Vtyjkyuk

I have a following expression for expectation of some random variable $X$:

$E(X(t)|lambda)=frac1p(1-e^-lambda pt) $

We assume that $p$ is fixed and $lambda$ has a following mixing gamma distribution:

$f(lambda|r,alpha)=fracalpha^r lambda^r-1 e^-lambda alphaGamma (r), lambda>0$

I need to develop final formula for $E(X(t)$ - to do this I believe I need to calculate the following:

$E(X(t))=E(X(t)|lambda) f(lambda)=frac1p-frac1p int_0^infty e^-lambda pt fracalpha^r lambda^r-1 e^-lambda alphaGamma (r) dlambda $

Can someone please confirm if above is correct? I'm not sure if my reasoning is fine, only final solution without derivation has been presented in the paper, namely:
$E(X(t))=frac1p-fracalpha ^rp(alpha+pt)^r$

Thank you

beginalignE(X(t))&=colorredint_0^inftyE(X(t)|lambda) f(lambda), colorreddlambda\&=frac1p int_0^infty (1-e^-lambda pt) fracalpha^r lambda^r-1 e^-lambda alphaGamma (r) dlambda \
&=frac1p left( int_0^infty fracalpha^rlambda^r-1e^-lambda alphaGamma(r), dlambda - int_0^infty fracalpha^rlambda^r-1e^-lambda (alpha+pt)Gamma(r), dlambda right)\
&=frac1p left( 1 - fracalpha^r(alpha+pt)^rint_0^infty frac(alpha+pt)^rlambda^r-1e^-lambda (alpha+pt)Gamma(r), dlambda right)\
&=frac1p left( 1 - fracalpha^r(alpha+pt)^rright)endalign

where I have used the property that $int_0^infty fracbeta^rlambda^r-1e^-lambda betaGamma(r), dlambda =1$ since the function that is being integrated is a density function.