Compute: $\lim_{n\to\infty} n^{p+1} \int_{0}^{1} e^{-nx} \ln (1+x^p) \space dx $ Find the following limit for any $p$ natural number:
$$\lim_{n\to\infty} n^{p+1} \int_{0}^{1} e^{-nx} \ln (1+x^p) \space dx $$ 
If i'm not wrong, without much effort one may see that this integral may be rewritten as Gamma function and the limit is $p!$ . I'm just curious about more different ways one might go here.  
 A: The limit is $\Gamma(p+1)$. To see that, first integrate by parts. We have 
\begin{align}
I_n&:=n^{p+1}\int_0^1e^{—nx}\ln(1+x^p)dx\\
&=n^{p+1}\left(\left[\frac{-e^{-nx}}n\ln(1+x^p)\right]_0^1-\int_0^1\frac{-e^{-nx}}n\frac{px^{p-1}}{1+x^p}dx\right)\\
&=-n^pe^{-n}+pn^p\int_0^1e^{-nx}\frac{x^{p-1}}{1+x^p}dx,
\end{align}
so we need to compute the limit of $J_n:=n^p\int_0^1e^{-nx}\frac{x^{p-1}}{1+x^p}dx$. We use the substitution $t=nx$ (then $dt=ndx$) to get 
\begin{align}J_n&=n^p\int_0^ne^{-t}\frac{t^{p-1}}{n^{p-1}\left(1+\left(\frac tn\right)^p\right)}dt\frac 1n \\
&=\int_0^ne^{-t}\frac{t^{p-1}}{1+\left(\frac tn\right)^p}dt.
\end{align}
This quantity converges to $\Gamma(p)$. Indeed, $\int_0^ne^{—t}t^{p-1}dt\to \Gamma(p)$ and 
\begin{align}
\left|J_n-\int_0^ne^{—t}t^{p-1}dt\right|&\leq \int_0^ne^{-t}t^{p-1}\frac{\left(\frac tn\right)^p}{1+\left(\frac tn\right)^p}dt\\
&=\int_0^ne^{—t}t^{p-1}\frac{t^p}{n^p+t^p}dt\\
&\leq \int_0^{+\infty}e^{—t}t^{p-1}\frac{t^p}{n^p+t^p}dt,
\end{align}
and we conclude by the monotone convergence theorem. 
A: $$\begin{eqnarray*}
n^{p+1} \int_{0}^{1} dx\, e^{-nx} \ln (1+x^p) 
&=& n^{p+1} \int_0^n \frac{dz}{n} e^{-z} \log\left(1 + \left(\frac{z}{n}\right)^p \right) \\
&=& n^p \int_0^n dz\, e^{-z} 
    \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \left(\frac{z}{n}\right)^{p k} \\
&=& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{1}{n^{p (k-1)}}
    \int_0^n dz\, e^{-z} z^{p k} \\
&=& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{1}{n^{p (k-1)}}
    \gamma(p k + 1,n) \\
&=& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{1}{n^{p (k-1)}}
    \left[(p k)! + O\left(e^{-n}n^{p k}\right)\right] \\
&=& p! - \frac{(2p)!}{2 n^p} + O(n^{-2p})
\end{eqnarray*}$$
Above we use the asymptotic expansion for the lower incomplete gamma function,
$$\begin{eqnarray*}
\gamma(s,n) &=& \int_0^n dt\, e^{-t} t^{s-1} \\
&=& \Gamma(s) - \int_n^\infty dt\, e^{-t} t^{s-1} \\
&=& \Gamma(s) - e^{-n}n^{s-1} + O(e^{-n}n^{s-2}) 
    \hspace{5ex} (\textrm{integrate by parts}). 
\end{eqnarray*}$$
