Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$


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

*Compute  $\lim_{n\to\infty} I_n(a)$
My attempt:


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

*first case $a\in(0,1) \Rightarrow \lim_{n\to\infty} I_n(a) =0$. $I_n(a)\le\frac{1}{n+1})\text{case 2 . }a\in(1,\infty) \Rightarrow$ ???????
I don't believe the limit is $\infty$ because $\frac{x^n }{x^n+1}\le 1$.
I would appreciate some hints.
 A: 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 $x\in [0,1]$, $0\le \frac{x^n}{1+x^n}\le x^n$ and for $x\in[1,a]$, $\frac{1}{1+x^n}\le \frac1{x^n}$.  Therefore, 
$$\left|\int_0^1 \frac{x^n}{1+x^n}\,dx\right|\le \frac1{n+1}$$
and
$$\left|\int_1^a \frac{1}{1+x^n}\,dx\right|\le \frac{1-a^{1-n}}{n-1}$$
Can you finish now?
A: 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 - u\right)^{\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 - u\right)^{\frac{1}{n} - 1}\:du \\
 &= \frac{1}{n}\left[\int_{0}^{1} u^{-\frac{1}{n}}\left(1 - u\right)^{\frac{1}{n} - 1}\:du  - \int_{0}^{\dfrac{1}{a^n + 1}} u^{-\frac{1}{n}}\left(1 - u\right)^{\frac{1}{n} - 1}\:du\right] \\
 &= \frac{1}{n}\left[B\left(1 - \frac{1}{n}, \frac{1}{n} \right) - B\left(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[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(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[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n},  \frac{1}{a^n + 1}\right)\right]
\end{align}
From here you can attempt your direct questions. 
A: 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.
