- #1

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

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- Thread starter LS1088
- Start date

- #1

- 3

- 1

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

- #2

UltrafastPED

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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?

- #3

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

- #4

UltrafastPED

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Are you familiar with change of base for logarithms?

Just apply it "backwards".

Just apply it "backwards".

- #5

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I see three problems here:Certainly ln(u) = x ln(x) would simplify things a bit ... your integrand is now e^u!

- That transformation is not a one-to-one onto mapping unless x is restricted to [1/e,∞).

- 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.

- It still isn't integrable in the elementary functions.

- #6

UltrafastPED

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I see three problems here:

- That transformation is not a one-to-one onto mapping unless x is restricted to [1/e,∞).

- 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.

- 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!

- #7

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- #8

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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?

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?

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- #9

UltrafastPED

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Great job!

- #10

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Nope. x^x has a branch point at x=0. Your series is about x=0. It's radius of convergence is zero.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?

- #11

- 1,796

- 53

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.

[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]

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.

Code:

```
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
```

Last edited:

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