I asked my Maths teacher recently how would you integrate the following, $$\int {x^x}^2 \, \mathrm{d}x$$ and she wasn't quite sure, I read you need to use as $x \to \infty$ but this was only briefly encountered, could someone please explain how you would go about doing this?
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The indefinite integral $$\int x^{x^2} \, \mathrm{d}x$$ has no closed form in terms of elementary functions.
You can, however evaluate the definite integral, for some limits. For example, we have $$\int_0^1 x^{x^2} \, \mathrm{d}x \approx 0.896489$$
The derivation of this can be seen in Jack's answer!
Zain Patel
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1Of course, that makes sense! Thanks a lot! – Lo-urc Jul 06 '15 at 16:58
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@Zain Patel: how do you know that $x^{x^2}$ has no closed form in terms of elementary functions? I agree that $e^{x^2}$ has none, and that $x^{x^2}$ most probably has none. – Damian Reding Jul 06 '15 at 17:13
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@Mathgemini It's a consequence of Lioville's theorem, see here for an excellent explanation: http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – Zain Patel Jul 06 '15 at 17:15
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I wrote two articles on Liouville's theorem and as far as I know it only enables us to decide the existence of elementary antiderivatives for functions of the form $fe^g$ with $f, g$ rational. And $x^{x^2}$ is not of this form... – Damian Reding Jul 06 '15 at 17:19
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Right. I skipped over that bit. I don't know enough advanced mathematics to prove that this has no elementary anti-derivative, do you have any ideas? :-) – Zain Patel Jul 06 '15 at 17:27
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2I don't know much about this beyond Liouville's theorem either, but I'm sure there is some advanced proof technique via Differential Galois Theory. Just thought you might perhaps know a simple reason why $x^{x^2}$ has no closed form integral ;) – Damian Reding Jul 06 '15 at 17:33
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$$\begin{eqnarray*}\int_{0}^{1}x^{x^2}\,dx &=& \int_{0}^{1}\exp\left(x^2 \log x\right)\,dx = \sum_{n\geq 0}\frac{1}{n!}\int_{0}^{1}x^{2n}\log^n x\,dx\\&=&\sum_{n\geq 0}\frac{1}{n!}\cdot\frac{n!(-1)^n}{(2n+1)^{n+1}}=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{n+1}}\tag{1}\end{eqnarray*} $$ and the last series converges extremely fast. It is just a variation on the classical sophomore's dream.
Jack D'Aurizio
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$\qquad$ Even the simpler-looking $\displaystyle\int e^{x^2}~dx$ and $\displaystyle\int x^x~dx$ do not possess a closed form expression, much less the integral you mentioned. See Liouville's theorem and the Risch algorithm for more information.
Lucian
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