Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am suppose to find $\int(\ln x)^3$ by using the proof (I have to prove this part first) of

$$\int (\ln x)^n dx = x(\ln x)^n - n \int (\ln x)^{n-1} dx$$

I can not prove it and I do not know how to work with n powers like that, to me it doesn't even look right.

share|improve this question
    
To prove the last formula (do you see how to apply it?), try integration by parts. The $u$ and $dv$ that work are somewhat surprising; for now let me just say that if your $dv$ is hard to integrate, then you're doing too much work. –  Dylan Moreland Jun 8 '12 at 19:21
    
I do not understand how to do it with n as a power though, it doesn't make sense. –  user138246 Jun 8 '12 at 19:25
    
@Jordan: You have seen formulas like $\frac{d}{dx}x^n=nx^{n-1}$? Does that make sense? –  Jonas Meyer Jun 8 '12 at 19:26
    
Would it be easier to think of the case $n=3$, at first? In essence it is no easier than the general case, but the concreteness might help. –  Dylan Moreland Jun 8 '12 at 19:30
2  
(Joke, sort of): If one is asked to prove an explicit integration formula, one way is to differentiate the "answer." –  André Nicolas Jun 8 '12 at 20:27

2 Answers 2

We have

$$\mathrm I(n)=\int (\log x)^ndx$$

integrate by parts with

$$dx=du$$ $$\log^n x = v$$

$$x=u$$ $$n (\log x)^{n-1} \frac{1}{x}dx=dv$$

Spoiler ahead

$$\mathrm I(n)=x\log^n x-\int n (\log x)^{n-1}\dfrac{1}{x}x dx$$


$$\mathrm I(n)=x\log^n x-n\int (\log x)^{n-1} dx$$


$$\mathrm I(n)=x\log^n x-n \mathrm I (n-1)$$

share|improve this answer
    
You are using ln and log as the same thing but they are not the same, does this not change the answer? –  user138246 Jun 8 '12 at 19:34
2  
@Jordan Many mathematicians use $\log$ instead of $\ln$ for the base $e$ logarithm. This is probably because it can be considered "the" logarithm. –  Argon Jun 8 '12 at 19:35
    
I is the integral? I do not really follow what happened in the last step. –  user138246 Jun 8 '12 at 19:42
    
"$I(n)$" means "that integral, with an n in the exponent". So "$I(n - 1)$" means "that integral, with an $n - 1$ in the exponent"... –  The Chaz 2.0 Jun 8 '12 at 19:52
2  
Oh so it means the integrand to the n power? So with this formula does that mean that the power just reduces by one each time I use the formula? –  user138246 Jun 8 '12 at 19:59

Integration by parts is a useful technique. You can use $u=(\ln x)^n$ and $v'=1$ in the formula $\int uv' dx = uv-\int vu' dx$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.