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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.

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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
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3 Answers

up vote 3 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 $L$.

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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
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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.

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Thanks a million, Davide! –  Conan Wong Jul 20 '12 at 7:27
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hint: $\frac {\sin(nt)}{\pi t}\to \delta(t)\ $ as $n\to\infty\ $ (see here) (here we have only half of it!)

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Thanks Raymond! –  Conan Wong Jul 20 '12 at 7:28
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