# How to prove $\lim\limits_{x\to\infty}\int_a^bf(t)\sin(xt)\,dt=0$

I need help to prove:

Suppose $f\in C$. Prove that $$\lim\limits_{x\to\infty}\int_a^bf(t)\sin(xt)\,dt=0$$

My idea is to use substitute of $xt=u$ and prove $$\lim\limits_{x\to\infty}\dfrac1{x}\int_{ax}^{bx}f\left(\dfrac{y}{x}\right)\sin(y)\,dy=0$$ But I am not sure how to do it. Thanks.

• This is the famous Riemann Lebesgue Lemma en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma – Paramanand Singh Jun 6 '15 at 4:57
• @user585440 The Riemann-Lebesgue Lemma guarantees this. If $f$ is differentiable, and $f'$ is integrable on $[a,b]$, then integration by parts shows, $\int_a^b f(t)\sin(xt) dt=\left. -\frac{f(t)\cos(xt)}{x}\right|_a^b+\frac{1}{x}\int_a^b f'(t)\cos (xt)dt$. Taking a limit as $x\to \infty$ yields $0$. – Mark Viola Jun 6 '15 at 5:04
• $f$ differentiable is not assumed. – user239780 Jun 6 '15 at 5:51

## 4 Answers

Since $f(t)$ is uniform continuous on $[a,b]$, $\forall \epsilon>0,\space\exists \delta>0, \space \forall t_1,t_2\in [a,b], |t_1-t_2|<\delta$, there is $|f(t_1)-f(t_2)|<\epsilon. \space$ Let $N=\left[\dfrac{b-a}{\delta}\right]+1,$ \begin{align} \left|\int_a^bf(t)\sin(xt)dt-\sum\limits_{i=0}^{N-1}\int_{a+i\delta}^{a+(i+1)\delta}f(t_i)\sin(xt)dt\right|&=\left|\sum\limits_{i=0}^{N-1}\int_{a+i\delta}^{a+(i+1)\delta}(f(t)-f(t_i))\sin(xt)dt\right| \\ &\leqslant\epsilon\int_a^b|\sin(xt)|dt \\ &\leqslant(b-a)\epsilon\tag{1} \end{align} where $t_i\in[a+i\delta,a+(i+1)\delta]$.

Next let $y=xt$ \begin{align} \left|\int_c^d\sin(xt)dt\right|&=\left|\dfrac1{x}\int_{cx}^{dx}\sin(y)dy\right| \\ &=\left|\dfrac1{x}\int_{cx}^{2n\pi}\sin(y)dy+\sum\limits_{k=n}^{m}\dfrac1{x}\int_{2k\pi}^{2(k+1)\pi}\sin(y)dy+\dfrac1{x}\int_{2(m+1)\pi}^{dx}\sin(y)dy\right| \\ &\leqslant\dfrac{2n\pi-cx+dx-2(m+1)\pi}{x} \\ &\leqslant\dfrac{4\pi}{x} \end{align} where $n=\left[\dfrac{cx}{2\pi}\right]+1,m=\left[\dfrac{dx}{2\pi}\right]-1$.

So there is \begin{align} \left|\sum\limits_{i=0}^{N-1}\int_{a+i\delta}^{a+(i+1)\delta}f(t_i)\sin(xt)dt\right|&=\left|\sum\limits_{i=0}^{N-1}f(t_i)\int_{a+i\delta}^{a+(i+1)\delta}\sin(xt)dt\right| \\ &\leqslant\dfrac{4\pi}{x}NM\tag{$|f(t_i)|\leqslant M$} \\ &\leqslant\dfrac{4\pi M(b-a)}{\delta x} \end{align} From $(1)$, we have \begin{align} \left|\int_a^bf(t)\sin(xt)dt\right|&\leqslant\left|\sum\limits_{i=0}^{N-1}\int_{a+i\delta}^{a+(i+1)\delta}f(t_i)\sin(xt)dt\right|+(b-a)\epsilon \\ &\leqslant\dfrac{4\pi M(b-a)}{\delta x}+(b-a)\epsilon \end{align} And $$\varlimsup\limits_{x\to\infty}\left|\int_a^bf(t)\sin(xt)dt\right|\leqslant (b-a)\epsilon$$ Since $\epsilon$ is arbitrary small, we have $$\varlimsup\limits_{x\to\infty}\left|\int_a^bf(t)\sin(xt)dt\right|=0 \hspace{4 mm} \text{and} \hspace{4 mm} \lim\limits_{x\to\infty}\int_a^bf(t)\sin(xt)dt=0$$

This is essentially a simplification of the idea presented in hermes answer.

