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∫ e^(x^x) dxby LS1088
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#1
Oct1913, 10:58 AM

P: 3

Just curious. Is it possible to compute this? if yes then how?



#2
Oct1913, 11:32 AM

Sci Advisor
Thanks
PF Gold
P: 1,911

First rewrite x^x as e^[x ln(x)], which puts all of your eggs ... I mean xs ... in one basket.
Next would be to try different substitutions for x which will simplify this ... then probably integrate by parts. Certainly ln(u) = x ln(x) would simplify things a bit ... your integrand is now e^u! What would be your next step? 


#4
Oct1913, 01:31 PM

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Thanks
PF Gold
P: 1,911

∫ e^(x^x) dx
Are you familiar with change of base for logarithms?
Just apply it "backwards". 


#5
Oct1913, 01:58 PM

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#6
Oct1913, 02:24 PM

Sci Advisor
Thanks
PF Gold
P: 1,911

I simply showed where the problem was easiest to attack, IMHO, and outlined some followon steps which would be required. Clearly other transforms would have to be tried  or somebody needs to get really clever! 


#7
Oct1913, 03:25 PM

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P: 15,201

The only way to attack this integral is to use numerical integration. It cannot be expressed in terms of the elementary functions, or in terms of any special function that I know of.



#8
Oct1913, 06:07 PM

P: 1,666

Let's try anyway.
Well, we know: [tex]e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!}[/tex] then should not: [tex]e^{x^x}=\sum_{n=0}^{\infty}\frac{(x^x)^n}{n!}=\sum_{n=0}^{\infty}\frac{ x^{nx}}{n!}[/tex] Now [itex]x^{nx}=e^{nx\log(x)}[/itex] So that [tex]e^{nx\log(x)}=\sum_{k=0}^{\infty} \frac{(nx\log(x))^k}{k!}=\sum_{k=0}^{\infty} \frac{n^k x^k \log(x)^k}{k!}[/tex] and surprisingly, [tex]\int x^k \log(x)^k dx=\text{Gamma}[1+k,(1+k) \text{Log}[x]] \text{Log}[x]^{1+k} ((1+k) \text{Log}[x])^{1k}[/tex] Let that just be [itex]g(n,x)[/itex] Then a possible antiderivative for this integral is: [tex]\int e^{x^x}dx=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{\infty} \frac{n^k g(n,x)}{k!}[/tex] Won't that work? 


#9
Oct1913, 07:10 PM

Sci Advisor
Thanks
PF Gold
P: 1,911

Looks good to me ... series expansion is good for otherwise unsolvable integrals ... then you can evaluate term by term for a numerical estimate.
Great job! 


#10
Oct1913, 08:41 PM

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#11
Oct2013, 06:19 AM

P: 1,666

I don't enjoy disagreeing with you but I believe this series does converge. It's a logarithmic series and is perhaps convergent in a punctured disc surrounding the branchpoint similar to Puiseux series for algebraic functions which have the form [itex]\displaystyle\sum_{k=p}^{\infty} c_n (z^{1/d})^n[/itex] which are convergent series representation for multivalued functions. However, I cannot initially prove it converges. May I instead present empirical evidence that suggest it may indeed have a nonzero radius of convergence? Of course numerical data is not a proof. Could you prove it does not converge? The numerical data below suggest otherwise however. First Test: Compute [itex]e^{x^x}[/itex] for x=3/2 accurate to 100 decimal places and compute [itex]\sum_{n=0}^{N} \frac{x^{nx}}{n!}[/itex] for N in the range of 1 to 35 likewise accurate to 100 decimal places. Notice I use an exact value of 3/2 so Mathematica can compute these values to arbitrary precision. The difference between the two follow. The results certainly look like the series is converging to the actual answer. [tex] \left( \begin{array}{c} 3.4413 \\ 1.7538 \\ 0.720418 \\ 0.245808 \\ 0.0714256 \\ 0.0180321 \\ 0.00401918 \\ 0.000801266 \\ 0.000144412 \\ 0.00002374 \\ \text{3.5864952178824044$\grave{ }$*${}^{\wedge}$6} \\ \text{5.011362544981476$\grave{ }$*${}^{\wedge}$7} \\ \text{6.512345832811623$\grave{ }$*${}^{\wedge}$8} \\ \text{7.90869733129278$\grave{ }$*${}^{\wedge}$9} \\ \text{9.013488214172771$\grave{ }$*${}^{\wedge}$10} \\ \text{9.676624489944749$\grave{ }$*${}^{\wedge}$11} \\ \text{9.818446293455492$\grave{ }$*${}^{\wedge}$12} \\ \text{9.443737583323711$\grave{ }$*${}^{\wedge}$13} \\ \text{8.63362720890291$\grave{ }$*${}^{\wedge}$14} \\ \text{7.520496283659276$\grave{ }$*${}^{\wedge}$15} \\ \text{6.255521977752776$\grave{ }$*${}^{\wedge}$16} \\ \text{4.9787601794491584$\grave{ }$*${}^{\wedge}$17} \\ \text{3.798597269079535$\grave{ }$*${}^{\wedge}$18} \\ \text{2.7829742952309837$\grave{ }$*${}^{\wedge}$19} \\ \text{1.960927906765562$\grave{ }$*${}^{\wedge}$20} \\ \text{1.3307991314972$\grave{ }$*${}^{\wedge}$21} \\ \text{8.71061004614173$\grave{ }$*${}^{\wedge}$23} \\ \text{5.5057436035672$\grave{ }$*${}^{\wedge}$24} \\ \text{3.3645298811674956$\grave{ }$*${}^{\wedge}$25} \\ \text{1.9899879616469038$\grave{ }$*${}^{\wedge}$26} \\ \text{1.1403572216891765$\grave{ }$*${}^{\wedge}$27} \\ \text{6.337461968469386$\grave{ }$*${}^{\wedge}$29} \\ \text{3.418759758195726$\grave{ }$*${}^{\wedge}$30} \\ \text{1.7917305430523684$\grave{ }$*${}^{\wedge}$31} \\ \text{9.130174261597168$\grave{ }$*${}^{\wedge}$33} \\ \end{array} \right)[/tex] Second Test: Plot the real or imaginary principle sheet of the function and the series expression for 35 terms in the annular disc [itex]0.1<r<2[/itex]. Color one red, the other blue. Superimpose them onto one another. The results are below and so identical that I cannot distinguish between the two plots on top of one another. Usually when this test is run and the results differ, the plot will be a patchwork of red and blue. In this case, one plot completely covers the other plot. Third Test: Numerically integrate both expressions from 0.1 to 2. In the Mathematica code below, the function mye[x] is the first 35 terms of the series:



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