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

 Apr 14, 2014
 #1
avatar+2353 
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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 

 Apr 14, 2014
 #2
avatar+890 
0

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

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