Suppose that $f$ is continuous on $[0,1]$. ($f'(x)$ may or may not exist).

How can I show that

$$\lim_{n\rightarrow\infty} \int\limits_0^1 \frac{nf(x)}{1+n^2x^2} dx = \frac{\pi}{2}f(0)\;?$$

My attempt was to recognize that $\int_0^1\frac{n}{1+n^2x^2}dx=\tan^{-1}(nx)$. But I can't simply use integration by parts on the original integral as $f'(x)$ might not exist. Any hints on how to solve this?

Help much appreciated...

  • 2
    $\begingroup$ This fits into the standard theory of approximation to the identity. First show that the family of kernels $$K_n(x) = \frac{1}{\tan^{-1} n}\frac{n}{1+n^2 x^2} $$ satisfies all the desired property: 1. $\int_{0}^{1} K_n = 1$ and 2. $\int_{\delta}^{1} K_n \to 0$ as $n \to \infty$ for any $0 < \delta < 1$. This tells us that the mass of $K_n$ concentrates near $x = 0$ as $n\to\infty$. Now you can rely on the standard argument. $\endgroup$ Mar 14, 2013 at 4:49
  • 3
    $\begingroup$ Hi sos440 thanks for the comment but this is a problem in an elementary real analysis book. Is there a simpler way of going about it? I've never heard of kernels before...or is it necessary? And may I know what you mean by standard argument? thanks again $\endgroup$ Mar 14, 2013 at 4:55

4 Answers 4


Here's the crux of the argument. As it sounds like this is homework, I'll leave it to you to write some of the "close to" and "approximately" statements as proper inequalities.

As you point out, $$\lim_{n\rightarrow\infty} \int_0^1 \frac{n}{1+n^2x^2} \, dx = \frac{\pi}{2}.$$ Furthermore, here's what the graphs of these functions look like.

enter image description here

Now, since $f$ is continuous at zero, we can pick $\delta>0$ so that $f(x)$ is close to $f(0)$ for $x$ in $[0,\delta)$. Given that $\delta$, we can pick $n$ large enough so that $$\int_0^{\delta} \frac{n}{1+n^2x^2} \, dx \approx \frac{\pi}{2}$$ and $$\int_{\delta}^1 \frac{n}{1+n^2x^2} \, dx \approx 0.$$ Taken together you get that $$\int_0^1 f(x) \frac{n}{1+n^2x^2} \, dx \approx f(0)\frac{\pi}{2}.$$

  • $\begingroup$ Hi Mark. Thanks for this, it's very useful. For the record, it's not homework, it's a quiz problem that I did not manage to answer. Sadly the professor did not provide any solutions. $\endgroup$ Mar 14, 2013 at 5:24
  • $\begingroup$ But if you think about it, by applying your solution we can very well conclude that the above expression can also be equal to $f(1)\pi/2$. Am I correct? $\endgroup$ Mar 14, 2013 at 5:25
  • $\begingroup$ Edit: I realized that as $\frac{n}{1+n^2x^2} \rightarrow 0$ as $x \rightarrow 1$ then indeed, the above comment is void. Thanks a lot for your help and for taking the time to plot the graph! $\endgroup$ Mar 14, 2013 at 5:27
  • $\begingroup$ @user1237300 No problem! $\endgroup$ Mar 14, 2013 at 5:27

To simplify my life, let $\phi_n(x) = \frac{n}{1+n^2x^2}$. The idea is that outside a small neighborhood of $0$, $\phi_n(x)$ is small, so that part of the integral can be ignored. Inside the small neighborhood, $f(x)$ can be replaced by $f(0)$ since it is continuous.

