I'm asked to show that $f(x)=\frac{1}{1+x^2}$ is Lebesgue integrable over the real line and that its integral is $\pi$.

I can bound this function by the measurable function $\phi=\sum_{n=0}^{\infty}\frac{1}{1+n^2}\chi_{I_n}$ where $I_n = (n-1,n]\cup [n,n+1)$ and $\chi$ means characteristic function. This function $\phi$ can be written as the limit of its increasing sequence of partial sums, so I know that its integral is $2\sum_{n=0}^{\infty}\frac{1}{1+n^2}$ which converges. This ensures that the integral of $f$ is bounded, and hence $f$ is Lebesgue integrable.

I'm a bit stuck on how to show that $\int f$ is $\pi$ however. I suspect that I'm intended to write $\int f$ as a limit of series like that above. By cutting up those characteristic functions ever more finely, I can express the integral of $f$ as $$\lim_{k\to \infty} \frac{2}{k} \sum_{n=0}^{\infty} \frac{1}{1+(n/k)^2}$$ but I don't at all know how to evaluate that limit (though apparently it does evaluate to $\pi$).

So, I suppose my question is whether this is a good way to go about this (the overall point is to use the theory of Lebesgue measure and Lebesgue integration), and whether this last limit I've written is somehow easily seen to converge to $\pi$? Or, should I try to express the function as a limit of some other functions?

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    $\begingroup$ I wonder if this is a "trick" question. Do you have theorems that show if a function is "improperly" Riemann integrable it is Lebesgue integralable and the integrals agree (in the limit for the former)? If you do, just state the theorem and evaluate it in the usual way. $\endgroup$ – Josh Apr 29 '11 at 0:08
  • $\begingroup$ @Josh As more people weigh in I'm getting the feeling that your suggestion (and the answers below) are probably the expected way to go about it. There are fairly easy ways to see that the Riemann integral coincides with the Lebesgue one. I guess I'm just overthinking the question and putting unnecessary constraints on myself. $\endgroup$ – matt Apr 29 '11 at 0:14

I don't know that this is what they intend, but here are a few observations to get you started:

(1) On a finite interval, continuous functions which are Lebesgue integrable are Riemann integrable, and the two integrals agree. Thus, you can actually calculate $\int_a^b \frac{dx}{1+x^2}$ with the fundamental theorem of calculus.

(2) If $\int_X \left| f(x) \right| dx<\infty$, then dominated convergence allows you to say that $\int_X f dx= \lim \int_X f\chi_I dx=\lim \int_I f dx$ where the limit is over some increasing collection of subsets (intervals $I$) which exhaust your space.

These two together allow you to solve the problem essentially as you would in a regular calculus class.

If there is some particularly clever and elementary (but completely ad hoc) way to see this integral without either the fundamental theorem of calculus or complex analysis, I don't know it.

  • $\begingroup$ Just to compare and contrast, I have shown that if a continuous function on $\mathbb{R}$ is $L^1$, then the Lebesgue integral is the improper Riemann integral (and the fact that you showed that the function is $L^1$ is quite important here). ncmathsadist showed that this condition cannot be dropped: there are functions whose improper integrals converge but which are NOT Lebesgue integrable. Hopefully this makes you feel better about this solution actually having something to do with Lebesgue integration. $\endgroup$ – Aaron Apr 29 '11 at 0:57
  • $\begingroup$ Btw: Dominated convergence (and for that matter checking, whether the function is integrable -- using matts method above -- first) is not needed in this case, since we can argue by monotone convergence instead of dominated convergence, which simplifies things a bit... $\endgroup$ – Sam Apr 29 '11 at 5:17
  • $\begingroup$ @Sam Yeah, monotone convergence will do in this case (I had forgotten about that, it's been many years since I've done any measure theory). However, if you want to show the method works for more than just positive functions, you need something extra. We have that usual improper integrals will work for either (1) continuous $L^1$ functions or (2) continuous positive functions. However, the only integrals you have in the latter case but not the former are divergent, and so somewhat uninteresting. $\endgroup$ – Aaron Apr 29 '11 at 11:45

This doesn't really answer the question, but, if you're still interested in evaluating

$\displaystyle\lim_{k\to\infty} 2\sum_{n=1}^{\infty}\frac{k}{k^2+n^2},$

I think this method might work.

