## Riemann-Lebesgue

The purpose of this post is to discuss the Riemann-Lebesgue Lemma, why it makes sense, and some interesting applications among which we have ${\int_{-\infty}^\infty \frac{\sin(x)}{x}dx = \pi}$, the proof of which, in my opinion, explains a lot more clearly why it should equal ${\pi}$.

Recently I have grown quite fond of the Riemann-Lebesgue Lemma, the Fourier Transform and Fourier Series, through studying Tauberian Theorems in Analytic Number Theory, and Partial Differential Equations.

The Riemann-Lebesgue lemma states that

$\displaystyle \lim_{n\rightarrow\infty}\int_A^B f(x)\,\sin(n\,x)dx = 0\,. \ \ \ \ \ (1)$

for quite a general set of functions ${f}$.

In the case of continuous functions the interpretation is that ${\sin(n x)}$ has periods of length proportional to ${1/n}$, in which, being ${f}$ continuous, the variation in ${f}$ is really small, while the integral of ${\sin(nx)}$ in this periods is ${0}$. Thus we will be adding roughly ${\epsilon/n}$ for each cycle, being ${\epsilon}$ the change in ${f}$ during the cycle. Adding these up leaves us with something of the order of ${\epsilon}$, which can be made arbitrarily small as ${n\rightarrow\infty}$.

Lemma 1 Let ${f\colon [A,B]\rightarrow \mathbb{R}}$ be continuous, then we have

$\displaystyle \lim_{n\rightarrow\infty}\int_A^B f(x)\,\sin(n\,x)dx = 0\,. \ \ \ \ \ (2)$

Proof: Observe that ${f}$ is uniformly continuous on the compact set ${[A,B]}$. Thus, given an arbitrary ${\epsilon > 0}$ there is ${\delta > 0}$ such that ${|f(x)-f(y)|\leq \epsilon}$ whenever ${|x-y|\leq \delta}$.

Next consider ${N \geq \frac{2\pi}{\delta}}$ so that ${\frac{2\pi}{N} \leq \delta}$, we partition the integration intervals into pieces of size less than ${\delta}$

$\displaystyle \int_A^B f(x)\,\sin(N\,x)dx = \int_A^{\frac{2\pi}{N}\,\lceil \frac{N\,A}{2\pi} \rceil}\clubsuit + \sum_{k=\lceil \frac{N\,A}{2\pi} \rceil}^{\lfloor \frac{N\,B}{2\pi} \rfloor - 1} \int_{\frac{2\pi\,k}{N}}^{\frac{2\pi\,k}{N}+2\pi/N}{\clubsuit} + \int_{\frac{2\pi}{N}\,\lfloor \frac{N\,B}{2\pi} \rfloor}^B\clubsuit\,,$

where the ${\clubsuit}$ are a short-hand for ${f(x)\,\sin(N\,x)dx}$.

Here it is clear that the first and the last integrals tend to ${0}$ as ${N\rightarrow\infty}$ because

$\displaystyle \left|\int_A^{\frac{2\pi}{N}\,\lceil \frac{N\,A}{2\pi} \rceil}f(x)\,\sin(N\,x)dx\right|\leq \int_A^{\frac{2\pi}{N}\,\lceil \frac{N\,A}{2\pi} \rceil} |f(x)|dx \leq \left(\frac{2\pi}{N}\,\left\lceil \frac{N\,A}{2\pi} \right\rceil-A\right)\|f\|_\infty\,,$

which $\rightarrow 0$, and similarly for the other one. Thus, for a certain ${N_\epsilon\in\mathbb{Z}_{\geq 1}}$ with ${N_\epsilon\geq\frac{2\pi}{\delta}}$ we may assume that ${N\geq N_\epsilon}$ implies that these two integrals have absolute value less than ${\epsilon}$.

As for the case of ${ \int_{\frac{2\pi\,k}{N}}^{\frac{2\pi\,k}{N}+2\pi/N}f(x)\,\sin(N\,x)dx\,, }$ with ${k}$ integer, note that ${\int_{\frac{2\pi\,k}{N}}^{\frac{2\pi\,k}{N}+2\pi/N}\sin(N\,x)dx = 0}$ (it goes through a whole cycle!) implies

\begin{aligned} \left|\int_{\frac{2\pi\,k}{N}}^{\frac{2\pi\,k}{N}+2\pi/N}f(x)\,\sin(N\,x)dx\right| &= \left|\int_{\frac{2\pi\,k}{N}}^{\frac{2\pi\,k}{N}+2\pi/N}\Big(f(x)-f\left(\frac{2\pi\,k}{N}\right)\Big)\,\sin(N\,x)dx\right| \\ &\leq \int_{\frac{2\pi\,k}{N}}^{\frac{2\pi\,k}{N}+2\pi/N}\Big|f(x)-f\left(\frac{2\pi\,k}{N}\right)\Big|dx \\ &\leq \frac{2\pi}{N}\,\epsilon\,, \end{aligned}

where the last inequality follows from the fact that the distance to the left-most point of the interval is less than ${\delta}$ and so ${\Big|f(x)-f\left(\frac{2\pi\,k}{N}\right)\Big|\leq \epsilon}$.

