if for n>1, Pn = $$\int_{1}^{e}$$$$(log x)^{n}$$ dx , Then P10 - 90P8 equals to.

Guest Apr 14, 2014

#1**0 **

I am guessing that there must be an easy result to this question, however I have not yet found it.

I have a WolframAlpha pro account and I did find the following;

the indefinite integral of (logx)^n is given by;

$$\int log^n(x)dx = (-log(x))^-^nlog^n(x)T(n+1,-log(x)) + C$$

Where T(alpha,x) is the incomplete gamma function (and C is off course the constant)

Since I did not feel like computing this myself, I gave WA a shot at your question and after it probably blew up servers in four countries around the world it gave me this;

$$\int_{1}^{e} log(x)^1^0 - 90* \int_{1}^{e} log(x)^8 = -9e$$

which to me was as meaningful as 42 (reference: )

To make a long story short (this is funny if you watch s16e06 of south park).

I have the answer, but I dont know how (I'm guessing it must be easier than actually filling in that dreadful formula)

Reinout

reinout-g Apr 14, 2014