- #1
LS1088
- 3
- 1
Just curious. Is it possible to compute this? if yes then how?
I see three problems here:Certainly ln(u) = x ln(x) would simplify things a bit ... your integrand is now e^u!
I see three problems here:
- That transformation is not a one-to-one onto mapping unless x is restricted to [1/e,∞).
- It might simplify the integrand, but it makes an absolute mess of dx.
That the transformation is not one-to-one onto makes it rather tough to deal with dx. Even if x is restricted to [1/e,∞), I get ##dx = du\,/\,(\operatorname W(\ln(u))+1)##, where W is the (non-elementary) Lambert W function.
- It still isn't integrable in the elementary functions.
Nope. x^x has a branch point at x=0. Your series is about x=0. It's radius of convergence is zero.Let's try anyway.
Well, we know:
[tex]e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!}[/tex] then should not:
[tex]e^{x^x}=\sum_{n=0}^{\infty}\frac{(x^x)^n}{n!}=\sum_{n=0}^{\infty}\frac{x^{nx}}{n!}[/tex]
...
Won't that work?
Nope. x^x has a branch point at x=0. Your series is about x=0. It's radius of convergence is zero.
In[68]:=
NIntegrate[Exp[x^x], {x, 0.1, 2}]
NIntegrate[mye[x], {x, 0.1, 2}]
Out[68]=
13.451772502215917
Out[69]=
13.451772502215917