Evaluating the limit $\lim_{x\to0}\frac{1}{x^3}\int_{0}^{x}\sin(\sin(t^2))dt$

Question

$$\lim_{x\to0}\frac{1}{x^3}\int_{0}^{x}\sin(\sin(t^2))dt$$

This is a compound question from me.

1. I don't know how to begin evaluating this limit. My guess would be that I would have to find the value of this Riemann's integral and then plug the result into the limit. Is this the right direction to head?

Which brings me to...

2. I am also stuck trying to resolve the integral. I tried integrating by substitution, trying with both $u = t^2$ and $u = \sin(t^2)$, but both have lead me to finding that $t$ or $dt$ popping back into the equation sooner or later and I'm not quite sure how to handle that. Any hints as to how I can integrate that function?

Thank you.

Answer 1

Using the fundemental theorem of calculus, we can apply L'Hopital to the limit to get:

$$\lim_{x\to 0} \frac{\sin(\sin(x^2))}{3x^2}$$

To avoid further L'Hopital shenanigans, do the following manipulation:

$$= \frac{1}{3} \lim_{x\to 0} \frac{\sin(\sin(x^2))}{\sin(x^2)}\cdot\frac{\sin(x^2)}{x^2}$$

Both limits go to $1$ because they are of the form $\frac{\sin z}{z}$ as $z\to 0$. Thus the product is $\frac{1}{3}$

Answer 2

Denote $f(x)$ the function $\int_0^x(....)dt$

By Fundamental theorem of integral calculous:

$$\lim_{x \to 0}f(x)=^{L'Hospital}\lim{x \to 0}\frac{\sin{(\sin{x^2})}}{3x^2}$$.

Apply again L'Hospital's Rule to find the limit.

Answer 3

The expression screams L'Hopital's Rule:

\begin{align}\lim_{x\to0}\frac1{x^3}\int_0^x \sin\left(\sin(t^2)\right)\,dt&\overset{\text{L'H}}=\lim_{x\to0}\frac{\frac{d}{dx}\int_0^x\sin\left(\sin(t^2)\right)\,dt}{\frac{d}{dx}x^3}\\ &=\lim_{x\to0}\frac{\sin\left(\sin(x^2)\right)}{3x^2}\\ &\overset{\text{L'H}}=\lim_{x\to0}\frac{\cos\left(\sin (x^2)\right)\cdot\cos(x^2)\cdot2x}{6x}\\ &=\frac13 \end{align}

Answer 4

Without L'Hopital: \begin{align*} \dfrac{1}{x^{3}}\int_{0}^{x}\sin(\sin t^{2})dt=\dfrac{1}{x^{3}}\left(x\sin(\sin x^{2})-\int_{0}^{x}t\cos(\cos t^{2})2tdt\right). \end{align*} Note that \begin{align*} \dfrac{1}{x^{3}}(x\sin(\sin x^{2}))=\dfrac{\sin(\sin x^{2})}{\sin x^{2}}\dfrac{\sin x^{2}}{x^{2}}\rightarrow 1. \end{align*} On the other hand, \begin{align*} \int_{0}^{x}t\cos(\cos t^{2})2tdt=\dfrac{2}{3}x^{3}\cos(\cos x^{2})-\dfrac{2}{3}\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt. \end{align*} And we have \begin{align*} -\dfrac{\dfrac{2}{3}x^{3}\cos(\cos x^{2})}{x^{3}}=-\dfrac{2}{3}\cos(\cos x^{2})\rightarrow-\dfrac{2}{3}, \end{align*} whereas fot the integral, by the change of variable $u=t^{4}$, we obtain that \begin{align*} \int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt=\dfrac{1}{2}\int_{0}^{x^{4}}u\sin(\sin u^{1/2})\dfrac{du}{u^{3/4}}=\dfrac{1}{2}\int_{0}^{x^{4}}u^{1/4}\sin(\sin u^{1/2})du, \end{align*} and that \begin{align*} \dfrac{1}{x^{3}}\int_{0}^{x}t^{3}\sin(\sin t^{2})2tdt&=\dfrac{1}{2}\cdot x\cdot\dfrac{1}{x^{4}}\int_{0}^{x^{4}}u^{1/4}\sin(\sin u^{1/2})du\\ &=\dfrac{1}{2}\cdot x\cdot\eta_{x}^{1/4}\sin(\sin\eta_{x}^{1/2})\\ &\rightarrow 0, \end{align*} where $\eta_{x}\in[0,x]$ is chosen by Integral Mean Value Theorem, therefore the whole limit is $1-2/3=1/3$.

Answer 5

Another way to do it.

Compose Taylor series to get $$\sin \left(\sin \left(t^2\right)\right)=t^2-\frac{t^6}{3}+\frac{t^{10}}{10}+O\left(t^{13}\right)$$ $$\int\sin \left(\sin \left(t^2\right)\right)\,dt=\frac{t^3}{3}-\frac{t^7}{21}+\frac{t^{11}}{110}+O\left(t^{14}\right)$$ $$\int_0^x\sin \left(\sin \left(t^2\right)\right)\,dt=\frac{x^3}{3}-\frac{x^7}{21}+\frac{x^{11}}{110}+O\left(x^{14}\right)$$

To give you an idea, using $x=\frac \pi 6$, which is far away from $0$, the above formula gives for the integral $0.047342792$ while the numerical integration gives $0.047342690$

So, back to your problem $$\frac 1 {x^3}\int_0^x\sin \left(\sin \left(t^2\right)\right)\,dt=\frac{1}{3}-\frac{x^4}{21}+\frac{x^8}{110}+O\left(x^{11}\right)$$ which shows the limit, how it is approached and give a shortcut evaluation of the definite integral for small values of $x$.

Answer 6

Use the substitution $t=z^{1/3}$ to obtain $$\int_{0}^{x}\sin\sin t^2\,dt=\frac{1}{3}\int_{0}^{x^3}\frac{\sin\sin z^{2/3}}{z^{2/3}}\,dz=\int_{0}^{x^3}f(t)\,dt$$ where $$f(t) =\frac{\sin\sin t^{2/3}}{3t^{2/3}}$$ has a removable discontinuity at $t=0$. Redefining $f(0)=1/3$ the function $f$ becomes continuous at $0$.

By fundamental theorem of calculus $$\frac{1}{x^3}\int_{0}^{x^3}f(t)\,dt\to f(0)=\frac{1}{3}$$ as $x\to 0$. And hence the desired limit is $1/3$.