Definite Integral of $\exp(ax+bx^c)$

Question

Can you please provide some thoughts / ideas / help in computing this definite integral? Any help will be great...I am so stuck with this one.

$$\int_0^{\infty}\exp\left(ax+bx^c\right)\,dx,$$

where $a < 0$, $b < 0$ and $c > 0$.

It looks like this one might not have a clean analytical solution but is there any standard form that this reduces to?

Thanks a lot for your help
Trambak

Answer 1

In general, there will be no non-recursive expressions. For example, even the simplest case:

$$\int_0^{\infty} \!\!e^{-x^k} \, dx = \frac{1}{k}\Gamma\left(\frac{1}{k}\right) .$$

Here $\Gamma : \mathbb{C} \to \overline{\mathbb{C}}$ denotes Euler's Gamma function, defined by

$$\Gamma(z) := \int_0^{\infty} e^{-t} \, t^{z-1} \, dt \, . $$

Of course, there are some special values of $k$ which give closed form expressions, e.g. $k = 2$ gives $\sqrt{\pi}/2$, but in general you have no hope of finding a nice expression.

(If there were then it'd be in the calculus books by now!)

Answer 2

I tried the integration by parts bit. Here is how it looks:

say $$I = \int_0^{\infty}e^{ax+bx^c}dx$$ $$ = \left(\dfrac{e^{ax+bx^c}}{a}\right)_{0}^{\infty} - \int_0^{\infty}\dfrac{bc}{a}x^{c-1}e^{ax+bx^c}dx$$ $$ = -\dfrac{1}{a} -\dfrac{1}{a}\int_0^{\infty}(a+bcx^{c-1})e^{ax+bx^c}dx + I$$

For $a < 0$, $b < 0$ and $c > 1$, this thing results in the trivial identity $0=0$. For $c=1$, it is easily computable. Am I completely off here?

Thanks Trambak