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Suppose $f\colon [0, 1] \to \mathbb{R}$ is a continuous function, Find the following limit: $$\lim\limits_{n\to\infty}\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx$$ I know the answer is equal $\frac{\pi}{2}f(0) $, but I don't prove that.

Any ideas or insight would be greatly appreciated.

pie
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user62498
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    $=\int_0^1 f(x),d\arctan(nx)\to\int_0^1 f(x),d\frac{\pi}2I(x>0)=\frac {\pi}2f(0)$. – A.S. Mar 01 '16 at 20:59
  • @Dear A.S,would you explain your answer please – user62498 Mar 01 '16 at 21:05
  • The first step is just variable change, the second is convergence in distribution and the third is evaluation of a Riemann-Stieltjes integral w.r.t. step function. – A.S. Mar 01 '16 at 21:07

3 Answers3

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With the change $y=nx$ you get $$\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx = \int _{0}^{n}\frac{f(\frac{y}{n})}{1+y^2}\,dx.$$ Now Lebesgue theorem should allow you to conclude.

C. Dubussy
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I get that the limit is $\frac12 (1+\pi \coth(\pi))f(0) $. Since this disagrees with what is presumably the correct answer, I would like to know where and what my error is.

Thanks.

$\begin{array}\\ \lim\limits_{n\to\infty}\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx &=\lim\limits_{n\to\infty}\frac1{n}\sum_{k=0}^{n-1}\frac{nf(k/n)}{1+n^2(k/n)^2} \qquad\text{(the error is probably here, but I'm not sure why)}\\ &=\lim\limits_{n\to\infty}\sum_{k=0}^{n-1}\frac{f(k/n)}{1+k^2}\\ &=\lim\limits_{n\to\infty}\sum_{0\le k\lt n^a}\frac{f(k/n)}{1+k^2} +\lim\limits_{n\to\infty}\sum_{n^a\ge k\le n-1}\frac{f(k/n)}{1+k^2} \qquad\text{where }\frac12 < a <1\\ &=\lim\limits_{n\to\infty}\sum_{0\le k\lt n^a}\frac{f(k/n)}{1+k^2} \qquad\text{since }\lim\limits_{n\to\infty}\sum_{n^a\le k\le n-1}\frac{f(k/n)}{1+k^2}=0\\ &=f(0)\lim\limits_{n\to\infty}\sum_{0\le k\lt n^a}\frac{1}{1+k^2}\\ &=f(0)\sum_{k=0}^{\infty}\frac{1}{1+k^2}\\ &=\frac12 (1+\pi \coth(\pi))f(0) \qquad\text{(according to Wolfy)}\\ &\approx 2.076674f(0) \end{array} $

marty cohen
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    Your mesh size is dependent on $n$. I don't think it's kosher. You basically collapsed two limiting processes into one. – A.S. Mar 01 '16 at 21:19
  • Your mesh need to be $o(\frac 1 n)$ to capture changing variations of $f_n$. – A.S. Mar 01 '16 at 21:33
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Since $f $ is continuous at $[0,1]$ then $\forall \epsilon >0, \ \exists \delta> {0} $ such that if $|x|<\delta \ , |f(x)- f(0) |<\epsilon $. Since Continuous functions attend both maximum and minimum value on compact sets, let $M = 2\sup_{x\in [0,1]} |f(x) |$

$$\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx= \int_{0}^{\delta} \frac{nf(x)}{1+n^2x^2}dx+\int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx $$ $$= \int_{0}^{\delta} \frac{n(f(x)- f(0))}{1+n^2x^2}dx+ \int_{0}^{\delta} \frac{nf(0)}{1+n^2x^2}dx +\int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx $$

$$=f(0) \arctan(n\delta)+ \int_{0}^{\delta} \frac{n(f(x)- f(0))}{1+n^2x^2}dx \int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx $$

Since $ \displaystyle \lim_{x \to \infty } \arctan(x) =\frac{\pi}{2}$ then for $\epsilon>0, \ \exists \alpha \in \mathbb{R} $ such that if $ x >\alpha $ then $|\arctan(x)- \frac{\pi}{2}| < \epsilon$ choose $N$ such that $N\delta >\alpha$ for all $n \ge N $

$$\bigg|\frac{\pi f(0)}{2}-f(0) \arctan(n\delta)- \int_{0}^{\delta} \frac{n(f(x)- f(0))}{1+n^2x^2}dx- \int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx \bigg|$$

$$\le 2M\left|\frac{\pi }{2}- \arctan(n\delta) \right|+\epsilon \arctan(n\delta) < 2M\epsilon + \epsilon^2 $$

Which implies that$\displaystyle \lim_{n \to \infty }\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi f(0)}{2} $

pie
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