Evaluating the following limit

Question

Compute the following limit: $\lim\limits_{n\to \infty}2n \int \limits_0^1\dfrac{x^{n-1}}{1+x}\,dx.$

The value is 1. I have done this by squeezing it. Is there any other way to evaluate this ?

Answer 1

Using $f(x) = x^n$, the integral becomes $$ 2\int_0^1 \frac{nx^{n-1}}{1+x}\,dx = 2\int_0^1 \frac{f'(x)}{1+x}\,dx $$ Partial integration lead to,

$$ 2\int_0^1 \frac{f'(x)}{1+x}\,dx = 2\left [\frac{f(x)}{1+x} \right ]_0^1 + 2\int_0^1\frac{f(x)}{(1+x)^2}\,dx = 1 + 2\int_0^1\frac{f(x)}{(1+x)^2}\,dx $$

And the limit should now be trivial.

Answer 2

More than likely, this is totally off-topic but too long for a comment.

Sooner or later, you will learn about hypergeometric functions which make $$I_n=\int \dfrac{x^{n-1}}{1+x}\,dx=\frac{x^n}{n}-\frac{x^{n+1} }{n+1}\, _2F_1(1,n+1;n+2;-x)$$ Using the bounds $$J_n=\int_0^1 \dfrac{x^{n-1}}{1+x}\,dx=\frac{1}{2} \left(\psi \left(\frac{n+1}{2}\right)-\psi \left(\frac{n}{2}\right)\right)$$ where appears the digamma function.

Using their asymptotics $$J_n=\frac{1}{2 n}+\frac{1}{4 n^2}+O\left(\frac{1}{n^4}\right)$$ which is a "quite good" approximation even for small values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 0.750000 & 0.693147 \\ 2 & 0.312500 & 0.306853 \\ 3 & 0.194444 & 0.193147 \\ 4 & 0.140625 & 0.140186 \\ 5 & 0.110000 & 0.109814 \\ 6 & 0.090278 & 0.090186 \\ 7 & 0.076531 & 0.076481 \\ 8 & 0.066406 & 0.066377 \\ 9 & 0.058642 & 0.058623 \\ 10 & 0.052500 & 0.052488 \end{array} \right)$$

Answer 3

You can use this result to get: Let $f(x)$ be continuous in $[0,1]$. Then $$ \lim_{n\to\infty}\int_0^1nx^nf(x)dx=f(1). $$

Answer 4

Let $x=u^{1/n}$, so that $dx={1\over n}u^{1/n}du/u$ and $x^n=u$, and thus

$$2n\int_0^1{x^{n-1}\over1+x}dx=2\int_0^1{du\over1+u^{1/n}}\to2\int_0^1{du\over1+1}=1$$