3
$\begingroup$

How can I see that

\begin{align*} \lim_{x\searrow0}\frac{\dfrac{\pi}{6}-\dfrac{1}{3}\displaystyle\int_0^{1/x^3}\dfrac{\sin t}{t}\,\mathrm{d}t}{x}=0? \end{align*}

I tried L'Hôpital's rule (since both the numerator and the denominator tend to zero and both are differentiable for $x>0$), but it doesn't work, since the resulting expression possesses no limit. I'm pretty sure the answer is relatively simple, I just ran out of ideas.

Please share your views on this with me. Thank you.

$\endgroup$
2
$\begingroup$

Since $\int_0^{+\infty}\frac{\sin t}t\mathrm dt=\frac{\pi}2$, the only task is to show that $$\lim_{x\downarrow 0}\frac 1x\int_{1/x^3}^{+\infty}\frac{\sin t}t\mathrm dt=0.$$ This can be done by an integration by parts (integrate $\sin t$, differentiate $\frac 1t$).

$\endgroup$
2
$\begingroup$

Write the numerator as $\int_{1/x^3}^{+\infty} \frac{\sin t}{t}dt$ and integrate by parts.

$\endgroup$
1
$\begingroup$

You can simply recall the (complex analysis) proof of

$$\int_0^{\infty}\frac{\sin t}{t}dt=\frac{\pi}{2}$$

and estimate the remainder term

$$\int_R^{\infty}\frac{\sin t}{t}dt$$

to see in what degree it is approaching zero when $R\to\infty$. Then the answer will be straightforward.

$\endgroup$
1
$\begingroup$

Thank you so much for your help. Based on your guidance, here's how I did it (it might be more complicated than necessary, but I have a rigorous proof style, so excuse me for the unnecessary confusion):

For a given $x>0$, fix $b>1/x^3$. Then,

\begin{align*} \int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t=\int_0^{b}\frac{\sin t}{t}\,\mathrm{d} t-\int_0^{1/x^3}\frac{\sin t}{t}\,\mathrm{d} t. \end{align*} Therefore,

\begin{align*} \lim_{b\nearrow+\infty}\left(\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t\right)=\frac{\pi}{2}-\int_0^{1/x^3}\frac{\sin t}{t}\,\mathrm{d} t. \end{align*}

Hence, the original limit in question can be rearranged as follows:

\begin{align*} \lim_{x\searrow0}\left(\frac{1}{3x}\lim_{b\nearrow+\infty}\left(\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t\right)\right)=\lim_{x\searrow0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\underbrace{\int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t}_{(\heartsuit)}\right)\right). \end{align*}

Now, for a given $x>0$ and $b>1/x^3$, $(\heartsuit)$ can be expanded, indeed, using integration by parts:

\begin{align*} \int_{1/x^3}^{b}\frac{\sin t}{t}\,\mathrm{d} t=-\frac{\cos t}{t}\Bigg|_{t=1/x^3}^{t=b}-\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t=-\frac{\cos b}{b}+x^3\cos(1/x^3)-\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t. \end{align*}

Hence,

\begin{align*} \lim_{x\searrow0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}(\heartsuit)\right)\right)=\lim_{x\searrow0}\left(\frac{1}{3x}\left(-\underbrace{\lim_{b\nearrow+\infty}\left(\frac{\cos b}{b}\right)}_{=0}\right)\right)+\underbrace{\lim_{x\searrow0}\left(\frac{x^2}{3}\cos(1/x^3)\right)}_{=0}-\lim_{x\searrow0}\left(\underbrace{\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t\right)}_{(\spadesuit)}\right). \end{align*}

But note that

\begin{align*} \lim_{x\searrow0}\,\left|\,(\spadesuit)\,\right|=\lim_{x\searrow0}\,\left|\,{\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t\right)}\,\right|\underbrace{=}_{\text{$|\cdot|$ cuous}} \lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\,\left|\,\int_{1/x^3}^{b}\frac{\cos t}{t^2}\,\mathrm{d}t\,\right|\right)\right)\leq\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\left|\,\cos t\,\right|}{t^2}\,\mathrm{d}t\right)\right)\leq\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{1}{3x}\int_{1/x^3}^{b}\frac{\mathrm{d}t}{t^2}\right)\right)=\lim_{x\searrow 0}\left(\lim_{b\nearrow+\infty}\left(\frac{x^3-\dfrac{1}{b}}{3x}\right)\right)=\lim_{x\searrow0}\frac{x^2}{3}=0. \end{align*}

Consequently, $\lim_{x\searrow0}\,\left|\,(\spadesuit)\,\right|=0$, implying that $\lim_{x\searrow0}(\spadesuit)=0$, too, completing the proof.

Thanks to everyone for the hints once again! @njguliyev @DavideGiraudo @shallpion

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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