Is $\int_0^\infty x^{a-1} (1-x)^{b-1} e^{t-cx} dx$ integrable? I am trying to evaluate the  integral below. Is it even integrable? (Online integral solvers  e.g. WolframAlpha could not solve the indefinite or the definite integral.)
$$\int_0^\infty x^{a-1} (1-x)^{b-1} \, e^{t-cx} dx$$
where $a > 0$, $b > 0$, $c > 0$, and $t$ is any real number.
 A: Mathematica found an answer as:
$$\left.e^t \left(-(-1)^b c^{-a-b+1} \Gamma (a+b-1) \, _1F_1(1-b;-a-b+2;-c)+\frac{\Gamma (a) \Gamma (b) \, _1F_1(a;a+b;-c)}{\Gamma (a+b)}-\frac{(-1)^b \Gamma (b) \Gamma (-a-b+1) \, _1F_1(a;a+b;-c)}{\Gamma (1-a)}\right),\\\Re(b)>0\land \Re(c)>0\land \Re(a)>0\right]$$
Which doesn't look too "closed" to me but may be a starting point for others.
A: $\int_0^\infty x^{a-1}(1-x)^{b-1}e^{t-cx}~dx$
$=e^t\int_0^\infty x^{a-1}(1-x)^{b-1}e^{-cx}~dx$
$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx+e^t\int_1^\infty x^{a-1}(1-x)^{b-1}e^{-cx}~dx$
$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx-(-1)^be^t\int_1^\infty x^{a-1}(x-1)^{b-1}e^{-cx}~dx$
$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx-(-1)^be^t\int_0^\infty(x+1)^{a-1}x^{b-1}e^{-c(x+1)}~d(x+1)$
$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx-(-1)^be^{t-c}\int_0^\infty x^{b-1}(x+1)^{a-1}e^{-cx}~dx$
$=\dfrac{e^t\Gamma(a)\Gamma(b)M(a,a+b,-c)}{\Gamma(a+b)}-(-1)^be^{t-c}\Gamma(b)U(b,a+b,c)$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Integral_representations)
A: You've asked about 
$$\int_0^\infty x^{a-1} (1-x)^{b-1} e^{t-cx} \: dx$$
And if $b$ is an integer,
$$ e^t \int_0^\infty x^{a-1} \left (\sum_{k=0}^{b-1} (-1)^k {b-1 \choose k} x^k \right) e^{-cx} \: dx$$
by applying the binomial theorem to $(1-x)^{b-1}$.  After a bit of rearrangement this becomes
$$ e^t \sum_{k=0}^{b-1} (-1)^k {b-1 \choose k} \int_0^\infty  e^{-cx} x^{a+k-1} \: dx$$
and this last integral can be written in terms of the gamma function.  This gives a formula for your function as a sum of $b$ terms.
