# Integral $\int_{0}^{\infty}x^{-x}dx$

I'm trying to find a closed form for this integral:$$\int_{0}^{\infty}x^{-x}dx$$

Here's the integrand graph:

Clearly it is convergent. My attempt is to obtain a closed form for the area under the curve. Is this possible?

• That depends on your notion of closed form. Is $$\sum_{n\geq 1}\frac{1}{n^n}$$ considered as a closed form? – Jack D'Aurizio Jul 13 '17 at 21:52
• @JackD'Aurizio No, that is most certainly not closed-form. en.wikipedia.org/wiki/Closed-form_expression – Franklin Pezzuti Dyer Jul 13 '17 at 21:53
• You could consider it a constant. Otherwise $\pi$ would not be closed form as all methods to calculate it are non-finite series expansions. – mathreadler Jul 13 '17 at 21:55
• Jack's first answer is probably the nearest you will get ... see en.wikipedia.org/wiki/Sophomore%27s_dream – Donald Splutterwit Jul 13 '17 at 21:58
• @Donald observe that this is a different case because the range of integration is different. – Masacroso Jul 13 '17 at 22:00

It is highly unlikely that you will be able to find a closed-form expression for your integral. Two special variants of your integral, however, are rather infamous as the "Sophomore's Dream" integrals. They are the integrals $$\int_0^1 x^{x} dx$$ and $$\int_0^1 x^{-x} dx$$ And, as of yet, no closed-form has been obtained for either of them. However, your integral $$\int_0^\infty x^{-x} dx$$ Converges incredibly quickly, even more so than an exponential function, and so a very good approximation can be obtained rather quickly. Wolfram Alpha yields the approximation $$\int_0^\infty x^{-x} dx\approx 1.99545595750014...$$ And so $2$ should be a good enough approximation. Even the inverse symbolic calculator doesn't yield anything for the approximation given by WA, so I doubt that it has any kind of closed form using the elementary functions or any other known constants.

• One can approximate the remainder to a further extent by using the Euler-Maclaurin formula, which quickly relates the given integral to Sophomore's dream. – Simply Beautiful Art Jul 13 '17 at 22:13
• Strange that it's so close to 2. – eyeballfrog Jul 13 '17 at 22:26
• Not really that strange when you consider how many of these simple-looking integrals it's possible to construct by defining arbitrary operations like tetration. Some of them are bound to be close to integers. As best as I can tell, there's no real physical significance of $x^{-x}$ or of this integral. – Michael L. Jul 13 '17 at 22:31

We have the following approximation:

Let $S$ be one of the Sophomore's dreams:

$$S=\int_0^1x^{-x}~\mathrm dx=\sum_{n=1}^\infty n^{-n}$$

Then we have the integral $I$ in question,

$$I=\int_0^\infty x^{-x}~\mathrm dx=S+\int_1^\infty x^{-x}~\mathrm dx$$

The second integral has a quick trapezoidal sum approximation:

$$\int_1^\infty x^{-x}~\mathrm dx\approx-\frac12+\sum_{n=1}^\infty n^{-n}$$

Thus, we have

$$I\approx2S-\frac12=2.08257$$

The error is approximately $0.08711$

$\int\limits_0^\infty x^{-x}dx=\int\limits_0^1 x^{-x}dx+\int\limits_0^1 x^{-2+1/x}dx$

The second integral is unpleasant to develope into a series.

But if we can use the solution $\,z_0\,$ for $\,\int\limits_0^1 (x^{xz} - x^{-2+1/x})dx=0$

with $\,z=z_0\approx 1.45354007846425$ , then we get:

$$\displaystyle \int\limits_0^\infty x^{-x}dx=\sum\limits_{n=1}^\infty \frac{1+(-z_0)^{n-1}}{n^n}$$