Find $\,\lim\limits_{t\to0^+}t\!\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x$

Question

Question:

Find $\;\displaystyle\lim_{t\to0^+}\,t\!\!\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x$.

Attempted solution:

Let: $$f(x)=\begin{cases}\frac{\cos x-1}{x^{3/2}}&\text{if }x\in(0,\pi/2)\\0&\text{otherwise}\end{cases}$$

$\lim\limits_{x\to0^+}f(x)=0\;$ so $\,f(x)\,$ is continuous on $[0,\pi/2]$.

So, $\;\displaystyle\left|\int_{t^2}^t\frac{\cos x-1}{x^{3/2}}\,\mathrm{d}x\right|\leq\int_0^t \frac{1-\cos x}{x^{3/2}}\,\mathrm{d}x\;,$
for $t\in[0,\pi/2]$, which tends to zero as $t\to0^+$.

$$\begin{align}t\int_{t^2}^t\frac{\cos x}{x^{3/2}}\,\mathrm{d}x&=t\left[\int_{t^2}^t\frac{\cos x-1}{x^{3/2}}\,\mathrm{d}x+\int_{t^2}^t\frac{1}{x^{3/2}}\,\mathrm{d}x\right]\\&=O(t)+2 (1-\sqrt{t})\end{align}$$

So the answer is $2$.

I just wanted to make sure is my approach correct.

Answer 1

Your approach is correct. I would personally write $o(t)$ rather than $O(t)$ there but, no matter. Only one thing I object to: your inequality with $|\int_{t^2}^t\cdots|\le\int_0^t\cdots$ is not immediately true, or maybe not true at all, when $t>1$, but as $t\to0^+$ you only care about $0<t<1$ and then the inequality holds for sure.

P.S. the formatting required a lot of fixing from myself and another user. Try the MathJax tutorial.

Answer 2

You also can use L'Hopital's Rule to calculate. So \begin{eqnarray} &&\lim_{t\to0^+}t\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x=\lim_{t\to0^+}\frac{\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x}{t^{-1}}\\ &=&\lim_{t\to0^+}\frac{\frac{\cos(t)}{t^{3/2}}-\frac{\cos(t^2)}{t^{3}}\cdot2t}{-t^{-2}}=-\lim_{t\to0^+}\bigg[t^{1/2}\cos(t)-2\cos(t^2)\bigg]\\ &=&2. \end{eqnarray}

Answer 3

You could also proceed in the following way.

Since $\,\lim\limits_{x\to0}\dfrac{1\!-\!\cos x}{x^2}\!=\!\dfrac12\!<\!1\,,\,$ there exists $\,a\in(0,1)\,$ such that

$0<\dfrac{1-\cos x}{x^2}<1\quad$ for any $\;x\in(-a,a)\setminus\{0\}\;.$

Consequently ,

$1-x^2<\cos x<1\quad$ for any $\;x\in(-a,a)\setminus\{0\}\;,$

$\dfrac1{\sqrt{x^3}}-\sqrt x<\dfrac{\cos x}{\sqrt{x^3}}<\dfrac1{\sqrt{x^3}}\quad$ for any $\;x\in(0,a)\;,$

$\displaystyle\int_{t^2}^t\!\!\left(\!\dfrac1{\sqrt{x^3}}\!-\!\sqrt x\!\right)\!\mathrm dx<\!\!\int_{t^2}^t\!\dfrac{\cos x}{\sqrt{x^3}}\mathrm dx<\!\!\int_{t^2}^t\!\!\!\dfrac1{\sqrt{x^3}}\mathrm dx\quad\forall\,t\in(0,a)\,,$

$\displaystyle\left[-\dfrac2{\sqrt x}\!-\!\dfrac23\sqrt{x^3}\right]_{t^2}^t\!\!<\!\!\int_{t^2}^t\!\dfrac{\cos x}{\sqrt{x^3}}\mathrm dx<\!\left[-\dfrac2{\sqrt x}\right]_{t^2}^t\quad\forall\,t\in(0,a)\,,$

$\displaystyle t\left[-\dfrac2{\sqrt x}\!-\!\dfrac23\sqrt{x^3}\right]_{t^2}^t\!\!<t\!\!\int_{t^2}^t\!\dfrac{\cos x}{\sqrt{x^3}}\mathrm dx<\!t\left[-\dfrac2{\sqrt x}\right]_{t^2}^t\;\;\forall\,t\in(0,a)\,.$

Since
$\lim\limits_{t\to0^+}t\left[-\dfrac2{\sqrt x}\!-\!\dfrac23\sqrt{x^3}\right]_{t^2}^t\!\!=\lim\limits_{t\to0^+}\left(\!-2\sqrt t\!-\!\dfrac23\sqrt{t^5}\!+\!2\!+\!\dfrac23t^4\!\right)=2$
and $\;\lim\limits_{t\to0^+}t\left[-\dfrac2{\sqrt x}\right]_{t^2}^t\!\!=\lim\limits_{t\to0^+}\left(-2\sqrt t+2\right)=2\;\;,$

by applying Squeeze theorem, it follows that

$\displaystyle\lim\limits_{t\to0^+}\,t\!\!\int_{t^2}^t\!\dfrac{\cos x}{\sqrt{x^3}}\,\mathrm dx=2\,.$

Answer 4

Here is another way to solve. By the MVT for integrals https://en.wikipedia.org/wiki/Mean_value_theorem, there is $c\in(t^2,t)$ such that $$ \int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x=\cos (c)\int_{t^2}^t\frac{1}{x^{3/2}}\,\mathrm{d}x=-2\cos(c)\bigg(\frac1{t^{1/2}}-\frac1{t}\bigg) $$ from which one has $$ \lim_{t\to0^+}t\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x=-2\lim_{t\to0^+}t\cos(c)\bigg(\frac1{t^{1/2}}-\frac1{t}\bigg)=2.$$

Answer 5

I think the asymptotics in this case is not complicated. Near $x=0$ $$\frac{\cos x}{x^{3/2}}=\sum^\infty_{n=0}\frac{(-1)^nx^{2n-\frac32}}{(2n)!}.$$ Hence $$\int^t_{t^2}\frac{\cos x}{x^{3/2}}\,dx\sim\sum^\infty_{n=0}(-1)^n\frac{t^{\frac{4n-1}{2}}-t^{4n-1}}{(2n-\frac12)(2n)!}=\frac{2}{t}-\frac{2}{t^{1/2}}-\frac13t^{5/2}+\frac13t^4+\ldots\qquad \text{as} \quad t\rightarrow0. $$ This is in the sense of Poincaré: $f(t)\sim \sum^\infty_{n=0}a_nf_n(t)$ iff for any $N$, $f(t)-\sum^N_{n=1}a_n f_n(t)=o(f_N(t))$ as $t\rightarrow0$.

Answer 6

In my humble opinion, the simplest and fastest method consists in the use of the fundamental theorem of calculus, just as @xpaul answered.

Otherwise, using the trivial series expansion of the cosine and integrating termwise gives $$f(t)=t\!\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,dx=2\sum_{n=0}^\infty (-1)^n\,\frac{ t^{2 n} \left(\sqrt{t}-t^{2 n}\right)}{(4 n-1) (2 n)!}$$ Using the first terms $$f(t)=2-2 \sqrt{t}\left(1+ \frac{t^2}{6}+O\left(t^{7/2}\right)\right)$$