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
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!)
:P /teasing
– apnorton
Aug 12 '13 at 14:53
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 List of integrals)
$=-\dfrac{1}{a}~_1\Psi_0\left[\begin{matrix}(1,c)\\-\end{matrix};\dfrac{b}{(-a)^c}\right]$ (according to Fox Wright function)
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 List of integrals)
$=\dfrac{1}{(-b)^\frac{1}{c}c}~_1\Psi_0\left[\begin{matrix}\left(\dfrac{1}{c},\dfrac{1}{c}\right)\\-\end{matrix};\dfrac{a}{(-b)^\frac{1}{c}}\right]$ (according to Fox Wright function)
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
As Fly_by_Night says, you're not going to get a simple, catch-all closed form for this integral. In fact, plenty of simple integrals like $\int e^{-x^2}\mathrm{d}x$ have no closed form in terms of elementary functions. And that's one of the nice integrals of this type; it even has its own name. Plenty of other integrals with no closed form are unlikely to have any neat representation, so it would be unwise to assume our one does too. Regardless, we can turn it into a series so it's slightly easier to work with. We start with the Maclaurin series for $e^x$.
$$\begin{aligned}\int_0^\infty e^{ax+bx^c}\,\mathrm{d}x &=\int_0^\infty \sum_{n=0}^\infty\frac{\left(ax+bx^c\right)^n}{n!}\,\mathrm{d}x\\ &\quad\text{Binomial expansion}\\ &=\int_0^\infty\sum_{n=0}^\infty\left[\frac{1}{n!}\sum_{k=0}^n {n\choose k}a^{n-k}b^kx^{n+k(c-1)}\right]\,\mathrm{d}x\\ &\quad\text{Swapping integral and sum operators *}\\ &=\sum_{n=0}^\infty\sum_{k=0}^n \left[\frac{a^{n-k}b^k}{n!}{n\choose k}\int_0^\infty x^{n+k(c-1)}\,\mathrm{d}x\right]\\ &\quad\text{Power rule for integrals and using }{n\choose k}=\frac{n!}{(n-k)!k!}\\ &=\sum_{n=0}^\infty\sum_{k=0}^n \frac{a^{n-k}b^kx^{n+k(c-1)+1}}{(n-k)!k!\big(n+k(c-1)+1\big)}\bigg|_{x=0}^{x=\infty}\\ &\quad\text{Evaluating limits}\\ &=\lim_{x\to\infty}\sum_{n=0}^\infty\sum_{k=0}^n \frac{a^{n-k}b^k\left(x^{n+k(c-1)+1}-1\right)}{(n-k)!k!\big(n+k(c-1)+1\big)}\\ \end{aligned}$$
Now we've got a nice-big-ugly double sum. I'll not justify swapping of the integral and sum operators in *. Please leave a comment if you think it's wrong for me to do so.
I'll try to generate a graph of $S(a,b,c)$ when I edit this answer.
