Computing the Unsolvable: Is it Possible?

[LS1088]

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

 

[UltrafastPED]

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]

why is x^x equivalent to e^[x ln(x)]?

 

[UltrafastPED]

Are you familiar with change of base for logarithms?

Just apply it "backwards".

 

[D H]

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]

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]

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]

Let's try anyway.

Well, we know:

e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!} then should not:

e^{x^x}=\sum_{n=0}^{\infty}\frac{(x^x)^n}{n!}=\sum_{n=0}^{\infty}\frac{x^{nx}}{n!}

Now x^{nx}=e^{nx\log(x)}

So that

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!}

and surprisingly,

\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}

Let that just be g(n,x)

Then a possible antiderivative for this integral is:

\int e^{x^x}dx=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{\infty} \frac{n^k g(n,x)}{k!}

Won't that work?

 

[UltrafastPED]

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]

Let's try anyway.

Well, we know:

e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!} then should not:

e^{x^x}=\sum_{n=0}^{\infty}\frac{(x^x)^n}{n!}=\sum_{n=0}^{\infty}\frac{x^{nx}}{n!}

...

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]

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 \displaystyle\sum_{k=-p}^{\infty} c_n (z^{1/d})^n 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 e^{x^x} for x=3/2 accurate to 100 decimal places and compute \sum_{n=0}^{N} \frac{x^{nx}}{n!} 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.

<br /> \left(<br /> \begin{array}{c}<br /> 3.4413 \\<br /> 1.7538 \\<br /> 0.720418 \\<br /> 0.245808 \\<br /> 0.0714256 \\<br /> 0.0180321 \\<br /> 0.00401918 \\<br /> 0.000801266 \\<br /> 0.000144412 \\<br /> 0.00002374 \\<br /> \text{3.5864952178824044$\grave{ }$*${}^{\wedge}$-6} \\<br /> \text{5.011362544981476$\grave{ }$*${}^{\wedge}$-7} \\<br /> \text{6.512345832811623$\grave{ }$*${}^{\wedge}$-8} \\<br /> \text{7.90869733129278$\grave{ }$*${}^{\wedge}$-9} \\<br /> \text{9.013488214172771$\grave{ }$*${}^{\wedge}$-10} \\<br /> \text{9.676624489944749$\grave{ }$*${}^{\wedge}$-11} \\<br /> \text{9.818446293455492$\grave{ }$*${}^{\wedge}$-12} \\<br /> \text{9.443737583323711$\grave{ }$*${}^{\wedge}$-13} \\<br /> \text{8.63362720890291$\grave{ }$*${}^{\wedge}$-14} \\<br /> \text{7.520496283659276$\grave{ }$*${}^{\wedge}$-15} \\<br /> \text{6.255521977752776$\grave{ }$*${}^{\wedge}$-16} \\<br /> \text{4.9787601794491584$\grave{ }$*${}^{\wedge}$-17} \\<br /> \text{3.798597269079535$\grave{ }$*${}^{\wedge}$-18} \\<br /> \text{2.7829742952309837$\grave{ }$*${}^{\wedge}$-19} \\<br /> \text{1.960927906765562$\grave{ }$*${}^{\wedge}$-20} \\<br /> \text{1.3307991314972$\grave{ }$*${}^{\wedge}$-21} \\<br /> \text{8.71061004614173$\grave{ }$*${}^{\wedge}$-23} \\<br /> \text{5.5057436035672$\grave{ }$*${}^{\wedge}$-24} \\<br /> \text{3.3645298811674956$\grave{ }$*${}^{\wedge}$-25} \\<br /> \text{1.9899879616469038$\grave{ }$*${}^{\wedge}$-26} \\<br /> \text{1.1403572216891765$\grave{ }$*${}^{\wedge}$-27} \\<br /> \text{6.337461968469386$\grave{ }$*${}^{\wedge}$-29} \\<br /> \text{3.418759758195726$\grave{ }$*${}^{\wedge}$-30} \\<br /> \text{1.7917305430523684$\grave{ }$*${}^{\wedge}$-31} \\<br /> \text{9.130174261597168$\grave{ }$*${}^{\wedge}$-33} \\<br /> \end{array}<br /> \right)

Second Test:

Plot the real or imaginary principle sheet of the function and the series expression for 35 terms in the annular disc 0.1&lt;r&lt;2. 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