# Evaluating $\int e^{ax} x^b (1-x)^c \mathrm{dx}$

Edit: clarify question

The integrand looks kind of like a gamma density function, and kind of like a beta density function, so maybe it has a somewhat nice solution?

$$\int e^{ax} x^b (1-x)^c \mathrm{dx}$$

Wolfram alpha does not want to do it.

• It is definitely a thing. – copper.hat Aug 21 '12 at 15:34
• What are you asking exactly? – gt6989b Aug 21 '12 at 15:39
• It's not polite to make someone do something if they do not want to do it. – Graphth Aug 21 '12 at 16:33
• Hint: $\int e^{ax}x^b(1-x)^c~dx=\int_0^xx^b(1-x)^ce^{ax}~dx+C=\int_0^xt^b(1-t)^ce^{at}~dt+C=\int_0^1(xt)^b(1-xt)^ce^{axt}~d(xt)+C=x^{b+1}\int_0^1t^b(1-xt)^ce^{axt}~dt+C$ – Harry Peter May 27 '14 at 13:55

You can expand out the $(1-x)^c$ to get terms of the form $\int e^{ax}x^n dx$. Wolfram Alpha then gives a solution in terms of the incomplete Gamma function. This is a form that can be integrated by parts-set $dv=e^{ax}dx, u=x^n$ and step down the exponents, giving $\int e^{ax}x^n dx=\frac {x^n e^{ax}}a -\frac na \int x^{n-1}e^{ax}dx$

• This only works if $b$ and $c$ are both integral; I'm presuming (with admittedly no evidence either way) that the OP wasn't assuming integer values... – Steven Stadnicki Aug 21 '12 at 16:26
• You could try a binomial series, but that would only work if $x$ was sufficiently small. – Calvin McPhail-Snyder Aug 21 '12 at 17:16

$\int e^{ax}x^b(1-x)^c~dx$

$=\int_0^xx^b(1-x)^ce^{ax}~dx+C$

$=\int_0^xt^b(1-t)^ce^{at}~dt+C$

$=\int_0^1(xt)^b(1-xt)^ce^{axt}~d(xt)+C$

$=x^{b+1}\int_0^1t^b(1-xt)^ce^{axt}~dt+C$

$=\dfrac{x^{b+1}\Phi_1(b+1,-c,b+2;x,ax)}{b+1}+C$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)