2
$\begingroup$

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.

$\endgroup$
1

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ how did you came to second step, I didn't understand. $\endgroup$ Oct 28, 2015 at 7:30
  • $\begingroup$ how did you removed lim please explain $\endgroup$ Oct 28, 2015 at 7:32
  • $\begingroup$ @Git $d(x^n)$ converges to $du(x)$ in distribution. $\endgroup$
    – A.S.
    Oct 28, 2015 at 7:32
  • $\begingroup$ is u(x) is a function? $\endgroup$ Oct 28, 2015 at 7:33
  • 1
    $\begingroup$ This approach is quite likely passing largely over the OP's head (and is not necessary at all). $\endgroup$
    – Did
    Oct 28, 2015 at 7:51

Not the answer you're looking for? Browse other questions tagged .