Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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
Trambak

share|improve this question
    
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

3 Answers 3

Case $1$: $c\leq1$

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

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

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{b^nx^{cn}e^{ax}}{n!}dx$

$=\sum\limits_{n=0}^\infty\dfrac{b^n\Gamma(cn+1)}{(-a)^{cn+1}n!}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#definite_integrals)

Case $2$: $c\geq1$

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

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

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

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{a^nx^{\frac{n+1}{c}-1}e^{bx}}{cn!}dx$

$=\sum\limits_{n=0}^\infty\dfrac{a^n\Gamma\left(\dfrac{n+1}{c}\right)}{(-b)^\frac{n+1}{c}~cn!}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#definite_integrals)

share|improve this answer

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

share|improve this answer

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!)

share|improve this answer
1  
Last sentence is a perfect example of "proof by lack of discovery by really smart people." :P /teasing –  anorton Aug 12 '13 at 14:53
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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