# Math Help - "Super Integral" (Apparently)

1. ## "Super Integral" (Apparently)

Saw this one on the univ's notice board...

I don't know how hard this will be for you people, especially Krizalid, but I'm interested to see the answer.

$
\int\limits_1^{100} {x^x } dx
$

2. because there is no elementary function whose derivative is x^x.

I reckon someone should define one. That is why it is not easy by elementary means.

I did run it through the calculator and got a huge answer.

$1.784\times{10^{199}}$

Maple nor the Integrator will give an indefinite integral for this. They just spit back the same thing.

3. If you wanted to get a numerical answer, I would just try a Riemann sum or even Simpson's rule.

4. Originally Posted by galactus
because there is no elementary function whose derivative is x^x.

I reckon someone should define one. That is why it is not easy by elementary means.

I did run it through the calculator and got a huge answer.

$1.784\times{10^{199}}$

Maple nor the Integrator will give an indefinite integral for this. They just spit back the same thing.
Now how could I have forgotten you, Galactus? Why did I only think of Krizalid?

Well you did get an answer...

I'm interested to see if anyone else has some other method of solving it...

5. Just for fun, I generated a graph with Maple using 1 to 10. For 1..100, the y value was too large.

Look how huge it is just using 1 to 10, let alone 1 to 100.

Afterall, this is quite the exponential. 2^2, 3^3, ....., 100^100.

6. Originally Posted by galactus
Just for fun, I generated a graph with Maple using 1 to 10. For 1..100, the y value was too large.

Look how huge it is just using 1 to 10, let alone 1 to 100.

Afterall, this is quite the exponential. 2^2, 3^3, ....., 100^100.
I then assume this is a question for very advanced students? It's like challenging questions that they post on the notice board sometimes.

But I still feel there must be an easier way of solving it. You know when you just have a feeling that there is a better, more elegant method?

7. We could rewrite it as $x^{x}=e^{xln(x)}$, but I don't know what good it'll do.

The Big K may have a trick up his sleeve.

8. You could try and write the function out as a Taylor series. We have $f(x)=x^x$ and $f'(x) = x^x(\ln{x}+1)$. I don't have the time right now... anyone want to volunteer?

Problem is that there are no elementary functions whose derivative is $x^x$.

9. Originally Posted by galactus
The Big K may have a trick up his sleeve.
That's what I'm thinking...

10. There is an identity with $x^x$, but again it is non-elementary.

$\int_0^1 x^x ~ dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n^n}$.

11. Hacker got a typo, it should be $\sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} }}
{{(n + 1)^{n + 1} }}}$
or $\sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}
{{n^n }}}.$
This identity is called as "Sophomore's Dream." One can prove it easily with galactus' trick (post #7) after series expansion and integration.

----

As for the problem, I dunno, I think I would've thought in numerical methods to tackle it too.

12. One trouble with Taylor expansion is where to center it at... 1? 50? 100?

The derivatives of the function are rather beautiful...

$f(x)=x^x$
$f'(x)=x^x(1+\ln{x})$
$f''(x)=x^x(1+\ln{x}+\frac{1}{x})$
$f^{(3)}(x)=x^x(1+\ln{x}+\frac{1}{x}-\frac{1}{x^2})$
$f^{(4)}(x)=x^x(1+\ln{x}+\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3})$

I think you get the idea...

Let's choose $a=1$ for simplicity...

$f(1)=1$
$f'(1)=1$
$f''(1)=2$
$f^{(3)}(1)=1$
$f^{(4)}(1)=3$

$x^x = 1 + (x-1) + (x-1)^2 + \frac{(x-1)^3}{6} + \ldots$

This is obviously not enough of an expansion for $x^x$. More work needs to be done, but this is a start.

13. Originally Posted by Krizalid
Hacker got a typo, it should be $\sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} }}
{{(n + 1)^{n + 1} }}}$
or $\sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}
{{n^n }}}.$
This identity is called as "Sophomore's Dream." Mr F says: This was actually an expression applied by Borwein et al (Borwein, J.; Bailey, D.; and Girgensohn Experimentation in Mathematics: Computational Paths to Discovery. Wellesley, MA: A K Peters, 2004.) ....

One can prove it easily with galactus' trick (post #7) after series expansion and integration.

----

As for the problem, I dunno, I think I would've thought in numerical methods to tackle it too.
Actually it's $\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{{\color{red}n+1}}}}
{{n^n }}} = 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + ....$

Johann Bernoulli first calculated $\, \int_{0}^{1} x^x \, dx\,$ in a paper he published in 1697. The series converges extremely rapidly (an attribute noted by Bernoulli) .... it only takes a few terms to calculate the integral accurately to ten decimal places.

Krizalid is quite right in what he says regarding the proof:

$x^x = 1 + x \ln x + \frac{1}{2!} \, x^2 (\ln x)^2 + \frac{1}{3!} \, x^3 (\ln x)^3 + \frac{1}{4!} \, x^4 (\ln x)^4 + ......$

and each of these integrals is simple to do. Especially if you know the reduction formula

$\int x^m (\ln x)^n \, dx = \frac{1}{m+1} \, x^{m+1} (\ln x)^n - \frac{n}{m+1}\, \int x^m (\ln x)^{n-1} \, dx$

14. Originally Posted by mr fantastic
Actually it's $\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{{\color{red}n+1}}}}
{{n^n }}} = 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + ....$
Ahhh, time ago I don't see the identity.

$\int_0^1x^k\ln^kx\,dx$ can be computed via substitution $u=-\ln x,$ it yields an integral with gamma function-type.

15. Originally Posted by janvdl
Saw this one on the univ's notice board...

I don't know how hard this will be for you people, especially Krizalid, but I'm interested to see the answer.

$
\int \limits_0^{100} {x^x} \, dx
$
Regarding the direction this thread has taken to date: The technical infelicity is the integral limits .....

Perhaps $\int \limits_0^{100} {x^x } \, dx - \int\limits_0^{1} {x^x } \, dx$ .....

Or a cunning substitution that gives the integral limits and yields the answer ..... $u = \ln x$ doesn't seem to quite cut it ......

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