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

{\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\,,


\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\,.


\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}}.

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