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

Question

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.

Answer 1

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.

Answer 2

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:

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

Answer 3

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.

Answer 4

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, $$ \begin{align} \lim_{n\to\infty}n^{p+1}\int_{0}^{1}e^{-nx}\log(1+x^{p})\,\mathrm{d}x &=\int_0^\infty x^pe^{-x}\,\mathrm{d}x\\ &=\Gamma(p+1) \end{align} $$