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.
$\displaystyle
\int\limits_1^{100} {x^x } dx
$
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.
$\displaystyle 1.784\times{10^{199}}$
Maple nor the Integrator will give an indefinite integral for this. They just spit back the same thing.
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?
You could try and write the function out as a Taylor series. We have $\displaystyle f(x)=x^x$ and $\displaystyle f'(x) = x^x(\ln{x}+1)$. I don't have the time right now... anyone want to volunteer?
Problem is that there are no elementary functions whose derivative is $\displaystyle x^x$.
Hacker got a typo, it should be $\displaystyle \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} }}
{{(n + 1)^{n + 1} }}}$ or $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}
{{n^n }}}.$ 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.
One trouble with Taylor expansion is where to center it at... 1? 50? 100?
The derivatives of the function are rather beautiful...
$\displaystyle f(x)=x^x$
$\displaystyle f'(x)=x^x(1+\ln{x})$
$\displaystyle f''(x)=x^x(1+\ln{x}+\frac{1}{x})$
$\displaystyle f^{(3)}(x)=x^x(1+\ln{x}+\frac{1}{x}-\frac{1}{x^2})$
$\displaystyle f^{(4)}(x)=x^x(1+\ln{x}+\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3})$
I think you get the idea...
Let's choose $\displaystyle a=1$ for simplicity...
$\displaystyle f(1)=1$
$\displaystyle f'(1)=1$
$\displaystyle f''(1)=2$
$\displaystyle f^{(3)}(1)=1$
$\displaystyle f^{(4)}(1)=3$
$\displaystyle x^x = 1 + (x-1) + (x-1)^2 + \frac{(x-1)^3}{6} + \ldots$
This is obviously not enough of an expansion for $\displaystyle x^x$. More work needs to be done, but this is a start.
Actually it's $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{( - 1)^{{\color{red}n+1}}}}
{{n^n }}} = 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + ....$
Johann Bernoulli first calculated $\displaystyle \, \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:
$\displaystyle x^x = 1 + x \ln x + \frac{1}{2!} \, x^2 (\ln x)^2 + \frac{1}{3!} \, x^3 (\ln x)^3 + \frac{1}{4!} \, x^4 (\ln x)^4 + ......$
and each of these integrals is simple to do. Especially if you know the reduction formula
$\displaystyle \int x^m (\ln x)^n \, dx = \frac{1}{m+1} \, x^{m+1} (\ln x)^n - \frac{n}{m+1}\, \int x^m (\ln x)^{n-1} \, dx$
Regarding the direction this thread has taken to date: The technical infelicity is the integral limits .....
Perhaps $\displaystyle \int \limits_0^{100} {x^x } \, dx - \int\limits_0^{1} {x^x } \, dx$ .....
Or a cunning substitution that gives the integral limits and yields the answer ..... $\displaystyle u = \ln x$ doesn't seem to quite cut it ......