Prove that $\lim\limits_{n \to +\infty} \int_0^1 f(t)\sin(nt^2)dt$ exists Problem: Suppose $f:[0,1] \to \mathbb{R}$ is continuous. Define $a_n=\int_0^1 f(t)\sin(nt^2)dt$. Prove that exists $\lim\limits_{n \to +\infty}a_n$.
Attempt:
I tried rewriting $a_n=\int_0^n \frac{f(\frac{\sqrt{y}}{\sqrt{n}})}{2\sqrt{ny}}\sin(y)dy$. Then I would define $I_k=[\sqrt{k\pi},\sqrt{(k+1)\pi}]$ and get that: $$\int_{[0,(k+1)\pi]}\frac{f(\frac{\sqrt{y}}{\sqrt{n}})}{2\sqrt{ny}}\sin(y)dy=\int_0^1 \frac{\sin(y)}{2\sqrt{n}}[\sum_{j=0}^k (-1)^j\frac{f(\frac{\sqrt{y+j\pi}}{\sqrt{n}})}{\sqrt{y+j\pi}}]dy$$
but I do not know if it helps.
 A: Let $t=\sqrt s$ in the integral defining $a_n.$ We get
$$a_n=\int_0^1 \frac{f(\sqrt s)}{2\sqrt s} \sin (ns) \,ds.$$
Since $\dfrac{f(\sqrt s)}{2\sqrt s}\in L^1[0,1],$ the Riemann Lebesgue lemma implies $a_n\to 0$ as $n\to \infty.$
A: Not only does $a_n$ converge, but we have that
$$\lim_{n\to\infty}2\sqrt{n}\cdot a_n=\sqrt{\frac{\pi}{2}}f(0)$$
and thus by extention
$$\lim_{n\to\infty}a_n=0$$
To get this we let $g(x)=f(\sqrt{x})$, since this is just as well an arbitrary real valued function. We note that the differentiable functions $k:[0,1]\to\mathbb{R}$ with $\sup_{x\in[0,1]}|f'(x)|<\infty$ are dense in the full set of continuous functions $[0,1]\to\mathbb{R}$, we we may assume $f$ is of this sort and then use the trial observation that if $\sup|g(x)-k(x)|<\epsilon$ then
$$\left|\int_0^1g(t^2)\sin(nt^2)dt-\int_0^1k(t^2)\sin(nt^2)dt\right|<\epsilon$$
so we can apply epsilon-delta from the density. Note that the compactness of the interval $[0,1]$ is very important for this step.
We can now apply integration by parts on $a_n$ to see that
\begin{align*}
&\frac{1}{2\sqrt{n}}\int_0^ng\left(y/n\right)\frac{\sin(y)}{\sqrt{y}}dy\\
&=\left.\frac{1}{2\sqrt{n}}\cdot g\left(y/n\right)\left(\int_0^y\frac{\sin(x)}{\sqrt{x}}dx\right)\right]_0^n-\frac{1}{2\sqrt{n}}\cdot\frac{1}{n}\int_0^ng'\left(y/n\right)\left(\int_0^y\frac{\sin(x)}{\sqrt{x}}dx\right)dy
\end{align*}
It is a well known fact (see, for example, Wolfram Alpha) that
$$\lim_{n\to\infty}\int_0^n \frac{\sin(x)}{x}dx=\sqrt{\frac{\pi}{2}}$$
as $n\to\infty$ we have
$$\left.g\left(y/n\right)\left(\int_0^y\frac{\sin(x)}{\sqrt{x}}dx\right)\right]_0^n\to \sqrt{\frac{\pi}{2}}\cdot g(1)$$
Turning our attention now to the second part of our expression, we note that
$$\frac{1}{n}\int_0^ng'\left(y/n\right)\left(\int_0^y\frac{\sin(x)}{\sqrt{x}}dx-\sqrt{\frac{\pi}{2}}\right)dy$$
tends to $0$ by a simple $\epsilon-\delta$ argument using the boundedness of $g'(y/n)$. This means that
\begin{align*}
\frac{1}{n}\int_0^ng'\left(y/n\right)\left(\int_0^y\frac{\sin(x)}{\sqrt{x}}dx\right)dy&\to \sqrt{\frac{\pi}{2}}\cdot\frac{1}{n}\int_0^ng'\left(y/n\right)dy\\
&=\sqrt{\frac{\pi}{2}}\cdot \left(g(1)-g(0)\right)
\end{align*}
Collecting these results, we get that
$$\lim_{n\to\infty}2\sqrt{n}a_n=\sqrt{\frac{\pi}{2}}\cdot g(0)$$
Since $g(0)=k(0)$ we, apply our density argument and we are done.
