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integration

Name: Anonymous 2007-11-28 1:02

how do i integrate x^x?

Name: Anonymous 2007-11-28 1:08

LOL WITH RESPECT TO WUT VARIABLE LOLOL

You can't use the same troll twice.

Name: Anonymous 2007-11-28 5:32

>>2
Sure you can. It just becomes delicious copy pasta.

Name: Anonymous 2007-11-28 8:39

>>1-3
Did the last discussion give any answer?

Name: Anonymous 2007-11-28 10:17

>>4
theres no such integer /thread

Name: Anonymous 2007-11-28 12:05

it can not be integrated in terms of elementary functions. failure.

Name: Anonymous 2007-11-28 12:56

>>3
2 here.  Good point.  I concede and allow this topic to unfold.

Name: Anonymous 2007-11-29 2:04

its a simple u substitution problem
do it, faggot

Name: Anonymous 2007-11-29 4:27

integral(x^x*dx) is the new 0.999... = 1 for the /sci/ philosophers

Name: Anonymous 2007-11-29 4:55

Just write x^x = e^(x log x).  Then the answer is obviously e^(x log x) / (x log x).

Hope this helps!

Name: Anonymous 2007-11-29 8:42

>>10
It's x ln x, and won't help, it is fallacious in the same manner that some prove that 0 = 1.
The only way you get even close to this shit is by expanding using Taylor, and you get something like:
http://files.liveadmaker.com/F/10044924/Image.gif

Also pretend I'm a kike and anything you disagree with me will be taken as a sign of anti-Semitism, and stfu about x^x already.

Name: Anonymous 2007-11-29 14:24

I can remember the professor writing down the integral of x^x on the blackboard at the beginning of each single class. At the exams some people still failed it.

Name: Anonymous 2007-11-29 18:05

step one: rewrite it as e^(x*ln(x))

step two: expand (x*ln(x)) into a taylor series for e^(variable)

step three: each term in theseries will have one x^n term and another n*ln(x) term. Integrate by parts, taking u=n*ln(x)

step four: if you can write the terms as a new series, then you're not an idiot and good job. apply whatever limits you want now and get your answer.

step five: ?????

step six: profit!

Name: Anonymous 2007-11-29 18:39

Mathematica could not find a formula for your integral. Most likely this means that no formula exists. [More information]

Name: Anonymous 2007-11-30 19:50

>>14
any douchebag can punch it into their old flawed software.
learn some math, idiot

Name: Anonymous 2007-12-02 15:54

>>15
You do realize that >>1 never gave a variable of integration, right?

Name: Anonymous 2007-12-02 15:57

>>16
HMM I WONDER WHAT IT IS THAT WOULD MAKE THIS INTERESTING.

Name: Anonymous 2007-12-02 19:31

>>17
Since this is a copy pasta troll, there isn't anything that would make it interesting.

Name: Anonymous 2007-12-03 1:28

>>15
learn some manners, cunt

Name: Anonymous 2008-03-27 15:45

x^x
=x ln(x)
=1/2*x^2*ln(x)-x^3/6*1/x
=1/2*x^2*ln(x)-x^2/6
=x^2/2*(ln(x)-3)
I think

Name: Anonymous 2008-03-27 18:06

>>20
x^x
=x ln(x)

uhhh wut?
x^x=/=ln(x^x)=xln(x)

Name: 4tran 2008-03-27 21:30

>>10
Nice try, but phail.

Name: Anonymous 2008-03-30 3:55

>>21
Yes, that should be e^(xln(x))

Name: Anonymous 2008-03-30 12:24

i think the only was is with power series expansion...right?

Name: Anonymous 2008-03-30 13:21

>>24
or tailor

Name: Anonymous 2008-04-02 5:54

yes taylor maclaurin series are the only way you can integrate that

Name: Anonymous 2008-04-02 20:28

S x^x dy = 0

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