Then we have \begin{eqnarray} \int_0^1 \phi_n(x)f(x)dx &=& \int_0^1 \phi_n(x)f(0)dx + \int_0^1 \phi_n(x)(f(x)-f(0))dx \\ &=& f(0) \arctan (n)+ \int_0^1 \phi_n(x)(f(x)-f(0))dx \end{eqnarray} So, we just need to bound the last term. $f$ is continuous, hence bounded on $[0,1]$, say $|f(x)| \le B$. Let $r>0$, then if $x \in [r,1]$ we have $\frac{n}{1+n^2x^2} \le \frac{1}{n r}$. This gives the following bound: \begin{eqnarray} |\int_0^1 \phi_n(x)(f(x)-f(0))dx | &\le& |\int_0^r \phi_n(x)(f(x)-f(0))dx | + | \int_r^1 \phi_n(x)(f(x)-f(0))dx |\\ &\le& \sup_{x \in [0,r)} |f(x)-f(0)|\int_0^1 \phi_n(x)dx + \frac{2B}{nr}\\ &\le& \arctan (n) \sup_{x \in [0,r)} |f(x)-f(0)| + \frac{2B}{nr} \\ &\le& \frac{\pi}{2} \sup_{x \in [0,r)} |f(x)-f(0)| + \frac{2B}{nr} \end{eqnarray} Choose $\epsilon>0$. Since $f$ is continuous at $0$, we can find an $r$ such that $\sup_{x \in [0,r)} |f(x)-f(0)| < \frac{\epsilon}{\pi}$, and choose $N$ such that $N > \frac{B}{\epsilon r}$. Then if $n \ge N$, we have $|\int_0^1 \phi_n(x)(f(x)-f(0))dx | < \epsilon$.

Hence $\lim_n \int_0^1 \phi_n(x)(f(x)-f(0))dx = 0$, and since $\lim_n \arctan n = \frac{\pi}{2}$, we have $\lim_n \int_0^1 \phi_n(x)f(x)dx = \frac{\pi}{2} f(0)$.


Here is an intuition behind the problem:

Suppose that a family of functions $K_n$ has the following property:

  1. $K_n \geq 0$ and $\int K_n = 1$. That is, $K_n$ has unit mass,
  2. $\int_{|x|\geq \delta} K_n \to 0$ as $n \to \infty$. That is, the mass of $K_n$ concentrates toward $0$ as $n$ grows.

We can give an intuitive interpretation of these conditions in terms of $K_n(x) \, dx$ as follows: Think of an integral $\int f(x) \, dx $ as the sum of infinitesimal masses $f(x) \, dx$, each of which is located on the interval $[x, x+dx]$. Then

  1. the total sum the infinitesimal masses $ K_n(x) \, dx $ is equal to one,
  2. $ K_n(x) \, dx \to 0$ if $x$ is away from $0$ as $n\to\infty$.

Thus we expect that $ K_n(0) \, dx \to 1$ and hence

\begin{align*} \lim_{n\to\infty} \int f(x) K_n(x) \, dx &= \lim_{n\to\infty} \sum f(x) K_n(x) \, dx \\ &= \sum f(x) \lim_{n\to\infty} K_n(x) \, dx = f(0). \end{align*}

This observation(?) suggests that we should divide the behavior of $K_n$ into

  • near-the-origin part where the mass of $K_n$ accumulates and
  • away-from-the-origin part where the mass of $K_n$ vanishes.

Now let us return the the question of devising a rigorous proof. Let $f$ be continuous on $[0, 1]$. In particular, $f$ is bounded by some constant $M > 0$ and continuous at $x = 0$. Thus for any $\epsilon > 0$, there exists $\delta > 0$ such that

$$ |x| < \delta \Longrightarrow |f(x) - f(0)| < \epsilon. $$

Now let

$$ K_n(x) = \frac{1}{\tan^{-1}n} \frac{n}{1+n^2 x^2}. $$

Then it is clear that

$$ \int_{0}^{1} K_n(x) \, dx = \frac{1}{\tan^{-1} n} \int_{0}^{n} \frac{dx'}{1+x'^2} = 1$$

and similarly

\begin{align*} \int_{\delta}^{1} K_n(x) \, dx &= \frac{1}{\tan^{-1} n} \int_{n\delta}^{n} \frac{dx'}{1+x'^2} \\ &\leq \frac{1}{\tan^{-1} n} \int_{n\delta}^{\infty} \frac{dx'}{1+x'^2} = \frac{\tan^{-1} 1/(n\delta)}{\tan^{-1} n} \to 0 \quad \text{as } n\to\infty. \end{align*}

Keeping these observations in mind, we make the following decomposition.

\begin{align*} \left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right| &\leq \left| \int_{0}^{1} f(x) K_n(x) \, dx - f(0) \right| \tan^{-1}n + \left|\tan^{-1} n - \frac{\pi}{2} \right| \left| f(0) \right|. \end{align*}

