Ytukyg

For $a>0$ we define
$$space I_n(a)=int_0^afrac{x^n}{x^n+1},mathrm{d}x , nin N.$$

1. Prove that $0le I_n(1) le frac{1}{n+1}$
   


2. Compute $lim_{ntoinfty} I_n(a)$
   

My attempt:

1. I regard $I_n(1)=int_0^1frac{x^n}{x^n+1}$. If $xin (0,1)$ then $x^nin(0,1)$ and $x^n+1in(1,2)$.
   $$x^n>0 Rightarrow x^n+1>1 Rightarrow 1>frac{1}{1+x^n }Rightarrow x^n>frac{x^n}{x^n+1}Rightarrow int_0^1frac{x^n }{x^n+1}dx<int_o^1 x^n mathrm{d}x\ Rightarrow int_0^1frac{x^n }{x^n+1}dx<frac{1}{n+1} \ 0lefrac{x^n}{x^n+1} \ text{In concusion } 0le I_n(1) le frac{1}{n+1}.$$




2. first case $ain(0,1) Rightarrow lim_{ntoinfty} I_n(a) =0$. $I_n(a)lefrac{1}{n+1})text{case 2 . }ain(1,infty) Rightarrow$ ???????

I don't believe the limit is $infty$ because $frac{x^n }{x^n+1}le 1$.
I would appreciate some hints.

Note that we have

$$begin{align}
int_0^a frac{x^n}{1+x^n},dx&=int_0^1 frac{x^n}{1+x^n},dx+int_1^a frac{x^n}{1+x^n},dx\\
&=int_0^1 frac{x^n}{1+x^n},dx+(a-1)-int_1^a frac{1}{1+x^n},dx
end{align}$$

For $xin [0,1]$, $0le frac{x^n}{1+x^n}le x^n$ and for $xin[1,a]$, $frac{1}{1+x^n}le frac1{x^n}$. Therefore,

$$left|int_0^1 frac{x^n}{1+x^n},dxright|le frac1{n+1}$$

and

$$left|int_1^a frac{1}{1+x^n},dxright|le frac{1-a^{1-n}}{n-1}$$

Can you finish now?

Certainly not the most compact approach, but:

begin{equation}
I_n(a) = int_{0}^{a} frac{w^n}{w^n + 1}:dw = int_{0}^{a}left[ 1 - frac{1}{w^n + 1}right]:dw = a - int_{0}^{a}frac{1}{w^n + 1}:dw
end{equation}

Now:
begin{equation}
J_n(a) = int_{0}^{a} frac{1}{w^n + 1}:dw
end{equation}

With $n geq 1$ and $x geq 0$

Here, let $t = a^n$ to arrive at:

begin{equation}
J_n(a) = frac{1}{n}int_{0}^{x^n} frac{1}{t + 1}t^{frac{1}{n} - 1}:dt
end{equation}

Now let $u = frac{1}{1 + t}$ to arrive at:

begin{align}
J_n(a) &= frac{1}{n}int_{0}^{a^n} frac{1}{t + 1}t^{frac{1}{n} - 1}:dt = frac{1}{n}int_{1}^{dfrac{1}{a^n + 1}} u left(frac{1 - u}{u} right)^{1 - frac{1}{n} }frac{-1}{u^2}:du \
&= frac{1}{n}int_{dfrac{1}{a^n + 1}}^{1} u^{-frac{1}{n}}left(1 - uright)^{frac{1}{n} - 1}:du
end{align}

Here, as $x geq 0$ and $n > 1$, we see that $dfrac{1}{a^n + 1} < 1$ and thus,

begin{align}
J_n(a) &= frac{1}{n}int_{dfrac{1}{a^n + 1}}^{1} u^{-frac{1}{n}}left(1 - uright)^{frac{1}{n} - 1}:du \
&= frac{1}{n}left[int_{0}^{1} u^{-frac{1}{n}}left(1 - uright)^{frac{1}{n} - 1}:du - int_{0}^{dfrac{1}{a^n + 1}} u^{-frac{1}{n}}left(1 - uright)^{frac{1}{n} - 1}:duright] \
&= frac{1}{n}left[Bleft(1 - frac{1}{n}, frac{1}{n} right) - Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{a^n + 1}right)right]
end{align}

Where $B(a,b)$ is the Beta function and $B(a,b,x)$ is the Incomplete Beta function

Using the relationship between the Beta and Gamma functions we arrive at:

begin{align}
J_n(a) &= int_{0}^{a} frac{1}{w^n + 1}:dw = frac{1}{n}left[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{a^n + 1}right)right]
end{align}

For $a geq 0$ and $n geq 1$. Returning to $I(a)$ we have

begin{align}
I_n(a) = a - J_n(a) = a - frac{1}{n}left[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(1 - frac{1}{n}, frac{1}{n}, frac{1}{a^n + 1}right)right]
end{align}

From here you can attempt your direct questions.

Lets consider the interval $(1, a)$. We have

$|frac{x^n}{1 + x^n} - 1| = |frac{1}{1+x^n}|$

By Bernoulli's Inequality, $x^n = (1 + (x - 1))^n geq 1 + n(x - 1)$. This implies that

$|frac{x^n}{1 + x^n} - 1| leq frac{1}{2 + n(x-1)}$

Now, fix $epsilon > 0$ small enough and choose $n$ big enough such that $frac{1}{2 + n(x-1)} < frac{epsilon}{2(a - 1)}$ for every $ x in (1 + frac{epsilon}{2}, a)$. We then have

$int_1^a |frac{x^n}{1 + x^n} - 1| dx = int_1^a |frac{1}{1+x^n}| dx = $

$int_{1}^{1 + frac{epsilon}{2}} |frac{1}{1+x^n}| dx + int_{1 +frac{epsilon}{2}}^{a} |frac{1}{1+x^n}| dx leq $

$int_{1}^{1 + frac{epsilon}{2}} 1 dx + int_{1 +frac{epsilon}{2}}^{a} frac{1}{2 + n(x - 1)} dx < epsilon $

Now separate the original integral in $(0, 1)$ and $(1, a)$, and conclude.