5
$\begingroup$

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

Here's the integrand graph:

enter image description here

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

$\endgroup$
16
  • 3
    $\begingroup$ That depends on your notion of closed form. Is $$\sum_{n\geq 1}\frac{1}{n^n}$$ considered as a closed form? $\endgroup$ – Jack D'Aurizio Jul 13 '17 at 21:52
  • 2
    $\begingroup$ @JackD'Aurizio No, that is most certainly not closed-form. en.wikipedia.org/wiki/Closed-form_expression $\endgroup$ – Franklin Pezzuti Dyer Jul 13 '17 at 21:53
  • 2
    $\begingroup$ You could consider it a constant. Otherwise $\pi$ would not be closed form as all methods to calculate it are non-finite series expansions. $\endgroup$ – mathreadler Jul 13 '17 at 21:55
  • 1
    $\begingroup$ Jack's first answer is probably the nearest you will get ... see en.wikipedia.org/wiki/Sophomore%27s_dream $\endgroup$ – Donald Splutterwit Jul 13 '17 at 21:58
  • 1
    $\begingroup$ @Donald observe that this is a different case because the range of integration is different. $\endgroup$ – Masacroso Jul 13 '17 at 22:00
8
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Jul 13 '17 at 22:13
  • $\begingroup$ Strange that it's so close to 2. $\endgroup$ – eyeballfrog Jul 13 '17 at 22:26
  • $\begingroup$ 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. $\endgroup$ – Michael L. Jul 13 '17 at 22:31
4
$\begingroup$

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 $0.08711$

$\endgroup$
0
$\begingroup$

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

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.