# Kindly solve this? which option is correct?

Mar 31, 2018

This is readily seen as not doable by elementary means, so I just solved it numerically and got:

I evaluated the integral for $n = 1 , 1.5 , 2 , . . . , 9.5 , 10 , 25 , 50 , 75 , 100$. By then it was clearly reaching $0.5$.

Apr 8, 2018

See below.

#### Explanation:

${\int}_{0}^{1} \frac{n {x}^{n - 1}}{1 + {x}^{2}} \mathrm{dx} \le {\int}_{0}^{1} n {x}^{n - 1} \mathrm{dx} = 1$

${\int}_{0}^{1} \frac{n {x}^{n - 1}}{1 + {x}^{2}} \mathrm{dx} \ge \frac{1}{2} {\int}_{0}^{1} n {x}^{n - 1} \mathrm{dx} = \frac{1}{2}$

or

$\frac{1}{2} \le {\int}_{0}^{1} \frac{n {x}^{n - 1}}{1 + {x}^{2}} \mathrm{dx} \le 1$

Now assuming that one of the answers is true, the most natural seems to be the fourth 4)

NOTE

for $x \in \left[0 , 1\right]$
$\frac{1}{2} \le \frac{1}{1 + {x}^{2}} \le 1$

Apr 9, 2018

$\frac{1}{2}$

#### Explanation:

As has already been shown in a previous solution,

${I}_{n} = {\int}_{0}^{1} \frac{n {x}^{n - 1}}{1 + {x}^{2}} \mathrm{dx}$

exists and is bounded :

$\frac{1}{2} \le {I}_{n} < 1$

Now integration by parts yields

${I}_{n} = {\left(\frac{\int n {x}^{n - 1} \mathrm{dx}}{1 + {x}^{2}}\right)}_{0}^{1} - {\int}_{0}^{1} {x}^{n} \times \left(- \frac{2 x}{1 + {x}^{2}} ^ 2\right) \mathrm{dx}$
$q \quad = {\left({x}^{n} / \left(1 + {x}^{2}\right)\right)}_{0}^{1} + 2 {\int}_{0}^{1} {x}^{n + 1} / {\left(1 + {x}^{2}\right)}^{2} \mathrm{dx}$
$q \quad = \frac{1}{2} + {J}_{n}$

Now , since $0 < {\left(1 + {x}^{2}\right)}^{-} 1 < 1$ in $\left(0 , 1\right)$

${J}_{n} = \frac{2}{n + 2} {\int}_{0}^{1} \frac{\left(n + 2\right) {x}^{n + 1}}{1 + {x}^{2}} ^ 2 \mathrm{dx}$
$q \quad \le \frac{2}{n + 2} {\int}_{0}^{1} \frac{\left(n + 2\right) {x}^{n + 1}}{1 + {x}^{2}} \mathrm{dx} = \frac{2}{n + 2} {I}_{n + 2}$

Since ${\lim}_{n \to \infty} {I}_{n}$ exists, we have

${\lim}_{n \to \infty} {J}_{n} = {\lim}_{n \to \infty} \frac{2}{n + 2} {I}_{n + 2} = {\lim}_{n \to \infty} \frac{2}{n + 2} \times {\lim}_{n \to \infty} {I}_{n + 2} = 0$

Hence

${\lim}_{n \to \infty} {I}_{n} = \frac{1}{2}$