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Some work on it reveals the possibility of using gamma function. Is there any easy way to compute it? $$\lim_{n\to\infty}\left(\frac{1}{n!} \int_0^e \log^n x \ dx\right)^n$$

Student_kid
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    "Some work..." what exactly did you do that makes you expect the gamma function would be useful? – Stahl May 18 '13 at 15:49
  • @Stahl I let $x=e^{-y}$. Since it's a high school question I expect to find a way that avoids the use of the incomplete gamma function. – Student_kid May 18 '13 at 15:51
  • That's fair - I only ask because it's easier to give suitable answers when one exhibits his or her previous work on the problem. – Stahl May 18 '13 at 16:02
  • I played around with this a bit and this very subtle limit may be involved: http://math.stackexchange.com/questions/160248/lim-n-rightarrow-infty-e-n-sum-limits-k-0n-fracnkk/. – Jonathan May 18 '13 at 16:11
  • It's kind of odd integration limit. Why $[0,e]$ and not $[0,1]$? It should be clear that, for odd $n$: $$\int_0^e dx \log^n{x} = \int_0^{1/e} dx \log^n{x}$$ – Ron Gordon May 18 '13 at 16:14
  • By partially integrating repeatedly you get $\frac{1}{n!} \int_0^e \log^n x , \mathrm{d}x = e(-1)^n \sum_{k=0}^n \frac{(-1)^k}{k!}.$ – Cocopuffs May 18 '13 at 16:40

2 Answers2

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Let $ x = e^{-y} $ to arrive at $ \displaystyle(-1)^n \int_{-1}^\infty y^n e^{-y} \ dy = (-1)^n \left(n! + \int_{-1}^0 y^n e^{-y} \ dy\right) $. This second integral is bounded between $ 0 $ and $ (-1)^n $. In fact, it is exactly equal to $ u_n e - n! $ where $ u_n $ has a recursive definition easily found though integration by parts, namely $ u_{n} = (-1)^{n} + nu_{n-1} $.

Hence, the overall integral has value $ (-1)^n u_n e $. Because $ u_n \sim \frac{n!}{e} $ the overall result is $ (-1)^{n^2} = (-1)^n $.

The latter result can be drawn from the fact that $\displaystyle \lim_{n \to \infty} \int_{-1}^0 y^n e^{-y} \ dy = \lim_{n \to \infty} u_n e - n! = 0 $.

A more interesting question is my opinion is $$ \sum_{n = 1}^\infty \left(\frac{1}{n!}\int_0^e \log^n x \ dx\right)^n - (-1)^n $$

Jon Claus
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I'm not totally sure of this, but Mathematica seemed to agree. Via integration by parts, $u=\log^{n}(x)\implies du = (n/x)\log^{n-1}(x)$ and $dv = dx\implies v = x$. Therefore,: $$ \int_{0}^{e}\log^{n}(x)dx = e - n\int_{0}^{e}\log^{n-1}(x) $$

By defining $a_{n} = \int_{0}^{e}\log^{n}(x)dx$, we see that $a_{n} = e - n a_{n-1}$. If $n$ is very large, the term iwth $e$ becomes unimportant, and effectively $a_{n} = -na_{n-1}$, which means that $a_{n} \sim (-1)^n n!$. Subbing this in, the limit $L$ becomes $$ L\sim\lim_{n\to\infty}\left(\frac{(-1)^n n!}{n!}\right)^n = \lim_{n\to\infty}\left((-1)^n\right)^n = \lim_{n\to\infty}(-1)^{n^2} $$

Which oscillates between $-1$ and $1$ for odd and even $n$, so the limit does not exist.

Mr. G
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  • You have a few mistakes. First, you can't just say the $ e $ becomes unimportant without justification. While $ a_n $ does increase towards infinite, at low values of $ n $ the $ e $ term matters a lot and this reflects on the sequence later. Also, $ \left((-1)^n\right)^n = (-1)^{2n} = 1 $. – Jon Claus May 18 '13 at 19:13
  • You're right, that was a really dumb mistake on my part. Also, I did make a mistake in that I forgot $ I_n = u_n e $ in my post, as opposed to $ u_n $. So the limit is indeed $ (-1)^{n^2} $. – Jon Claus May 18 '13 at 19:17