Choose $t>0$, and consider the Fourier series $\sum_{-\infty}^{\infty}c_ne^{inx}$ of $e^{tx}$, where

$c_n=\displaystyle\frac{1}{2\pi}\int_0^{2\pi}e^{(t-in)x}\,dx=\frac{(e^{2\pi t}-1)}{2\pi(t-in)}.$

Parseval's Theorem gives

$\displaystyle\frac{e^{4\pi t}-1}{4\pi t}=\frac{1}{2\pi}\int_0^{2\pi}e^{2tx}\,dx=\sum_{-\infty}^{\infty}\frac{(e^{2\pi t}-1)^2}{4\pi^2(t^2+n^2)}=\frac{(e^{2\pi t}-1)^2}{4\pi^2}\left(\frac{1}{t^2}+2\sum_{n=1}^{\infty}\frac{1}{t^2+n^2}\right).$

If I'm not mistaken above,

$\displaystyle\pi\frac{(e^{4\pi t}-1)}{(e^{2\pi t}-1)^2}-\frac{1}{t}=2\sum_{n=1}^{\infty}\frac{t}{t^2+n^2}.$

Now let $t\to\infty$, you get $\pi$.

  • 1
    $\begingroup$ Wow! That's a wonderfully clever solution. $\endgroup$ – Elchanan Solomon May 1 '11 at 4:01

The Lebesgue integral on the line and the Riemann integral coincide when the positive and negative parts of the function are integrable. To wit, suppose $f$ is a measurable function. Define $f^+(x) = f(x)$ if $f(x)\ge 0$. and 0 otherwise. Likewise, define $f^-(x) = -f(x)$ if $f(x) \lt 0$ and 0 otherwise. Then we know that $f = f^+ - f^-$ and $|f| = f^+ + f^-$.

The Lebesgue integral $$\int_{\infty}^\infty f(x)\, dx$$ for a measurable function $f$ exists provided that at least one of $\int f^-(x)\, dx$ or $\int f^+(x)\, dx$ is finite. In this case we have $$\int_{\infty}^\infty f(x)\, dx = \int_{\infty}^\infty f^+(x)\, dx - \int_{\infty}^\infty f^-(x)\, dx.$$
We should view this difference as being a difference between extended reals; it makes sense if $f$ is integrable.

The nub of the difference arises for an integral such as $$\int_0^\infty {\sin(x)\over x}\, dx.$$
This Riemann integral is defined as $$\lim_{T\to\infty} \int_0^T {\sin(x)\over x}\, dx.$$ and it converges in this sense to $\pi/2$.

Here we allow an ``unwarranted'' cancellation to occur. Both $\int \sin^+(x)/x\, dx$ and $\int \sin^-(x)/x\, dx$ are infinite, so the Lebesgue integral does not exist.

A bounded function on a bounded interval is Riemann integrable iff it is continuous almost everywhere in Lebesgue Measure (cf. Royden's Real Analysis).
The problem comes from cancellation, not from the irregularity of the integrand. Lebesgue integration, intrinsically, makes minimal smoothness demands on its integrands. These can be discontinuous everywhere.

  • $\begingroup$ A few typos: (i) Second display equation, second $f$ should be $f^+$, (ii) $\sin^-(x)$ should be $\sin^-(x)/x$, (iii) "eveywhere" is missing an "r", and (iv) "inegrands" is missing a "t". $\endgroup$ – cardinal Apr 29 '11 at 2:52

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