Evaluating $\lim_{n\to \infty} n \int_{0}^{1} \frac{x^n}{x+1} dx$

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 $$L$$?

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

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 : $$\lim_{n\to +\infty}{n\int_{0}^{1}{\frac{x^{n}}{x+1}}\,\mathrm{d}x}=\frac{1}{2}$$

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}

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

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

$$\displaystyle \frac {1}{2} \cdot \frac {n}{n + 1} $$\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 $$x \in [0,1]$$

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 $$nI_n \to \frac{1}{2}$$.

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 $$L = 1/2$$.

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, $$L=\lim_{n\to \infty} n\int_{0}^{1} \frac{x^n}{x+1} dx=\frac{1}{2}$$

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

• This is not correct. Feb 1 at 18:53