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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.

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A sketch: you can break the integral into $0$ to $\epsilon$ and $\epsilon$ to $1$ pieces for any $\epsilon > 0$. The second piece's limit will go to zero fast, and in the first you can use $\ln(1 + x^p)$ is between $(1 - \delta)x^p$ and $(1 + \delta)x^p$ for an $\delta$ depending on $\epsilon$. Then a change of variables in the first integral to $u = nx$ should be helpful. –  Zarrax Jul 13 '12 at 22:13
    
@Zarrax: yeah. That's one of the usual ways to go. Thanks. –  Chris's sis Jul 13 '12 at 22:22
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2 Answers

up vote 3 down vote accepted

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.

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Might as well just say $\Gamma(p+1)$. –  Zarrax Jul 13 '12 at 22:31
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$$\begin{eqnarray*} n^{p+1} \int_{0}^{1} dx\, e^{-nx} \ln (1+x^p) &=& n^{p+1} \int_0^n \frac{dz}{n} e^{-z} \log\left(1 + \left(\frac{z}{n}\right)^p \right) \\ &=& n^p \int_0^n dz\, e^{-z} \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \left(\frac{z}{n}\right)^{p k} \\ &=& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{1}{n^{p (k-1)}} \int_0^n dz\, e^{-z} z^{p k} \\ &=& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{1}{n^{p (k-1)}} \gamma(p k + 1,n) \\ &=& \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{1}{n^{p (k-1)}} \left[(p k)! + O\left(e^{-n}n^{p k}\right)\right] \\ &=& p! - \frac{(2p)!}{2 n^p} + O(n^{-2p}) \end{eqnarray*}$$ Above we use the asymptotic expansion for the lower incomplete gamma function, $$\begin{eqnarray*} \gamma(s,n) &=& \int_0^n dt\, e^{-t} t^{s-1} \\ &=& \Gamma(s) - \int_n^\infty dt\, e^{-t} t^{s-1} \\ &=& \Gamma(s) - e^{-n}n^{s-1} + O(e^{-n}n^{s-2}) \hspace{5ex} (\textrm{integrate by parts}). \end{eqnarray*}$$

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