[Razor]

Originally posted by CP5670
what was the original problem in that last quote?

try this one:

¥
ò ( e^(1-x)t - 1 ) log(t) / ( e^t - 1 ) dt
0

You have 2 unknowns there. You must give us a value for X if you want us to solve that.

By the way, you are insane. How the heck do you integrate a function when it's deffinition is from 0 to unlimited?

EDIT: Here is how I started solving that:

[CP5670]

actually the answer to that is a function of x. your image is a red x though.

By the way, you are insane. How the heck do you integrate a function when it's deffinition is from 0 to unlimited?

eh...take appropriate limits, how else?

[Martinus]

Originally posted by Tiara
...

Why is it that you guys think nonly CP knows his math?

 

[color=66ff00] It's usually only CP who throws a math problem up out of the blue, just for the hell of it.
We're more used to you running around 'axing' people and slitting the throats of those that notice your flagrant use of elven underwear.
[/color]

[Razor]

Originally posted by CP5670
actually the answer to that is a function of x.

You are starting to scare me now.

[CP5670]

alright I will give you the answer; see if you can prove why it is so. The integral equals log PF_-1(x) + g H(x) , where PF is the powerfactorial, H is the harmonic number and g is the euler constant; if x is a positive integer, PF_n(x+1) = 1¹ⁿ2²ⁿ3³ⁿ...x^xn and H(x+1) = 1/1+1/2 +1/3+...+1/x.

[Galemp]

I hate you.

[SKYNET-011]

Arrrgh maths!

[Razor]

Originally posted by CP5670
alright I will give you the answer; see if you can prove why it is so. The integral equals log PF_-1(x) + g H(x) , where PF is the powerfactorial, H is the harmonic number and g is the euler constant; if x is a positive integer, PF_n(x+1) = 1¹ⁿ2²ⁿ3³ⁿ...x^xn and H(x+1) = 1/1+1/2 +1/3+...+1/x.

Yeah...eh...I get it. Jesus Christ! I am never gonna take another math course again.