Integral x^-x.

1. Mar 24, 2008

gamesguru

I'm just curious to find how what this integral is, but more importantly how it's computed. I've looked at the method that the sophomore's dream integrals are computed with and they do not seem to work here, I get to a step where I'm evaluating a sum from x=0 to x=$$\infty$$.
Here's the integral, and I've tried it with CAS and it oddly appears to be ~1.995456≈2:
$$\int^\infty_0{x}^{-x}dx$$

Last edited: Mar 24, 2008
2. Mar 24, 2008

sutupidmath

well, if you are trying to exactly compute it, i do not think it has a closed form. It can only be approximatet.

3. Mar 24, 2008

gamesguru

I should have said this above, but I'm fairly confident that an answer that can be expressed with purely elementary functions and constants does not exist. However, I'm not yet certain whether or not it can be represented with a power series, just as the sophomore's dream integrals can be.

4. Mar 25, 2008

gamesguru

Well, I found the answer to be the following limit:
$$\int^\infty_0{x}^{-x}dx=\lim_{a\rightarrow\infty}a\sum_{n=0}^\infty (-a)^n \sum_{k=0}^n (-1)^k \frac{(\log a)^{n-k}}{(n-k)!(n+1)^{k+1}}\approx 1.99545595\approx2$$
When I plug this in for large $a$ in CAS, I get weird behavior, but for $a$ approximately 5 and summing from $n=0$ to $n\approx 100$ it gives accurate and fairly precise answers.

5. Mar 25, 2008

Pere Callahan

Did you try the residue theorem ...an indefinte exponential integral on the positive halfline ...