Is $\int_0^\infty x^{a-1} (1-x)^{b-1} e^{t-cx} dx$ integrable?

Question

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.

Answer 1

$\int_0^\infty x^{a-1}(1-x)^{b-1}e^{t-cx}~dx$

$=e^t\int_0^\infty x^{a-1}(1-x)^{b-1}e^{-cx}~dx$

$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx+e^t\int_1^\infty x^{a-1}(1-x)^{b-1}e^{-cx}~dx$

$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx-(-1)^be^t\int_1^\infty x^{a-1}(x-1)^{b-1}e^{-cx}~dx$

$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx-(-1)^be^t\int_0^\infty(x+1)^{a-1}x^{b-1}e^{-c(x+1)}~d(x+1)$

$=e^t\int_0^1x^{a-1}(1-x)^{b-1}e^{-cx}~dx-(-1)^be^{t-c}\int_0^\infty x^{b-1}(x+1)^{a-1}e^{-cx}~dx$

$=\dfrac{e^t\Gamma(a)\Gamma(b)M(a,a+b,-c)}{\Gamma(a+b)}-(-1)^be^{t-c}\Gamma(b)U(b,a+b,c)$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Integral_representations)

Answer 2

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.

Answer 3

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.