# 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 $$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$

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.

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

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?

Certainly not the most compact approach, but:

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

Now: $$$$J_n(a) = \int_{0}^{a} \frac{1}{w^n + 1}\:dw$$$$

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

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

$$$$J_n(a) = \frac{1}{n}\int_{0}^{x^n} \frac{1}{t + 1}t^{\frac{1}{n} - 1}\:dt$$$$

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.

• I believe that the limit is $a-1$. – Mark Viola Dec 17 '18 at 13:58
• @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. – user150203 Dec 18 '18 at 5:19

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.