Continous mixture distribution help? I've been staring at this example for hours now and I can't seem to understand it.
This is an example from Marcel Finan's C book.

The distribution of $X|\Lambda$ is exponential with parameter $\dfrac{1}{\Lambda}$. The distribution of $\Lambda$ is Gamma with parameters $\alpha$ and $\theta$. Find $f_X(x)$

Solution: First I try to slice things up:

Defn: $X$ is a mixture distribution if its pdf is
  $$f_x(x) = \int_{-\infty}^{\infty} f_{X|\Lambda}(x,\lambda)f_{\Lambda}(\lambda)d\lambda$$
  where $\Lambda$ is a continous random variable.

Also

$\bullet \Lambda$~Gamma($\alpha,\theta$) if its pdf is $$f(\lambda) = \dfrac{\lambda^{\alpha-1}e^{\frac{\lambda}{\theta}}}{\Gamma(\alpha)\theta^\alpha}$$ for $\lambda\geq0$ and $0$ otherwise.

But Finan's solution kinda messes me up:

$$f_X(x) = \int_{0}^{\infty} \lambda e^{-\lambda x}\dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \lambda^{\alpha-1}e^{-\lambda\theta}d\lambda = \dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} \lambda^{\alpha} e^{-\lambda(x+\theta)}d\lambda = \dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \dfrac{\Gamma(\alpha+1)}{(x+\theta)^{\alpha+1}}$$ 

My questions are:
1.) What does $X|\Lambda$ having exponential distribution looks like?
2.) Also, how did $\theta^{\alpha}$ became the numerator on the first equation when in the Gamma distribution, its on the denominator? 
3.) Can you please make a clarification on how did Finan arrive on the first equation? I can't seem to make a distinguish on the conditional distribution and the marginal density of $\Lambda$ on the definition of Mixture distribution. 
4.) And finally, how is $$\int_{0}^{\infty} \lambda^{\alpha} e^{-\lambda(x+\theta)}d\lambda= \dfrac{\Gamma(\alpha+1)}{(x+\theta)^{\alpha+1}}$$
Is this by the Gamma function?
 A: *

*$f_{X|\Lambda}(x,\lambda) = \lambda e^{\lambda x}$. Here we use the rate parameter to write the p.d.f. of exponential distribution. We call $\lambda$ rate because $E[X|\Lambda] = \lambda^{-1}$ gets larger when $\lambda$ gets smaller. You may also write the p.d.f. by the scale parameter $\beta = \lambda^{-1}$. You can check this on Wikipedia.

*Also, if you look at Wikipedia, you can find that there are two forms of the density function of Gamma distribution, one is given by shape-rate parameters and the other is given by shape-scale parameters. Since we use the rate parameter for the exponential distribution, we should use the shape-rate form for Gamma distribution, given by
$$f_{\Lambda}(\lambda) = {\frac {\theta ^{\alpha }}{\Gamma (\alpha )}}x^{\alpha -1}e^{-\theta x}.$$ 

*Now we are able to get the first equation. By simply plugging in the expressions into $f_X(x)$
$$f_X(x) = \int_{-\infty}^{\infty} f_{X|\Lambda}(x,\lambda)f_{\Lambda}(\lambda)d\lambda = \int_{0}^{\infty} \lambda e^{-\lambda x}\dfrac{\theta^{\alpha}}{\Gamma(\alpha)} \lambda^{\alpha-1}e^{-\lambda\theta}d\lambda$$

*Recall the definition of $\Gamma(t+1) = \int_{0}^{\infty} x^{t} e^{-x}d{x}$. Now let $s = \lambda(x+\theta)$, then
$$\int_{0}^{\infty} \lambda^{\alpha} e^{-\lambda(x+\theta)}d\lambda= \frac{1}{(x+\theta)^{\alpha+1}}\int_{0}^{\infty} s^{\alpha} e^{-s}ds = \dfrac{\Gamma(\alpha+1)}{(x+\theta)^{\alpha+1}}.$$
