Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume $f(t) \colon [0,1] \to \mathbb{R}$ is a smooth function with $f(0) = 1$. Find the value of $\int_0^1\frac{\sin{nt}}{t}f(t)\,\mathrm{d}t$ as $n$ approaches infinity.

I've tried approaching this integral in several ways but have made no headway.

I understand that this may have something to do with the fact that the integral of $\frac{\sin{t}}{t}\mathrm{d}t$ converges.

share|cite|improve this question
What exactly have you tried? – draks ... Jul 17 '12 at 12:37
I suggest writing $f=g+1$, so that the contribution of 1 can be evaluated explicitly. Then prove that g does not contribute to the limit: this may involve splitting the integral into small and large t, and some integration by parts – user31373 Jul 17 '12 at 12:38
Rather than writing "Find the value of WHATEVER as $n\to\infty$, I'd write "Find the limit of WHATEVER as $n\to\infty$." – Michael Hardy Jul 17 '12 at 22:15
Thanks Draks & Leonid. I tried integration by parts, but looking at the answers below, I realised that I made some mistakes in my working. – Conan Wong Jul 20 '12 at 7:26
up vote 4 down vote accepted

Since $f$ is continuous, $f(x)\to f(0)$ as $x\to 0$. This implies that for every $\varepsilon>0$, no matter how small, if $n$ is big enough, then $f\left(\dfrac u n\right)$ is between the points $f(0)\pm \varepsilon$.

$$ \begin{align} & \int_0^1 \frac{\sin nt}{t} f(t) \, dt = \int_0^1 \frac{\sin nt}{nt} f\left(\frac{nt}{n}\right) \Big( n\,dt\Big) \\[10pt] & = \int_{t=0}^{t=1} \frac{\sin u}{u} f\left(\frac{u}{n}\right) \, du = \int_{u=1}^{u=n} \frac{\sin u}{u} f\left(\frac{u}{n}\right) \, du\tag{1} \end{align} $$

If all of the values of $g(u)$ are between $f(0)\pm\varepsilon$ then $$ \int_0^n \frac{\sin u}{u} g(u)\,du\text{ is between }\int_0^n f\left(\frac{\sin u}{u}\right) (f(0)\pm\varepsilon)\, du =(1\pm\varepsilon)\int_0^n\frac{\sin u}{u}\,du.\tag{2} $$

We haven't yet established that the last expression in $(1)$ approaches a limit as $n\to\infty$, but it has a liminf and a limsup. The last integral in $(2)$ does have a limit $L$ as $n\to\infty$. What we've done entails that the liminf and limsup of the last expression in $(1)$ are between $(1\pm\varepsilon)L$. That liminf and that limsup are quantities that don't depend on $n$. And, for every $\varepsilon$, no matter how small, that liminf and that limsup must be between $(1\pm\varepsilon)L$. That can happen only if the liminf and the limsup are both equal to $L$.

share|cite|improve this answer
I'm wondering if there's a gap in my reasoning: If $h(u)$ is between $a$ and $b$, then $\int_0^n h(u)j(u)\,du$ is between $\int_0^n a j(u)\,du$ and $\int_0^n b j(u)\,du$..... – Michael Hardy Jul 18 '12 at 20:43
Thanks Michael for your answer & for all the editing. I can't see the gap in your reasoning (using Extreme Value Theorem for integrals?) but my math level is a bit low to comment, I think. – Conan Wong Jul 20 '12 at 7:28
My hesitation comes from the fact that the function is sometimes positive and sometimes negative. I'd forgotten about this; I'll look at it again. – Michael Hardy Jul 21 '12 at 20:20

We can write $$f(t)=1+\int_0^tf'(s)\,ds=1+t\int_0^1f'(yt)\,dy,$$ hence $$\int_0^1\frac{\sin(nt)}tf(t)\,dt=\int_0^1\frac{\sin(nt)}t+\int_0^1\sin(nt)\int_0^1f'(yt)\,dy\,dt.$$ Since $f$ is twice continuously differentiable, the map $t\mapsto \int_0^1f'(yt)\,dy$ is of class $C^1$, and by Riemann-Lebesgue lemma or an integration by parts, we have $$\lim_{n\to +\infty}\int_0^1\sin(nt)\int_0^1f'(yt)\,dy\,dt=0.$$ For the first term, do the substitution $s=nt$, $dt=\frac{ds}n$ to get $$\int_0^1\frac{\sin(nt)}t\,dt=\int_0^n\frac{\sin s}{\frac sn}\frac{ds}n=\int_0^n\frac{\sin s}s\,ds.$$ It converges to $\frac{\pi}2$, see here.

share|cite|improve this answer
Thanks a million, Davide! – Conan Wong Jul 20 '12 at 7:27

hint: $\frac {\sin(nt)}{\pi t}\to \delta(t)\ $ as $n\to\infty\ $ (see here) (here we have only half of it!)

share|cite|improve this answer
Thanks Raymond! – Conan Wong Jul 20 '12 at 7:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.