How to integrate this type of integral: $\lim_{n \rightarrow \infty}{\int_ 0^1}\frac{n x^{n-1}}{1+x}dx$

Question

What should be the strategy to solve these type of integrals. And how to solve this integral.

$\lim_{n \rightarrow \infty}{\int_ 0^1}\frac{n x^{n-1}}{1+x}dx$

I tried to solve it by substitution: multiplying and deviding by $x^n$ and substibution $t=x^n$ I got this

$\lim_{n \rightarrow \infty}{\int_ 0^1}\frac{1}{1+x^{1/n}}dx$

and don't know what to do now.

See also: Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ — Martin Sleziak, May 06 '17 at 05:47

score 3 · Accepted Answer · edited Oct 28 '15 at 07:54

Note that: $$\int_ 0^1\frac{n x^{n-1}}{1+x}dx=\left.\frac{x^n}{1+x}\right|_{0}^{1} +{\int_ 0^1}\frac{x^{n}}{(1+x)^2}dx=\frac{1}{2}+{\int_ 0^1}\frac{x^{n}}{(1+x)^2}dx$$ Furthermore: $$0\leq\int_ 0^1\frac{x^{n}}{(1+x)^2}dx\leq\int_ 0^1x^ndx=\frac{1}{n+1}$$

As $n\rightarrow \infty$, $\frac{1}{n+1}\rightarrow 0$ so the original integral $\rightarrow \frac{1}{2}$ by squeeze theorem.

A.S. · Answer 2 · 2015-10-28T07:31:11.753

2

$$=\lim_{n\to\infty}\int_0^1\frac 1 {1+x}\,d(x^n)=\int_0^1\frac 1 {1+x^2}\,du(x)=\frac 1 {1+1}=\frac 1 2$$ where $u(x)=I(x=1)$.

edited Oct 28 '15 at 07:31

answered Oct 28 '15 at 07:21

A.S.

4,004

how did you came to second step, I didn't understand. – Giteshwar Mali Oct 28 '15 at 07:30
how did you removed lim please explain – Giteshwar Mali Oct 28 '15 at 07:32
@Git $d(x^n)$ converges to $du(x)$ in distribution. – A.S. Oct 28 '15 at 07:32
is u(x) is a function? – Giteshwar Mali Oct 28 '15 at 07:33
It is, but it's more important that it's defined a distribution on $[0,1]$. – A.S. Oct 28 '15 at 07:34
1

This approach is quite likely passing largely over the OP's head (and is not necessary at all). – Did Oct 28 '15 at 07:51
@Did Which essentially different method is there? – A.S. Oct 28 '15 at 08:37
See other answer. – Did Oct 28 '15 at 08:49
@GiteshwarMali: you can move the limit under the integral sign in this case: http://math.stackexchange.com/questions/120768/moving-a-limit-inside-an-integral?rq=1 – Alex Oct 28 '15 at 09:01

How to integrate this type of integral: $\lim_{n \rightarrow \infty}{\int_ 0^1}\frac{n x^{n-1}}{1+x}dx$

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