Hence, collecting terms, we get

$\displaystyle \left|\int_A^B f(x)\,\sin(N\,x)dx \right| \leq \epsilon + \left(\frac{N\,B}{2\pi} - \frac{N\,A}{2\pi}\right)\,\frac{2\pi}{N}\,\epsilon + \epsilon = (B-A+2)\,\epsilon\,,$

for ${N\geq N_\epsilon}$. $\Box$

Example 1. We will prove that (assuming its existence, which may be shown by the alternating series test):

${\displaystyle {\int_{0}^\infty \frac{\sin(x)}{x}dx = \frac{\pi}{2}\,.}} \ \ \ \ \ (3)$

Let us denote this integral by ${I}$. Consider then

$\displaystyle I_N=\int_0^{2\pi\,N} \frac{\sin(x)}{x}dx\,, \qquad J_N = \int_0^{\pi\,N} \frac{\sin(x)}{x}dx\,, \ \ \ \ \ (4)$

which clearly satisfy ${I_N\rightarrow I}$ and ${J_N\rightarrow I}$. Setting ${x = N\,u}$ we get

$\displaystyle I_N=\int_{0}^{2\pi} \frac{\sin(N\,u)}{u}du\,,\qquad J_N = \int_0^{\pi} \frac{\sin(N\,u)}{u}du$

Note that ${I_N}$ almost fits into the Riemann-Lebesgue, but ${1/u}$ is not continuous on ${[0,2\pi]}$, it tends to infinity close to ${u=0}$. To remedy this problem (somewhat surprisingly) we look first at

$\displaystyle \frac{1}{u} - \frac{1}{\sin(u)}\,,$

here we run into trouble again, but at the points ${u=\pi}$ and ${u=2\pi}$ instead of ${u=0}$.

Why insist with ${\frac{1}{\sin(u)}}$ ? Well, in fact, it is well-known that

$\displaystyle \int_{0}^{2\pi} \frac{\sin(N\,u)}{\sin(u)}du = \begin{cases} 2\pi &\text{if }N\text{ is odd}\\ 0 &\text{otherwise}\,, \end{cases}$

for a proof look at ${(*)}$ below.

To finally eliminate the discontinuities, we note that

$\displaystyle f(u)=\frac{1}{\sin(u)} - \frac{1}{u} + \frac{1}{u-\pi} - \frac{1}{u-2\pi} \ \ \ \ \ (5)$

can be extended continuously (by considering the limits at ${0,\pi,2\pi}$) to ${u\in [0,2\pi]}$.

The choice should not be surprising as

$\displaystyle \frac{1}{\sin(u)} = \frac{1}{\sin(u)-\sin(\pi\,k)} \sim \frac{1}{(u-\pi\,k)\,\sin^\prime(\pi\,k)} = \frac{(-1)^k}{u-\pi\,k}\,,$

when ${u\rightarrow\pi\,k}$. In fact we are eliminating the contribution from the simple poles!

We are very lucky because

$\displaystyle \int_0^{2\pi}\frac{\sin(N\,u)}{u-\pi}du = \int_{-\pi}^{\pi} \frac{\sin(N\,u+N\,\pi)}{u}du = (-1)^N\int_{-\pi}^{\pi} \frac{\sin(N\,u)}{u}du = (-1)^N\,2J_N\,,$

while

$\displaystyle \int_0^{2\pi}\frac{\sin(N\,u)}{u-2\pi}du = \int_{-2\pi}^{0} \frac{\sin(N\,u+2N\,\pi)}{u}du = \int_{-2\pi}^{0} \frac{\sin(N\,u)}{u}du = I_N\,.$

Thus

$\displaystyle \int_0^{2\pi} f(u)\,\sin(N\,u)du = \int_{0}^{2\pi} \frac{\sin(N\,u)}{\sin(u)}du - I_N +(-1)^N\,2J_N-I_N\,,$

and considering ${N}$ odd we have

$\displaystyle \int_0^{2\pi} f(u)\,\sin(N\,u)du = 2\pi - 2I_N - 2J_N \,, \ \ \ \ \ (6)$

which shows ${I = \pi/2}$ by the Riemann-Lebesgue lemma. ${\square}$

The Riemann-Lebesgue lemma is more general actually.

