Integration of x^x. NB: not in syllabus (1 Viewer)

[daodao]

Hey everyone,

This question is just out of curiousity. It is not in our syllabus but was encountered during one of my lessons at school. The teacher did not know how to do it, but i'm just wondering if anyone has a clue?

The question is:
Integrate x^x dx.

Any suggestions???

daodao

 

[McLake]

*Tries "The Integrator"*

DAMN, it dosn't know the answer.

I remember being shown this, but don't remmeber the answer ...

 

[daodao]

lol.

I'm new to this. What is the "integrator".

 

[Affinity]

some webpage which does integrals for you..
try integrals.wolfram.com

 

[Heinz]

I changed x^x into e^xln(x) and thats where i stopped

 

[daodao]

thx affinity.

hmm...
it is preety tough isn't it?
i've asked around and no one seems to know how to do it...

 

[martin]

It's possible to do the derivative of x^x.

y= x^x
= e^(ln(x^x))
= e^(x*lnx)

so y' = e(x*lnx)*[x*1/x+1*lnx]
=(lnx+1)*x^x

but this doesn't help us with the integral because we have that lnx.

So if integrator doesn't work maybe there's no integral in terms of elementary functions (the same as integral of e^(-x^2)).

 

[Xayma]

Originally posted by daodao
hmm...
it is preety tough isn't it?
i've asked around and no one seems to know how to do it...

Just concerning this question when you graph x^x what exactly happens at the negative side, I know you will get a series of points alternating on the positive and negative side on the integers, and you will also get points on the negative side alternating between them as long as the fraction has a denominator that is odd, but what happens at the irrational numbers inbetween?

 

[daodao]

Yea we found the derivative in class, but no one could find the integral...

hmm there has to be a way...

 

[daodao]

Xayma,

I have no idea about graphing it at the moment.

I'm still going through depression from my half yearlies, so haven't been doing much work lol

but i'll find out soon.. i'll get back to you...

 

[CM_Tutor]

Originally posted by daodao
Yea we found the derivative in class, but no one could find the integral...

hmm there has to be a way...

Actually, there doesn't have to be a result in elementary functions - As Martin pointed out, there is not an answer in elementary functions for int e^-x²/2 dx - x^x could be expressed in terms of power series, and then integrated term-by-term but that is light years from the syllabus (at least, dealing with this one would be! )

PS: I tried the usual differentiate and hence integrate trick - ie differentiate x * x^x, and integrate the result, but it just leads to worse integrals...

 

[Giant Lobster]

What is x^x as a series?

 

[martin]

The normal series used for this sort of stuff is MacLaurin series.

f(x) = f(0) + f'(0)*x + f''(0)/2*x^2 + f'''(0)/3!*x^3 + ... + f(n dashes)(0)/n!*x^n+ ..

but note that f(0) = 0^0 is undefined so MacLaurin series doesn't exist.

So just because its easier to do I'll give a Taylor series centred around x=1.

f(x) = f(1) + f'(1)*(x-1) + f''(1)/2*(x-1)^2 + f'''(1)/3!*(x-1)^3 + ... + f(ndashes)(1)/n!*(x-1)^n+ ..

f(x)=x^x -> f(1)=1
f'(x)=(1+lnx)*x^x -> f'(1)=1
f''(x)=(1/x)x^x+(1+lnx)^2*x^x
=(1/x+(1+lnx)^2)*x^x -> f''(1)=2
f'''(x)=(-1/x^2+2(1+lnx)*1/x)*x^x + (1/x(1+lnx)+(1+lnx)^3)*x^x
= (3/x(1+lnx)+(1+lnx)^3-1/x^2)*x^x
-> f'''(1)=3+1+1-1=4

I can't be bothered doing more and can't see any obvious pattern

so x^x approximately
= 1 + 1*(x-1) + 2/2(x-1)^2 + 4/6(x-1)^3
= x + (x-1)^2 +2/3(x-1)^3

at this point we should check that it converges and that it converges to the right thing but that seems a bit hard so instead we'll put numbers into a calculator.

1.5^1.5 = 1.83711..
series approx. = 1.83333...

(-1)^(-1) = -1
series approx. = 2.3333
so it obviously doesn't work for nonpositive numbers (look at all the logs we used)

0.5^0.5 = 0.7071...
series approx. = 0.6666...

0.9^0.9 = 0.909535..
series approx. = 0.909333..

so it looks alright for positive numbers (pretty good close to 1) and we can do the integration term by term

integral(x^x) approx.
= integral(x+(x-1)^2+2/3(x-1)^3)
= 1/2x^2 + 1/3(x-1)^3+ 1/6(x-1)^4 + C

What can we tell from all this? Not much but around x=1 the integral acts like this polynomial so we could graph the successive taylor series approximations and get a good idea of what it looks like.

If anybody finds a mistake in here or has some ideas on general terms or radius of convergence please reply.

thanks,
Martin

 

[martin]

on the subject of graphing x^x, I just put (-Pi)^(-Pi) into MATLAB and got -0.0248 + 0.0118i, a complex number!

I guess this means it doesn't exist but its pretty confusing, isn't it?

 

[Affinity]

many elementary functions don't have elementary functions as integrals

 

[Xayma]

Well you are going to have points where a -ve rational number has an odd denominator, so would it be true to assume that all -ve rational numbers with an even denominator and -ve irrationals dont have points?

 

[+Po1ntDeXt3r+]

hmm .. i did the matlab thing

hope ppl can read it right.. cos i graphed the function on 3 planes for a visual representation

:S

 

[daodao]

lol
I have no idea what the power series is.
Am i suppose to know it, is it in the HSC syllabus?

Wow,
I've never seen a graph anything like that in my life.

So basically x^x is a 3d graph involving the complex plane and by integrating it you will obtain a complex number?

btw thanks +Po1ntDeXt3r+ for the graph.

daodao

 

[+Po1ntDeXt3r+]

Originally posted by daodao
So basically x^x is a 3d graph involving the complex plane and by integrating it you will obtain a complex number?

no it just means that u couldnt integrate that function(in a yr 12 context..) for certain values of x as the X^X values are imaginary

the graph was to show wat the function did in around the -ve and x=0 region

 

[Euler]

It's one thing to say that something can't be integrated, it's another thing to say why it can't be integrated.

You could be sitting there all day trying to find the primitive function of $e^{x^2}$ and you would get nothing for this question. Why? it's because someone has shown that such a primitive function doesn't exist.

From what I gather, there is no straightforward way to see why there is not a primitive function. It probably really depends on the function.

And no, I don't know why $e^{x^2}$ can't be integrated. Nor $x^x$.