Then dividing the integral term into two parts with one near from the origin and the other away from the origin, we observe that

\begin{align*} \left| \int_{0}^{1} f(x) K_n(x) \, dx - f(0) \right| &= \left| \int_{0}^{1} (f(x) - f(0)) K_n(x) \, dx \right| \\ &\leq \int_{0}^{1} \left| f(x) - f(0) \right| K_n(x) \, dx \\ &\leq \int_{0}^{\delta} \left| f(x) - f(0) \right| K_n(x) \, dx + \int_{\delta}^{1} \left| f(x) - f(0) \right| K_n(x) \, dx \\ &\leq \int_{0}^{\delta} \epsilon K_n(x) \, dx + \int_{\delta}^{1} 2M K_n(x) \, dx \\ &\leq \epsilon + 2M \int_{\delta}^{1} K_n(x) \, dx, \end{align*}

and hence

\begin{align*} \left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right| &\leq \frac{\pi}{2} \epsilon + \pi M \int_{\delta}^{1} K_n(x) \, dx + \left| f(0) \right| \tan^{-1} \frac{1}{n}. \end{align*}

Taking $\limsup_{n\to\infty}$, we have

\begin{align*} \limsup_{n\to\infty} \left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right| &\leq \frac{\pi}{2} \epsilon. \end{align*}

But since this is true for any $\epsilon > 0$, we must have

\begin{align*} \limsup_{n\to\infty} \left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right| = 0 \quad \Longleftrightarrow \quad \lim_{n\to\infty} \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx = \frac{\pi}{2} f(0). \end{align*}

  • $\begingroup$ The intuition?? $\endgroup$ Mar 14, 2013 at 5:22
  • $\begingroup$ @MarkMcClure, Yes, I thought that this explanation appeals our intuitive understanding of the integral (especially in measure-theoretic viewpoint). But if it wasn't, well... maybe that means I am not a good storyteller. :( $\endgroup$ Mar 14, 2013 at 5:28
  • $\begingroup$ Although, I didn't read your answer in detail, my point was that the inequalities and such look to go far beyond intuition - perhaps to the point of an answer. It can be difficult, of course, to judge the level of detail to provide in this context. $\endgroup$ Mar 14, 2013 at 5:30

Take $\epsilon>0$. There exists $\alpha >0$ such that $|f(x)-f(0)|\leq \epsilon/2$ for all $0\leq x\leq \alpha$. Then decompose your integral $L_n$ into three pieces as follows $$ L_n=\int_0^\alpha\frac{nf(0)}{1+n^2x^2}dx+\int_0^\alpha\frac{n(f(x)-f(0))}{1+n^2x^2}dx+\int_\alpha^1\frac{nf(x)}{1+n^2x^2}dx=I_n+J_n+K_n. $$ First, note that $$ I_n=f(0)\arctan(n\alpha)\longrightarrow \frac{\pi}{2}f(0). $$ Second, we have $$ |J_n|\leq \int_0^\alpha\frac{n|f(x)-f(0)|}{1+n^2x^2}dx\leq \frac{\epsilon}{2}\int_0^\alpha\frac{n}{1+n^2x^2}dx=\frac{\epsilon}{2}\arctan(n\alpha)\longrightarrow\frac{\pi\epsilon}{4}. $$ Third, denoting $\|f\|_\infty$ the maximum of $|f|$ on $[0,1]$, we get $$ |K_n|\leq \|f\|_\infty \int_\alpha^1\frac{n}{1+n^2x^2}dx\leq \|f\|_\infty \int_\alpha^1\frac{n}{1+n^2\alpha^2}dx \leq \frac{\|f\|_\infty}{n\alpha^2}\longrightarrow 0. $$ So $L_n-\frac{\pi}{2}f(0)$ is bounded by a sequence which converges to $\frac{\pi\epsilon}{4}<\epsilon$. It follows that there exists $N$ such that $$ |L_n-\frac{\pi}{2}f(0)|\leq \epsilon\qquad\forall n\geq N. $$ This proves the desired result, namely that $L_n$ converges to $\frac{\pi}{2}f(0)$.

  • $\begingroup$ Sorry, I clicked on the wrong button. I fixed it. $\endgroup$
    – Tomás
    Mar 14, 2013 at 13:52
  • $\begingroup$ @Tomás Ah! So my answer is correct, good. $\endgroup$
    – Julien
    Mar 14, 2013 at 13:59

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