Note that $$\int_{a+(k-1)\frac{2\pi}x}^{a+k\frac{2\pi}x}\sin(xt)\,\mathrm{d}t=0\tag{1}$$ With $n=\left\lfloor(b-a)\frac{x}{2\pi}\right\rfloor$, we have $\left|b-a-n\frac{2\pi}x\right|\le\frac{2\pi}x$ and \begin{align} \left|\int_a^bf(t)\sin(xt)\,\mathrm{d}t\right| &\le\left|\sum_{k=1}^n\int_{a+(k-1)\frac{2\pi}x}^{a+k\frac{2\pi}x}f(t)\sin(xt)\,\mathrm{d}t\right|\\ &+\left|\int_{a+n\frac{2\pi}x}^bf(t)\sin(xt)\,\mathrm{d}t\right|\\ &=\left|\sum_{k=1}^n\int_{a+(k-1)\frac{2\pi}x}^{a+k\frac{2\pi}x}\left(f(t)-f\left(a+k\tfrac{2\pi}x\right)\right)\sin(xt)\,\mathrm{d}t\right|\\ &+\left|\int_{a+n\frac{2\pi}x}^bf(t)\sin(xt)\,\mathrm{d}t\right|\\[12pt] &\le|b-a|\max_{|s-t|\le\frac{2\pi}x}|f(s)-f(t)|\\ &+\frac{2\pi}x\max_{x\in[a,b]}|f(x)|\tag{2} \end{align} Since $f$ is continuous on $[a,b]$ it is uniformly continuous, and therefore, $$\lim_{x\to\infty}\max_{|s-t|\le\frac{2\pi}x}|f(s)-f(t)|=0\tag{3}$$ Thus, $(2)$ and $(3)$ imply that $$\lim_{x\to\infty}\int_a^bf(t)\sin(xt)\,\mathrm{d}t\tag{4}$$

Proof of this result is given in any real-analysis textbook (especially in chapters dealing with Lebesgue Integral and Fourier Series). An approach from Tom Apostol's Mathematical Analysis is as follows.

If $f$ is a constant then the result is almost obvious. If $f$ is a step function (i.e. there is a partition $\mathcal{P} = \{x_{0}, x_{1}, x_{2},\ldots, x_{n}\}$ of $[a, b]$ such that $f$ is constant on each sub-interval $(x_{k - 1}, x_{k})$) then also the result is obvious by splitting the integral as sum of integrals over these sub-intervals.

Next we need a powerful result from theory of Lebesgue Integrals which says that if $f$ is Lebesgue integrable on $[a, b]$ then for any $\epsilon > 0$ we have a step function $g$ on $[a, b]$ such that $$\int_{a}^{b}|f(t) - g(t)|\,dt < \epsilon$$

Next we can write $f(t) = \{f(t) - g(t)\} + g(t)$ and split integral into two sums and each can be estimated to be arbitrarily small when $x \to \infty$ and the proof is done.

In the current question we have assumed that $f$ is continuous and finding the step function is very easy (via the theory of Darboux sums). Since $f$ is continuous therefore it is Riemann integrable on $[a, b]$. Thus for any $\epsilon > 0$ we can find a partition $\mathcal{P} = \{x_{0}, x_{1}, \ldots, x_{n}\}$ of $[a, b]$ with the lower Darboux sum $$s = \sum_{k = 1}^{n}m_{k}(x_{k} - x_{k - 1})$$ where $m_{k} = \inf \,\{f(x)\mid x \in [x_{k - 1}, x_{k}]\}$ such that $$\int_{a}^{b}f(t)\,dt - s < \epsilon$$ Thus the step function $g(x)$ can be defined by $g(x) = m_{k}$ when $x \in (x_{k - 1}, x_{k})$ and then $$\int_{a}^{b}g(t)\,dt = s$$ and thus $$0 \leq \int_{a}^{b}(f(t) - g(t))\,dt < \epsilon$$ I hope these ideas explained above would help you to write a complete proof without the use of theory of Lebesgue Integral.

• Note that quite the same proof works with Riemann integral. – Gabriel Romon Jun 6 '15 at 6:50
• @LeGrandDODOM When I begin the statement "In the current question ..." below the horizontal line I just use the Riemann integrals. Perhaps it was not that obvious. – Paramanand Singh Jun 6 '15 at 7:20

This result is a direct consequence of Riemann-Lebesgue Lemma.