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.

  • 9
    $\begingroup$ It is definitely a thing. $\endgroup$ – copper.hat Aug 21 '12 at 15:34
  • 2
    $\begingroup$ What are you asking exactly? $\endgroup$ – gt6989b Aug 21 '12 at 15:39
  • 1
    $\begingroup$ It's not polite to make someone do something if they do not want to do it. $\endgroup$ – Graphth Aug 21 '12 at 16:33
  • $\begingroup$ Hint: $\int e^{ax}x^b(1-x)^c~dx=\int_0^xx^b(1-x)^ce^{ax}~dx+C=\int_0^xt^b(1-t)^ce^{at}~dt+C=\int_0^1(xt)^b(1-xt)^ce^{axt}~d(xt)+C=x^{b+1}\int_0^1t^b(1-xt)^ce^{axt}~dt+C$ $\endgroup$ – 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 $\int e^{ax}x^n dx=\frac {x^n e^{ax}}a -\frac na \int x^{n-1}e^{ax}dx$

  • 2
    $\begingroup$ 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... $\endgroup$ – Steven Stadnicki Aug 21 '12 at 16:26
  • $\begingroup$ You could try a binomial series, but that would only work if $x$ was sufficiently small. $\endgroup$ – Calvin McPhail-Snyder Aug 21 '12 at 17:16

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





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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.