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I want to answer this question: If $$ \lim_{n \to\infty} n^k \int_0^{1/n} x^{x+k-1} dx = f(k) $$ for $k \in \mathbb N$, what is $$ \left[\frac{1}{f(5)}\right], $$ where square brackets denote the greatest integer function (i.e., ceil)?

I tried it by substituting $t = x + k - 1$, but got stuck.

tired
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user399511
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  • Related: http://math.stackexchange.com/questions/141347/finding-int-xxdx – innisfree Dec 16 '16 at 10:24
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    since $x$ is small on the whole range of integration we might taylor expand the integrand. To first order we get $$ f(k)/n^k\sim \int^{1/n}_0x^{k-1}=\frac{1}{n^{k}k} $$ can you conclude? – tired Dec 16 '16 at 10:53
  • Furthermore it is simple to show that higher order terms are $\mathcal{O}(n^{k+1})$ and the correct limit is indeed given by the above approximation – tired Dec 16 '16 at 10:56

2 Answers2

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As innisfree suggested in his comment, this is an ideal case for the squeeze theorem.

Since $n \in \mathbb{N}$, we have $x \in [0,1/n] \subseteq [0,1]$ so the larger the exponents, the smaller the number. Thus

$$\int_0^{1/n}x^{1/n + k - 1} \,dx < \int_0^{1/n} x^{x+k-1} \,dx < \int_0^{1/n} x^{0+k-1} \,dx$$

$$\frac{1}{(1/n+k)n^k} < \int_0^{1/n} x^{x+k-1} \,dx < \frac{1}{kn^k}$$

$$\frac{1}{1/n+k} < n^k\int_0^{1/n} x^{x+k-1} \,dx < \frac{1}{k}$$

and taking the limit as $n \to \infty$ we find that

$$\lim_{n \to \infty} n^k\int_0^{1/n} x^{x+k-1} \,dx = \frac{1}{k}$$

CodeLabMaster
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Try $y= nx $. The integral simplifies, $$ \lim_{n\to\infty}\int_0^1 \left(\frac yn \right)^{\frac yn} y^{k -1} dy = \int_0^1 y^{k -1} dy = 1/k $$ The first factor in the integrand gives 1 when we take the limit. Thus we find $1/f(5) = 5$.

innisfree
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