# Can $\int e^{-|x|^a}\,dx$ where 0<a<1 be given an analytical solution? [closed]

As the title suggests. Are there any solutions to solving this integral analytically or by using a power series? $$\int e^{-|x|^a}dx$$ where $$0?

Any help, suggestions, or leads given are greatly appreciated. Also are there any functions that you could suggest which could approximate this integral such that $$f(x) \leq \int e^{-|x|^a}\,dx \leq g(x)$$?

• I suppose you’re integrating over the whole real line. By symmetry, this is twice the integral over $(0,\infty)$. Now make the change of variables $t=x^a$. It seems like you should be able to reduce this to the integral definition of the Gamma function. Off the top of my head, $2\cdot \frac{1}{a}\Gamma(1/a)$ or something similar. Commented Aug 1 at 8:28
• It's possible when $1/a$ is an integer. Commented Aug 1 at 8:31
• Ohhh it looks like you want the indefinite integral (I wrote the above because you said you wanted to approximate the integral so I assumed you meant a definite one). But in general you won’t have a nice formula for this; it will involve the incomplete Gamma function. Commented Aug 1 at 8:37
• I also think the gamma functions the way, but can't quite find the solution. Commented Aug 1 at 8:37
• Ok then just apply what I said in the first comment. Commented Aug 1 at 8:39

Let $$x^a=t$$, therefore, we have: $$x=t^{\frac1a}$$ $$dx=\frac1a t^{\frac{1-a}{a}}dt$$
Thus the integral leads to: $$\int_0^{x(t)} \frac 1a e^{-t}t^{\frac{1-a}{a}}dt$$ For an easier approach, consider the exponent to be $$m$$, then the integral takes the form of: $$\int_0^{x(t)} (m+1) e^{-t}t^mdt$$ $$=(m+1)\int_0^{x(t)} e^{-t}t^mdt$$ Since the definite limits are not from $$0$$ to $$\infty$$, we cannot use the gamma function $$\Gamma$$ here. Moreover, we shall go through reduction method to find an analytical solution.
Just consider the integral here: $$I_m=\int e^{-t} t^m dt$$ We can go through by integration by parts and obtain: $$I_m=e^{-t} t^m-\int e^{-t}mt^{m-1}dt=e^{-t}t^m-mI_{m-1}$$ And this recursive process goes on, until we find and reach $$I_0$$ which is: $$I_0=\int e^{-x}dx=e^{-x}$$ This combined results gives a general formula of: $$I_m=e^{-x}\left(x^m-mx^{m-1}+\frac{m(m-1)}{2!}x^{m-2}-...(-1)^m\right)$$ $$I_m=e^{-x}\left(\sum_{k=0}^m\frac{(-1)^km!}{(m-k)!k!}x^{m-k}\right)$$ Wrapping up, we have our total integral to be: $$(m+1)\int_0^{x(t)} e^{-t}t^mdt=(m+1)\left[e^{-x}\left(\sum_{k=0}^m\frac{(-1)^km!}{(m-k)!k!}x^{m-k}\right)\right]_0^{x(t)}$$ You better set the rest by yours. And hopefully, get a nicer solution.
• Seems like it only works when $m$ is an integer. Commented Aug 2 at 1:41