Let the model distribution (likelihood) be exponential, i.e. $$ p(x \mid \lambda) := \text{Exp}(\lambda) := \lambda e^{-\lambda x} $$ and the prior distribution be gamma (shape-rate-parametrization), i.e. $$ p(\lambda \mid \alpha, \beta) := \text{Gamma}(\alpha,\beta) := \frac{\beta^{\alpha}}{\Gamma{(\alpha)}} \lambda^{\alpha-1}\exp{(-\beta\lambda)} $$ We were now asked to find the posterior distriubtion $$ p(\lambda \mid X,\alpha,\beta) = p(X \mid \lambda)p(\lambda \mid\alpha,\beta) = \text{Exp}(\lambda) \text{Gamma}(\alpha,\beta) $$ Now assuming $X = (x_k)_{k = 1}^{n}$ and i.i.d conditions, I calculated $$\text{Exp}(\lambda) = \prod_{k = 1}^{n} \lambda e^{-\lambda x_k} = \lambda^n e^{-\lambda \bar{X}},\quad \text{where } \overline{X} := \sum_{k = 1}^{n} x_k$$ and thus \begin{align} p(\lambda \mid X,\alpha,\beta) & = \lambda^n \exp\left(-\lambda \bar{X}\right) \frac{\beta^{\alpha}}{\Gamma(\alpha)} \lambda^{\alpha - 1} \exp(-\beta \lambda) \\ & = \lambda^{\alpha + n - 1} \frac{\beta^{\alpha}}{\Gamma(\alpha)} \exp\left(-\lambda \left(\beta + \overline{X}\right)\right) \end{align} and noticed $p(\lambda \mid X, \alpha, \beta) = \text{Gamma}(\alpha + n, \beta + \overline{X})$.
Until now, everything's fine.
How we were asked to find the marginal likelikhood $$ p(X \mid \alpha,\beta) = \int p(X \mid \lambda) p(\lambda \mid \alpha,\beta) \ d\lambda, $$ which I calculated to be (I used the substitution $u = \lambda(\beta + \bar{X})$) \begin{align} \int_{0}^{\infty} \lambda^{\alpha + n - 1} \frac{\beta^{\alpha}}{\Gamma(\alpha)} \exp\left(-\lambda \left(\beta + \overline{X}\right)\right) \ d\lambda & = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} \lambda^{\alpha + n - 1} \exp\left(-\lambda \left(\beta \overline{X}\right)\right) \ d\lambda \\ & = \frac{\beta^{\alpha}}{\Gamma(\alpha) (\beta + \overline{X})^{\alpha + n}} \int_{0}^{\infty} \lambda^{\alpha + n - 1} \exp\left(-u\right) \ d u \\ & = \frac{\beta^{\alpha} \cdot \Gamma(\alpha + n)}{\Gamma(\alpha) (\beta + \overline{X})^{\alpha + n}}. \end{align} Is this correct? Furthermore, can I find $\widehat{\alpha}$, $\widehat{\beta}$ such that $p(X|\alpha,\beta) = \text{Gamma}(\widehat{\alpha}, \widehat{\beta})$?
A similar problem arises with the predictive distribution $$ p(x \mid X,\alpha,\beta) = \int p(x \mid \lambda) p(\lambda \mid X,\alpha,\beta) d\lambda, $$ which I calculated to be (similar substitution as above) $$ \frac{\beta^{\alpha} \cdot \Gamma(\alpha + n)}{\Gamma(\alpha) (\beta + x + \overline{X})^{\alpha + n}}. $$ Is this correct? What are my $\widehat{\alpha}$ and $\widehat{\beta}$ now?