Lemma 2 Consider ${g\in L^1([A,B],\mathcal{M},m)}$ where ${\mathcal{M}}$ is the ${\sigma}$-algebra of the Lebesgue-measurable sets and ${m}$ is the Lebesgue measure. Then

$\displaystyle \lim_{n\rightarrow\infty}\int_A^B g(x)\,\sin(n\,x)dx = 0\,\, \ \ \ \ \ (7)$

where ${\int_A^B h(x) dx}$ is to be interpreted as ${\int_{[A,B]} h dm}$, the Lebesgue integral.

Proof: The continuous functions ${C^1[A,B]}$ are dense in ${L^1([A,B])}$ under the Lebesgue measure ${m}$.Thus it follows that, given a measurable function ${g\in L^1([A,B])}$ we have a sequence of continuous functions ${(f_n)}$ such that ${\displaystyle\int_A^B |g(x)-f_n(x)|dx \rightarrow 0}$ so that, decomposing ${g(x)\,\sin(N\,x)}$ as ${(g(x)-f_n(x))\,\sin(N\,x) + f_n(x)\,\sin(N\,x)}$ we have

$\displaystyle \left|\int_A^B g(x)\,\sin(N\,x) dx\right| \leq \int_A^B |g(x)-f_n(x)| dx + \left|\int_A^B f_n(x)\sin\left(N\,x\right)dx\,\right|\,,$

from where the result follows upon making ${n}$ and ${N}$ sufficiently large in order. $\Box$

Of course, the proofs read verbatim when we change ${\sin}$ for any bounded periodic-function ${S}$, since the principle of the proof stays the same.

Lemma 3 Consider ${g\in L^1([A,B],\mathcal{M},m)}$ where ${\mathcal{M}}$ is the ${\sigma}$-algebra of the Lebesgue-measurable sets and ${m}$ is the Lebesgue measure. Let ${S(x)}$ be a bounded periodic measurable function on ${\mathbb{R}_{\geq 0}}$, with period ${\tau}$, such that ${\int_0^\tau S(x)dx=0}$. Then

$\displaystyle \lim_{n\rightarrow\infty}\int_A^B g(x)\,S(n\,x)dx = 0\,\, \ \ \ \ \ (8)$

where ${\int_A^B h(x) dx}$ is to be interpreted as ${\int_{[A,B]} h dm}$, the Lebesgue integral.

Another possible extension is the case in which we have a continuously differentiable function ${g}$, in such a case we may say much more about the rate of decay.

Lemma 4 Consider ${g\in C^1([A,B])}$. Then

$\displaystyle \int_A^B g(x)\,\sin(n\,x)dx = O(1/n)\,\, \ \ \ \ \ (9)$

as ${n\rightarrow\infty}$

Proof: Applying integration by parts

\begin{aligned} \int_A^B g(x)\,\sin(nx)dx &= \int_A^B g(x)\,\left(-\frac{\cos(nx)}{n}\right)^\prime dx \\ &= \frac{g(A)\,\cos(nA) - g(B)\,\cos(nB)}{n} + \frac{1}{n}\,\int_A^B g^\prime(x)\,\cos(nx) dx\,, \end{aligned}

and therefore

$\displaystyle \left|\int_A^B g(x)\,\sin(nx)dx\right| \leq \frac{|g(A)|+|g(B)|+\|g^\prime\|_{L^1}}{n}\,, \ \ \ \ \ (10)$

thus proving the result. $\Box$

Example 2. We will prove that

$\displaystyle { \sum_{k=1}^\infty \frac{\sin(k\,x)}{k} = \frac{\pi-x}{2}\,,} \ \ \ \ \ (11)$

for ${x\in (0,2\pi)}$, and that this convergence is uniform on ${[\epsilon,2\pi-\epsilon]}$ for any ${\epsilon>0}$.

