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This is a nice calculus problem, which seems not to be easy. Compute the limit: $$\lim_{n\rightarrow\infty}n\left(\frac{1}{2}-(n-1)\int_{0}^{1}\frac{x^n}{x^2+1} dx\right)$$. A solution would really be appreciated!!

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do you mean $\lim_{n \to \infty}$ or $\lim_{x\to \infty}$? –  Integrals Mar 28 at 17:04
    
It's n, not x. Now it's ok! –  user137654 Mar 28 at 17:05
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Also your thoughts about the problem would be appreciated. –  egreg Mar 28 at 17:29
    
Substituting atan(x)=t,I have been trapped with summing an infinite series which I think may require some taylor expansions or its beyond my scope. –  Awesome Mar 28 at 17:35

2 Answers 2

up vote 9 down vote accepted

A solution without assuming the knowledge of limiting behaviour of special functions:

The catch of this limit is that $x^n$ becomes very flat and zero-ish everywhere except for a steep part at the end. Only there, the $1+x^2$ has any effect when $n$ is large. We can't expand $x$ around $0$ because the series diverges at $x=1$. But we can expand around $1$. Say $u=1-x$. The idea is that $u$ is small where $x^n$ matters at all (essentially we are doing asymptotic expansion).

$$\frac{1}{1+x^2}=\frac{1}{2-2u+u^2}=\frac{1}{2}\frac{1}{1-(u-u^2/2)}=\frac{1}{2}(1+((1-x)-(1-x)^2/2)+((1-x)-(1-x)^2/2)^2+\cdots)$$

We used the geometric expansion, $\frac{1}{1-q}=1+q+q^2+\cdots$ for $q=u-u^2/2$ that represents the lowest order effects of $\frac{1}{1+x^2}$ around $x=1$, where the contribution of $x^n$ is strongest.

We still need to integrate by $x$ (remember, expansion didn't affect the $x^n$ term, just the denominator). So we get

$$\int_0^1 \frac{x^n}{1+x^2}dx=\frac12\int_0^1 x^n+x^n(1-x)-x^n(1-x)^2/2+x^n(1-x)^2+\cdots dx$$

$$\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{(n+2)(n+1)}+\cdots\right)$$ Other terms don't matter because the power of $n$ terms in the denominator rises with power of $(1-x)$, which can be verified by noticing $\int_0^1 x^n (1-x)^m dx=B(n+1,m+1)=\frac{n!m!}{(n+m+1)!}$, where $B$ is the Beta function.

Now, plug this back in the limit:

$$\lim_{n\to \infty} n\left(\frac{1}{2}-\frac{1}{2}\left(\frac{n-1}{n+1}+\frac{n-1}{(n+2)(n+1)}+\cdots\right)\right)$$ Now we can see that the quadratic term in $n$ is the last one that matters: the next one would limit to $0$ by default. This limit is easily evaluated to $$\frac{1}{2}\lim_{n\to \infty} n\left(\frac{n+1}{n+1}-\frac{n-1}{n+1}-\frac{n-1}{(n+2)(n+1)}+\cdots\right)$$ $$\frac{1}{2}\lim_{n\to \infty} \left(\frac{2n}{n+1}-\frac{n(n-1)}{(n+2)(n+1)}+\cdots\right)$$ $$\frac{1}{2}\lim_{n\to \infty} \left(2-1\right)=\frac12$$

It's not pretty, but it uses the idea inspired by the behaviour of the functions in the limit, and expansions under the integral sign are pretty standard, you just have to be careful to stay within the convergence radius.

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Nice observation about the behavior of $x^n$ on that interval for large $n$. –  Random Variable Mar 28 at 18:36
    
How do you get $\frac{1}{2}(1+(1-x)-(1-x)^2/2)+...$? –  user137654 Mar 28 at 18:51
    
@user137654 Geometric series $1+q+q^2+q^3+\cdots$ where $q=u-u^2/2=(1-x)-(1-x)^2/2$. –  orion Mar 28 at 19:11
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Beautiful answer @orion, you really outdid yourself. –  alvonellos Mar 28 at 23:03

We can show that $$ \lim_{n\rightarrow\infty}n\left(\frac{1}{2}-(n-1)\int_{0}^{1}\frac{x^n}{x^2+1} dx\right)=\frac{1}{2} $$ by first calculating the integral inside of the limit. Once we have the integral result, we can calculate the limit. The integral is given by $$ I\equiv\int_0^1 \frac{x^n}{x^2+1} dx=\frac{1}{4}\left(\psi^{(0)}\big(\frac{3+n}{4}\big)-\psi^{(0)}\big(\frac{1+n}{4}\big)\right), \ \Re(n) > -1. $$ A similar integral is given here(Closed form of integral.) Note also the poly-gamma function is defined by $\psi^{(0)}$ and is given here http://en.wikipedia.org/wiki/Polygamma_function. Thus we have $$ \lim_{n\rightarrow\infty}n\left(\frac{1}{2}-(n-1)\int_{0}^{1}\frac{x^n}{x^2+1} dx\right)=\lim_{n\rightarrow\infty}n\left(\frac{1}{2}-(n-1)I\right)=\frac{1}{2}. $$ The polygamma function is defined as $$ \psi^{m}(z)=\frac{d^{m+1}}{dz^{m+1}}\ln \Gamma(z) $$ where $\Gamma(z)$ is the gamma function. We can also write $$ \psi^{(0)}(z)=\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)} $$ where $\psi(z)$ is the digamma function.

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Maybe you should specify what $\psi_0$ is. –  orion Mar 28 at 17:10
    
Yes, please! It would be nice to see a complete solution. –  user137654 Mar 28 at 17:11
    
It must be an easier way to approach this. –  user137654 Mar 28 at 17:12
    
@orion Special functions are defined quite frequently on this page, so I wasn't sure what to explicitly write out. Thanks for your help, I added it now! –  Integrals Mar 28 at 17:14
    
@user137654 Well, the integral is expressed in terms of special functions. So there is not much more we can do, if the integral result was nicer than yes I agree there would be an easier way to approach it. But this integral is widely known, so it is not that difficult. –  Integrals Mar 28 at 17:15

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