Compute $\lim_{n\to\infty}\int_0^n\frac{f(\frac{x}{n})}{1+x^2}dx$

Question

In the middle of an exercise I have to compute:

$$\lim_{n\to\infty}\int_0^n\frac{f(\frac{x}{n})}{1+x^2}dx$$

($f\in C[0,1]$) It is obvious that this has to be equal to: $$\int_0^{\infty}\frac{f(0)}{1+x^2}dx=\frac{\pi}{2}f(0)$$

But I am not sure how to prove it in a rigurous way, since you can have divergence problems or some weird situations.

Answer 1

Suppose $|f(x)|\le B$ on $[0,1]$, and given $\epsilon>0$, we can choose $\delta>0$ such that $|f(x)-f(0)|<\epsilon$ for $x\in [0, \delta]$. Now $$|\int_0^n \frac{f(x/n)-f(0)}{1+x^2}dx| \le \int_0^{\delta n} |\frac{f(x/n)-f(0)}{1+x^2}| + \int_{\delta n}^\infty \frac{2B}{1+x^2}$$ $$\le \int_0^{\delta n} \frac{\epsilon}{1+x^2} + \int_{\delta n}^\infty\frac{2B}{1+x^2}\le \epsilon\int_0^\infty\frac{1}{1+x^2}+\int_{\delta n}^\infty \frac{2B}{1+x^2}$$

Because $\int_0^\infty\frac{1}{1+x^2}dx<\infty$, for sufficiently large $n$, $\int_{\delta n}^\infty \frac{2B}{1+x^2}dx\le \epsilon$, hence finally

$$|\int_0^n \frac{f(x/n)-f(0)}{1+x^2}dx|\le (\int_0^\infty\frac{1}{1+x^2}+1)\epsilon$$

More generally, this can be used to show that $$\int_0^n f(x/n)g(x)dx\rightarrow f(0)\int_0^\infty g(x)dx$$ if $f$ is bounded on $[0, 1]$ and right continuous at $0$, and $\int_0^\infty |g(x)|dx<\infty$.

Answer 2

$\lim_{n\to\infty}\int_0^n\frac{f(\frac{x}{n})}{1+x^2}dx $

This looks like the usual split the integral and see what happens, so I'll do that.

For $0 < c < n$, let $I_n =\int_0^n\frac{f(\frac{x}{n})}{1+x^2}dx $, $J_n(c) =\int_0^{c}\frac{f(\frac{x}{n})}{1+x^2}dx $, $K_n(c) =\int_c^n\frac{f(\frac{x}{n})}{1+x^2}dx $, so $I_n =J_n(c)+K_n(c) $.

I want $K_n(c) $ to be small and $J_n(c)$ to be close to $\int_0^{c}\frac{f(0)}{1+x^2}dx $ so we can let $c\to\infty$ and get your result.

Looking at $J_n(c)$, we want to make $c/n \to 0$.

Let $M =\max_{0 \le x \le 1} |f(x)| $.

$\begin{array}\\ |K_n(c)| &=|\int_c^n\frac{f(\frac{x}{n})}{1+x^2}dx|\\ &\le|\int_c^n\frac{M}{1+x^2}dx|\\ &=M\int_c^n\frac{1}{1+x^2}dx\\ &=M\arctan(x)|_{x=c}^n\\ &=M(\arctan(n)-\arctan(c))\\ &=M\arctan(\frac{n-c}{1+nc})\\ &\le M(\frac{n-c}{1+nc})\\ &\le M(\frac{n-n^a}{1+n^{1+a}}) \qquad\text{if } c = n^a, 0 < a < 1\\ &\le M(\frac{n}{n^{1+a}})\\ &= M(\frac{1}{n^a})\\ \end{array} $

So we can make $K_n(c) $ small by making $c$ of order $n^a$.

For $J_n(c)$, since $f$ is continuous, $|f(z)-f(0)| \to 0 $ as $z \to 0$. In particular, $\max(|f(z)-f(0)|) \to 0 $ for $0 < z < \frac1{n^a}$ so, if $g(n, a) =\max(|f(z)-f(0)|)|_{x=0}^{\frac1{n^a}} $ then, for any $0 < \epsilon < 1,0 < a < 1$, $g(n, a) < \epsilon $ for large enough $n$.

Therefore, for any $\epsilon > 0, 0 < a < 1$ for large enough $n$,

$\begin{array}\\ |J_n(c)-\int_0^{c}\frac{f(0)}{1+x^2}dx| &\le\int_0^{c}\frac{|f(\frac{x}{n})-f(0)|}{1+x^2}dx\\ &\le\int_0^{c}\frac{\epsilon}{1+x^2}dx\\ &\le \frac{\epsilon\pi}{2}\\ &\to 0\\ \end{array} $