We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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

 Apr 14, 2014

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)



 Apr 14, 2014

Write your integrand as $$1.(\log(x))^{n}$$ and integrate by parts, integrating the 1 and differentiating the log.

That gets you a reduction formula which you have to use twice, once for P10 and then again for P9.

 Apr 14, 2014

16 Online Users