Compute $\lim_{n\to\infty}\int_0^n\frac{f(\frac{x}{n})}{1+x^2}dx$ 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.
 A: 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 $\int_0^\infty |g(x)|dx<\infty$.
A: $\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}
$
