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

  1. Prove that $0\le I_n(1) \le \frac{1}{n+1}$
  2. Compute $\lim_{n\to\infty} I_n(a)$

My attempt:

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

  2. 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.

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  • $\begingroup$ I started to make your layout readable please look at what I have done and edit your posting. $\endgroup$ – Nathanael Skrepek Dec 16 '18 at 13:39
  • $\begingroup$ I have a feeling this may help you - functions.wolfram.com/GammaBetaErf/Gamma/29 $\endgroup$ – user150203 Dec 17 '18 at 9:35
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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?

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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.

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    $\begingroup$ I believe that the limit is $a-1$. $\endgroup$ – Mark Viola Dec 17 '18 at 13:58
  • $\begingroup$ @MarkViola - I'm sure you're correct. I just realised where I made my error, the incomplete beta funciton should not have '0' as it's final value and should be '1'. I will amend now. Thank you for the pickup. Much appreciated. $\endgroup$ – user150203 Dec 18 '18 at 5:19
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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.

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