Gamma identity $\lim_{n\to \infty}n^{p+1}\int_{0}^{1}e^{-nx}\ln(1+x^{p})dx=\Gamma(p+1)$

I ran across what appears to be another Gamma identity.

Show that $$\lim_{n\to \infty}n^{p+1}\int_{0}^{1}e^{-nx}\ln(1+x^{p}) \,\mathrm dx=\Gamma(p+1)=p!$$

I tried several different subs and even the series for $$\ln(1+x^{p})$$ and nothing materialized.

What would be a good start on this one?

It looks similar to $$\Gamma(p+1)=n^{p+1}\int_{0}^{\infty}x^{p}e^{-nx} \,\mathrm dx,$$ but I was unable to hammer it into that form. I would think there is a clever sub of some sort that may work it into this last mentioned form.

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Should it be $\Gamma(p+1) = n^{p+1} \int_0^\infty x^p e^{-nx} \,\mathrm d x$? – cardinal Mar 31 '12 at 20:35
Yes, Cardinal. Thanks. Sorry about that. – Cody Mar 31 '12 at 21:20
It's a little funny that you accepted an almost identical answer to mine that was submitted over an hour after mine was posted. I'm not complaining, just found it a little amusing. I've also noticed that you have asked a lot of (interesting) questions but have never voted. You do know that you can vote and accept the same answer and also vote on any questions or answers that are helpful or interesting, right? – cardinal Apr 1 '12 at 13:56
I'm sorry. I do not like to slight anyone by voting for a best answer. I just clicked one because I thought that is what I am supposed to do. I liked them all and they're all helpful. No, sorry, I was under the impression only one green check mark. Each time I have tried to check several, the other vanishes. – Cody Apr 1 '12 at 17:07
No slighting involved, at all. :) You should accept (choose the checkmark) the answer that most helps you, whichever one that is. Whether it came first, last or in-between does not matter. That's what the checkmark is there for. :) Voting up or down is an additional way to participate and indicate helpfulness of an answer (or question). I was just checking to make sure you were aware of all of the feedback facilities on this site. Cheers. :) – cardinal Apr 1 '12 at 17:10

By nonnegativity of the integrand and the fact that $\log(1+u)\leq u$, we have that $$n^{p+1} \int_0^1 e^{-nx} \log(1+x^p) \,\mathrm d x \leq n^{p+1} \int_0^\infty e^{-nx} x^p \,\mathrm dx = \Gamma(p+1) \>.$$

For the other direction, note that for $u \geq 0$ we have the crude bound $\log(1+u) \geq u - u^2$, and so $$n^{p+1} \int_0^1 e^{-nx}\log(1+x^p) \mathrm d x \geq n^{p+1} \int_0^1 e^{-nx} (x^p - x^{2p}) \,\mathrm dx \>.$$

Substituting $t = n x$, the right-hand side becomes $$\int_0^n e^{-t} t^p \,\mathrm d t - n^{-p} \int_0^n e^{-t} t^{2p}\,\mathrm dt \to \Gamma(p+1)$$ as $n\to\infty$ since the second term converges to zero.

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@Didier: Thanks for the typo corrections! As I transitioned from paper to computer, I changed midstream from using $u$ to using $t$ and managed to miss a couple spots. – cardinal Mar 31 '12 at 21:49
You are welcome. Nice clean answer. – Did Mar 31 '12 at 21:50

Substituting $x\mapsto x/n$ yields \begin{align} n^{p+1}\int_{0}^{1}e^{-nx}\log(1+x^{p})\,\mathrm{d}x &=n^p\int_0^ne^{-x}\log(1+(x/n)^p)\,\mathrm{d}x\\ &=n^p\int_0^n\left((x/n)^p+O(x/n)^{2p}\right)e^{-x}\,\mathrm{d}x\\ &=\int_0^nx^pe^{-x}\,\mathrm{d}x+\int_0^\infty x^{2p}e^{-x}\,\mathrm{d}x\;O\left(n^{-p}\right) \end{align} Thus, \begin{align} \lim_{n\to\infty}n^{p+1}\int_{0}^{1}e^{-nx}\log(1+x^{p})\,\mathrm{d}x &=\int_0^\infty x^pe^{-x}\,\mathrm{d}x\\ &=\Gamma(p+1) \end{align}

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(+1) This looks essentially identical to mine, with the additional cleanliness of big-O. :) – cardinal Mar 31 '12 at 21:39

As known,

$$\Gamma(p)=\int_0^{\infty}x^{p-1}e^{-x}dx,$$

If we now put $x=\ln{\frac{1}{z}}$, we will get:

$$\Gamma(p)=\int_0^{1} (\ln{\frac{1}{z}})^{p-1}dz.$$

Now, using the well known limit:

$$\ln{\frac{1}{z}}=\lim_{n\to\infty} n(1-z^{\frac{1}{n}})$$

So, we get

$$\Gamma(p)=\int_0^{1}\lim_{n\to\infty} (n(1-z^{\frac{1}{n}}))^{p-1}$$

By theorem, you allowed to get the limit out, and:

$$\Gamma(p)=\lim_{n\to\infty}\int_0^{1} (n(1-z^{\frac{1}{n}}))^{p-1}$$

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By "theorem"? Can you be more clear? – Patrick Da Silva Mar 31 '12 at 20:12
@Patrick: Dominated convergence theorem or Arzela's lemma(to those whom not familiar with measure theory) – Salech Alhasov Mar 31 '12 at 20:27

Thanks for the inputs. This reminds me of another Gamma identity. Looks like the same idea.
$$\Gamma(p+1)=\lim_{n\to \infty}n^{p+1}\int_{0}^{1}x^{n-1}(1-x)^{p}dx=\lim_{n\to \infty}n^{p+1}B(n,p+1)$$.

Thus, we conclude that $$\lim_{n\to \infty}n^{p+1}\frac{\Gamma(n)}{\Gamma(n+p+1)}=1$$.

This leads to a proof of the Taylor series for $$ln(\Gamma(1+x))=-\gamma x+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k}x^{k}, \;\ |x|<1$$.

by using $$\lim_{a\to 0^{+}}\left(\frac{1-e^{-a\nu}}{a}\right)^{x}={\nu}^{x}$$

Thanks all.

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