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Evaluating $$L=\lim_{n\to \infty} n\int_{0}^{1} \frac{x^n}{x+1} dx.$$ Let $$I_n=\int_{0}^{1}\frac{x^n}{x+1} dx=\int_{0}^{1} x^{n-1}dx-I_{n-1}$$ When $n$ infinitely large $I_{n-1}\sim I_{n}$, then $$I_n\sim \frac{1}{2n} \implies L= \frac{1}{2}.$$

The question is how else one can evaluate $L$?

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  • $\begingroup$ Substitute $x=\mathrm{e}^{-t}$ with $0<t<+\infty$ and employ Watson's lemma. $\endgroup$
    – Gary
    Feb 2 at 5:03
  • $\begingroup$ Related $\endgroup$ Mar 13 at 16:02

7 Answers 7

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Using an integration by parts, we have : \begin{aligned} n\int_{0}^{1}{\frac{x^{n}}{x+1}\,\mathrm{d}x}&=n\left[\frac{x^{n+1}}{\left(n+1\right)\left(x+1\right)}\right]_{0}^{1}+\frac{n}{n+1}\int_{0}^{1}{\frac{x^{n+1}}{\left(1+x\right)^{2}}\,\mathrm{d}x}\\ &=\frac{n}{2\left(n+1\right)}+\frac{n}{n+1}\int_{0}^{1}{\frac{x^{n+1}}{\left(1+x\right)^{2}}\,\mathrm{d}x} \end{aligned}

Since $ \left\lvert\int_{0}^{1}{\frac{x^{n+1}}{\left(1+x\right)^{2}}\,\mathrm{d}x}\right\rvert\leq\int_{0}^{1}{x^{n+1}\,\mathrm{d}x}=\frac{1}{n+2}\underset{n\to +\infty}{\longrightarrow}0$, then : $$ \lim_{n\to +\infty}{n\int_{0}^{1}{\frac{x^{n}}{x+1}}\,\mathrm{d}x}=\frac{1}{2} $$

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Using $\lim\limits_{n\to \infty} \sqrt[n]{t} = 1$ for $t > 0$ and Dominance convergence:

\begin{align} nI_n &\underset{t\,:=\,x^n}{=} \int_0^{1} \frac{\sqrt[n]{t}}{1 + \sqrt[n]{t}}\mathrm d t \underset{n\to \infty}{\to} \int_{0}^1 \frac12 \mathrm dt = \frac12 \end{align}

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$I_n=n \displaystyle{\int_{0}^{1}}\dfrac{x^n}{x+1}dx$:

$\displaystyle \frac {1}{2} n \int_{0}^{1} x^ndx < I_n <$ $\displaystyle \frac {1}{2} n \int_{0}^{1} x^{n-1} dx$;

$\displaystyle \frac {1}{2} \cdot \frac {n}{n + 1} <I_n<$ $\displaystyle \frac {1}{2} \cdot \frac {n}{n}$;

Take the limit.

Used:

$\dfrac{x^n}{1+1}\le \dfrac{x^n}{x+1}\le \dfrac{x^n}{x+x}$ for $x \in (0,1]$;

Define a continuos function $f$: $f(x)=0$ for $x=0$, $f(x)= \dfrac {x^n}{x+x} = \displaystyle \frac {1}{2} x^{n-1}$ for $x>0$.

Then

$\dfrac{x^n}{1+1}\le \dfrac{x^n}{x+1}\le f(x)$ for $x \in [0,1]$

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Of course, $$nI_n = n \int_0^1 \frac{x^n}{x+1}dx \ge n \int_0^1 \frac{x^n}{2}dx = \frac{n}{2(n+1)} \to \frac{1}{2}$$ and for any $\varepsilon > 0$, $$ n I_n = n\int_0^{1-\varepsilon} \frac{x^n}{1+x}dx + n\int_{1-\varepsilon}^1 \frac{x^n}{1+x}dx $$ $$ \le n \int_0^{1-\varepsilon}(1-\varepsilon)^n dx + n \int_{1-\varepsilon}^1 \frac{x^n}{2-\varepsilon}dx $$ $$= n(1-\varepsilon)^{n+1} + \frac{n}{(n+1)(2-\varepsilon)}(1 - (1-\varepsilon)^{n+1}) \to \frac{1}{2-\varepsilon}.$$ Since $\varepsilon > 0$ was arbitrary, we can conclude by squeeze theorem that $nI_n \to \frac{1}{2}$.

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Assume your limit exists. From your recursion formula, $I_n + I_{n-1} = 1/n$. Since $I_n>0$ for all $n$, this in particular implies that $I_n \to 0$ as $n\to\infty$. Multiplying the above by $n$ we get $$nI_n + (n-1)I_{n-1} + I_{n-1} = 1$$ and letting $n\to\infty$ yields $2L = 1$, hence $L = 1/2$.

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Let $x=\sqrt{t}$, then $$I_n=\int_{0}^{1}\frac{x^n}{x+1} dx=\frac{1}{2} \int_0^{1} \frac{t^{n/2-1/2} dt}{1+\sqrt{t}}$$ Note that $$\frac{t^{n/2-1/2}}{1+\sqrt{t}}=\frac{1}{2}\left(\frac{1-t^{n/2}}{1-t} -\frac{1-t^{n/2-1/2}}{1-t}\right)$$ By the definition of Harmonic Numbers $$H_m=\int_{0}^{1} \frac{1-z^m}{1-z} dz$$ So we get $$I_n=\frac{1}{2} [H_{n/2}-H_{n/2-1/2}]$$ Further as $$H_m\sim \gamma+\ln m+\frac{1}{2m}+O(1/m^2)$$ Then $$I_n\sim -\frac{1}{2} \ln(1-1/n) \sim \frac{1}{2n}$$ Eventually, $$L=\lim_{n\to \infty} n\int_{0}^{1} \frac{x^n}{x+1} dx=\frac{1}{2}$$

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Given that $\lim_{n\to \infty}\frac{x^n}{x+1}=0$, except for $x=1$, $$\lim_{n\to \infty} n\int_{0}^{1} \frac{x^n}{x+1} dx=\lim_{n\to \infty}n\int_{0}^{1} \frac{x^n}{1+1} dx=\frac12 $$

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  • 8
    $\begingroup$ This is not correct. $\endgroup$
    – Kroki
    Feb 1 at 18:53

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