the integral of x^x (calculus)

The integral of x^n is (1/n)x^(n+1) +C (where C is any constant).

The way to understand is that if you take the derivative of this, you will get the original (x^n).

For example the integral of x^2 is (1/3)x^3 + C.

[You should check that the derivative of (1/3)x^3 + C is equal to x^2 before you trust what I am saying.]

The expression x^x is equivalent to e^(x logx), where logx is the natural logarithm of x. You will be able to figure out the integral in this form.

thanks for the ideas, but i remember that the lecturer proved it using taylor. i will be greatfully if you give me a link to a website that proves it the same way my calculus lecturer did, since i didn't find any thing helpful in google.

x^x = e^(log(x^x)) = e^(x logx) = e^u {u = x logx}

You should be able to take it from there I think.

Tough questions
This is a lot harder than the posts above imply. Here is a link that discusses it somewhat. The answer you are looking for may be half way down the page. No guarantees about accuracy.

Shoot!

I sure read that wrong.... Untill just now, I was seeing X * X.

I am going to walk sheepishly away now.

Re: Tough questions


Thank you but i need the FULL PROOF. There, he gave only the integral without typing the way he developed for having this expression, eventually.
Anyway, if you know where I can find the full proof, I will be thankful!

I was doing derivatives, don't ask me why.

In general
Teh integral of x^n dx = x^(n+1)/(n+1) + C


Don't forget the C if it is an indefinite integral.

Re: In general


Welcome Tex! You might want to look at the question again. He's looking for the integral of x^x, not x^n.

Engineer
Good point.

Since the integral of the log is the log of the inegral, your approah is about the only one. Then by parts I suppose.

I'll use a ~ as the integral sign

I = ~ x^x dx

ln I = ~ x ln x dx

u = lnx dv = xdx

du = dx/x v = x^2/2

ln I = x^2/2 * ln x - ~ xdx/2
= x^2/2 ln x - x^2/4

I = exp [x^2/2 lnx - x^2/4] + C


Right?

I thought it was alot harder than that, even Integrate won't solve this one

http://support.wolfram.com/mathematica/mathematics/calculus/integratefails.html

It is possible that Integrate may not find a symbolic solution to your integral. When this occurs, it will return the unevaluated integral, as can be seen in the following example.


In[1]:= Integrate[x^x, x]

Out[1]= Integrate[x^x, x]

Integrate[x^x, x] cannot be integrated in terms of established functions. While it is possible to invent a function and define it as the solution to this integral, this is not what the Integrate function does.

Even if you use Lambert functions on Power Towers this is a very complex integral that can't be expressed in simple elementary terms:

http://mathworld.wolfram.com/PowerTower.html (2/3 of the way down):

The best you can do is express some definite integrals of it like



= 0.7834305107...

or look up a sophmore's dream

http://mathworld.wolfram.com/SophomoresDream.html





Integrating term by term then gives

But a very good approximation I realised,

using log x is very close to 1024 * (the tenth root of x -1) ( as 2 ^ 10 = 1024)

So as x ^ x is e ^ x log x, this is approximately

Integrate e ^ (1024 * x * [(x^0.1) - 1])
or

Integrate e ^ 1024*(x^1.1 -x)

Alternately you could use

ln(x) ~ 6*(x-1)/ ( x + 1 + 4*(x^0.5))

so solve Integrate e ^ 6x*(x-1)/ ( x + 1 + 4*(x^0.5))