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How can I calculate area under $\frac{1}{x^x}$ on any interval, I tried the Archimedes method, but I get

$$\frac1n\sum \frac 1{X_n^{X_n} }$$

and that's very complex to calculate because of the roots, is there an easier method to calculate this?

Тyma Gaidash
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michal Mravec
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    $1/x^x$ does not have an antiderivative that can be expressed in elementary terms. Also, you should probably restrict to $x > 0$ to avoid complex numbers and to ensure that your function is defined. – Eric Towers Jan 13 '21 at 17:13
  • I don’t think there is a closed form for the area under this curve over an arbitrary interval. I am not sure a closed form exists for any interval (though $(0,\infty)$ may be an exception). – Clayton Jan 13 '21 at 17:13
  • I doubt you will do better than approximations. For example $\int\limits_0^1 \frac{1}{x^x} , dx \approx 1.291286$ and $\int\limits_0^\infty \frac{1}{x^x} , dx \approx 1.995456$ – Henry Jan 13 '21 at 17:13
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    https://math.stackexchange.com/questions/2585634/prove-that-int-0-infty-frac1xx-dx2 – Arash Jan 14 '21 at 00:56

1 Answers1

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Sorry for the short answer, but here is the link with my work. It needs a better form though. You can also use the definition of the Riemann Sum to find another value.

(My Answer)

Using the definition of the Riemann Sum, we can rewrite this as:

$$\mathrm{\int_{\Bbb R^+}x^{-x}dx=\lim_{b,n\to \infty}\frac bn \sum_{k=0}^n\left(\frac{bk}{n}\right)^{-\left(\frac{bk}{n}\right)},n\gg b}$$

Here is proof of this result: Graph

Тyma Gaidash
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