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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 e^{ax+bx^c}~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

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I don't think there will be a general standard form since this highly depends on the value of $c$. – Patrick Li Sep 28 '12 at 21:18

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!)

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Last sentence is a perfect example of "proof by lack of discovery by really smart people." :P /teasing – apnorton Aug 12 '13 at 14:53
@anorton :o) The problem is because of the limited array of functions that we call elementary. Outside of trigonometric, exponential, logarithmic, and a jumble of all of these, there isn't too much left. Why should $\sin(\operatorname{e}^x)$ be allowable as a closed-form, when other functions aren't? Indeed, why not include $\Gamma$ in the set of allowable elementary functions, and then all of the problems go away, $\Gamma(z)$ would trivially be a closed form in itself. – Fly by Night Aug 12 '13 at 19:27

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

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Case $1$: $c\leq1$

Then $\int_0^\infty e^{ax+bx^c}~dx$

$=\int_0^\infty e^{ax}e^{bx^c}~dx$


$=\sum\limits_{n=0}^\infty\dfrac{b^n\Gamma(cn+1)}{(-a)^{cn+1}n!}$ (can be obtained from

$=-\dfrac{1}{a}~_1\Psi_0\left[\begin{matrix}(1,c)\\-\end{matrix};\dfrac{b}{(-a)^c}\right]$ (according to

Case $2$: $c\geq1$

Then $\int_0^\infty e^{ax+bx^c}~dx$

$=\int_0^\infty e^{ax^\frac{1}{c}}~e^{bx}~d\left(x^\frac{1}{c}\right)$

$=\dfrac{1}{c}\int_0^\infty x^{\frac{1}{c}-1}e^{ax^\frac{1}{c}}~e^{bx}~dx$


$=\sum\limits_{n=0}^\infty\dfrac{a^n\Gamma\left(\dfrac{n+1}{c}\right)}{(-b)^\frac{n+1}{c}~cn!}$ (can be obtained from

$=\dfrac{1}{(-b)^\frac{1}{c}c}~_1\Psi_0\left[\begin{matrix}\left(\dfrac{1}{c},\dfrac{1}{c}\right)\\-\end{matrix};\dfrac{a}{(-b)^\frac{1}{c}}\right]$ (according to

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