bigg)^{y+\alpha}\Gamma(y+\alpha) $

$ =\frac{1}{y! \Gamma(\alpha) \beta^\alpha}\bigg(\frac{\beta}{\beta+1}\bigg)^{y}\bigg(\frac{\beta}{\beta+1}\bigg)^{\alpha}\Gamma(y+\alpha) $

$ =\frac{(y+\alpha -1)!}{y! (y+\alpha-1-y)!}\bigg(\frac{\beta}{\beta+1}\bigg)^{y}\bigg(\frac{1}{\beta+1}\bigg)^{\alpha} $

$ =\begin{pmatrix}y + \alpha -1 \\ y \end{pmatrix} \left( \frac{\beta}{\beta+1}\right)^y \left( \frac{1}{\beta+1} \right)^\alpha$

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    @user51966 Now, this is final solution. – 2017-02-01