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∫ e^(x^x) dx

by LS1088
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LS1088
#1
Oct19-13, 10:58 AM
P: 3
Just curious. Is it possible to compute this? if yes then how?
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UltrafastPED
#2
Oct19-13, 11:32 AM
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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?
Superposed_Cat
#3
Oct19-13, 12:46 PM
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why is x^x equivalent to e^[x ln(x)]?

UltrafastPED
#4
Oct19-13, 01:31 PM
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PF Gold
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∫ e^(x^x) dx

Are you familiar with change of base for logarithms?

Just apply it "backwards".
D H
#5
Oct19-13, 01:58 PM
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Quote Quote by UltrafastPED View Post
Certainly ln(u) = x ln(x) would simplify things a bit ... your integrand is now e^u!
I see three problems here:
  1. That transformation is not a one-to-one onto mapping unless x is restricted to [1/e,∞).

  2. It might simplify the integrand, but it makes an absolute mess of dx.
    That the transformation is not one-to-one onto makes it rather tough to deal with dx. Even if x is restricted to [1/e,∞), I get ##dx = du\,/\,(\operatorname W(\ln(u))+1)##, where W is the (non-elementary) Lambert W function.

  3. It still isn't integrable in the elementary functions.
UltrafastPED
#6
Oct19-13, 02:24 PM
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Quote Quote by D H View Post
I see three problems here:
  1. That transformation is not a one-to-one onto mapping unless x is restricted to [1/e,∞).

  2. It might simplify the integrand, but it makes an absolute mess of dx.
    That the transformation is not one-to-one onto makes it rather tough to deal with dx. Even if x is restricted to [1/e,∞), I get ##dx = du\,/\,(\operatorname W(\ln(u))+1)##, where W is the (non-elementary) Lambert W function.

  3. It still isn't integrable in the elementary functions.
Well, how would you attack this integral? There was never a guarantee that it could be done in terms of elementary functions, nor was the integration range specified by the OP.

I simply showed where the problem was easiest to attack, IMHO, and outlined some follow-on steps which would be required. Clearly other transforms would have to be tried - or somebody needs to get really clever!
D H
#7
Oct19-13, 03:25 PM
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P: 15,067
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.
jackmell
#8
Oct19-13, 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])^{-1-k}[/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?
UltrafastPED
#9
Oct19-13, 07:10 PM
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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!
D H
#10
Oct19-13, 08:41 PM
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Quote Quote by jackmell View Post
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]

...

Won't that work?
Nope. x^x has a branch point at x=0. Your series is about x=0. It's radius of convergence is zero.
jackmell
#11
Oct20-13, 06:19 AM
P: 1,666
Quote Quote by D H View Post
Nope. x^x has a branch point at x=0. Your series is about x=0. It's radius of convergence is zero.
Hi DH,

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 branch-point 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 multi-valued functions. However, I cannot initially prove it converges. May I instead present empirical evidence that suggest it may indeed have a non-zero 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 patch-work 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:

In[68]:=
NIntegrate[Exp[x^x], {x, 0.1, 2}]
NIntegrate[mye[x], {x, 0.1, 2}]

Out[68]=
13.451772502215917

Out[69]=
13.451772502215917
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