1. First show that ${\sum_{k=1}^{N-1} \cos(k\,x) = - \frac{1}{2} - \frac{\cos\left(N x\right)}{2} + \frac{\sin\left(N x\right) \cos\left(x/2\right)}{2 \,\sin(x/2)} }$.
2. Integrate it in ${x}$ from ${0}$ to ${x}$ to get

$\displaystyle \sum_{k=1}^{N-1} \frac{\sin(k\,x)}{k} = -\frac{x}{2} - \frac{\sin(N\,x)}{2\,N} + \frac{1}{2}\,\int_0^x \sin\left(N y\right)\,\frac{\cos\left(y/2\right)}{\sin\left(y/2\right)}dy\,.$

3. To make the integrand of the form ${\sin(N y)\,g(y)}$ with ${g(y)}$ differentiable when ${y \leq x < \pi}$, we must remove the pole of ${{\cos\left(y/2\right)}/{\sin\left(y/2\right)}}$ at ${y=0}$. Consider

$\displaystyle g(y) = \frac{\cos\left(y/2\right)}{\sin\left(y/2\right)} - \frac{2}{y}\,,$

then ${g(z)}$ is holomorphic on ${|z|<2\pi}$ because we removed the pole, in particular it is continuously differentiable on ${[0,2\pi-\epsilon]}$ for any ${\epsilon > 0}$.

4. Thus

$\displaystyle \int_0^x \sin(N y) g(y) dy = \int_0^x \sin\left(N y\right)\,\frac{\cos\left(y/2\right)}{\sin\left(y/2\right)}dy - 2\,\int_0^x \frac{\sin(N y)}{y}dy$

is ${O(1/N)}$ when ${x\in [0,2\pi-\epsilon]}$.

5. Observe that, by what we know from Example 1, we have

$\displaystyle \int_0^x \frac{\sin(N y)}{y}dy = \int_0^{N\,x} \frac{\sin(y)}{y}dy = \frac{\pi}{2} - \int_{N\,x}^\infty \frac{\sin(y)}{y}dy\,,$

and here ${\int_{N\,x}^\infty \frac{\sin(y)}{y}dy = O(1/N)}$ when ${x \geq \epsilon > 0}$. Indeed

\begin{aligned} \int_{N\,x}^\infty \frac{\sin(y)}{y}dy &= \int_{x}^\infty \frac{\sin(N y)}{y}dy \\ &= \int_{x}^\infty \frac{\left(-\cos(N y)/N\right)^\prime}{y}dy \,, \end{aligned}

thus, integrating by parts, we get ${\left|\displaystyle\int_{N\,x}^\infty \frac{\sin(y)}{y}dy\right| \leq \frac{1+1/\epsilon}{N} }$.

Proof of ${(*)}$. Somewhat surprisingly, the computation of

$\displaystyle I_{n,m} = \int_{0}^{2\pi} \left(\frac{\sin(nx)}{\sin(x)}\right)^mdx\,,$

boils down to a matter of evaluating a particular coefficient of a polynomial!

Let us write ${\sin(t) = \frac{e^{i\,t} - e^{-i\,t} }{ 2\,i}}$, so that we get

$\displaystyle I_{n,m} = \int_{0}^{2\pi} \left(\frac{e^{n\,ix} - e^{-n\,ix} }{e^{ix} - e^{-ix}}\right)^mdx\,.$

Here observe that ${ \displaystyle\frac{a^n-b^n}{a-b} = a^{n-1} + a^{n-2}\,b + \ldots + a\,b^{n-2} + b^{n-1}\,,}$ so that

$\displaystyle I_{n,m} = \int_{0}^{2\pi} \left(\sum_{j=0}^{n-1} \left(e^{ ix}\right)^j \left(e^{-ix}\right)^{n-1-j} \right)^mdx= \int_{0}^{2\pi} \left(\sum_{j=0}^{n-1} e^{(2j-(n-1))\,ix} \right)^mdx\,.$

Note that we have

$\displaystyle \int_0^{2\pi} e^{n\,i t } dt = \int_{0}^{2\pi} e^{n\,i t } dt = \begin{cases} 2\pi, & \text{if } n = 0\,, \\ 0, &\text{otherwise } \,, \end{cases}$

and therefore, by factoring ${e^{-(n-1)\,ix}}$ from each of the ${m}$ factors, we get

$\displaystyle I_{n,m} = 2\pi\,[z^{m\,(n-1)}]\Bigg\{ \Big(\sum_{j=0}^{n-1} z^{2j}\Big)^m\Bigg\}\,,$

where the notation ${[z^k]\{p(z)\}}$ denotes the coefficient of ${z^k}$ in ${p(z)}$.

For instance, for ${m=1}$, we observe

$\displaystyle I_{n,1} = \begin{cases} 2\pi, & \text{if }n\text{ is odd}\,, \\ 0, &\text{otherwise } \,, \end{cases}$

as the term ${z^{n-1}}$ appears in the sum ${\sum_{j=0}^{n-1} z^{2j}}$ if and only if ${(n-1)}$ is even.

Remark. The integral ${I_{n,1}}$ may also be computed by induction and the use of trigonometric identities. However, I feel that the proof above is much more illuminating.

Exercise. Use the result from Example 2. to prove that ${\displaystyle\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}}$.