# Limit of a definite integral

We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$ Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5+u^6}$$ Then by partial fractions, which I did manually and chcecked with WolframAlpha afterwards, it becomes \begin {align} 2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\left(\frac{1}{u^5}-\frac{1}{u^4}+\frac{1}{u^3}-\frac{1}{u^2}+\frac{1}{u}-\frac{1}{1+u}\right) du =\\ \lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}2\left(\log{u}-\log{(1+u)}+\frac{1}{u}-\frac{1}{2u^2}+\frac{1}{3u^3}-\frac{1}{4u^4}\right) du =\\ \lim_{x \to 0}\int_{\sin x}^{x}\left(\log{t}-2\log{(1+\sqrt{t})}+\frac{2}{\sqrt{t}}-\frac{1}{t}+\frac{2}{3t^{3/2}}-\frac{1}{2t^2}\right) dt \\\end{align} Fianlly we obtain the following limit: $$\lim_{x \to 0}\left(\log {x}-\log {\sin x}+2\log {(1+\sqrt{x})}-2\log {(1+\sqrt{\sin x})}+\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}-\frac{1}{x}+\frac{1}{\sin x}+\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}-\frac{1}{2x^2}+\frac{1}{\sin^2 x}\right)$$

Here's where I stuck. It gets messy when I try to calculate $\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}$ and $\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}$. The rest is rather doable - de l'Hospital's rule is useful with $-\frac{1}{x}+\frac{1}{\sin x}$ which is $0$ in limit, so as logarithm expressions (obviously) and Taylor expansion helps with $-\frac{1}{2x^2}+\frac{1}{\sin^2 x}$ which, in turn, equals $1/6$ when $x$ approaches $0$.

Did I make any mistakes? I hope not, but even so I'm not certain what to do with this horrible limit. I'd be glad if anyone could point out what to do.

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When using a substitution to transform an integral, don't forget to transform the limits of integration: $\int_a^b\rightarrow\int_\sqrt{a}^\sqrt{b}$ after the substitution $\sqrt{t}=u$. –  David H Aug 2 '14 at 11:21

Your proof is nice but there's more simple:

Let consider this simplified integral

$$\int_{\sin x}^x\frac{dt}{t^3}=-\frac12t^{-2}\Bigg|_{\sin x}^x=-\frac12\left(\frac1{x^2}-\frac1{\sin^2x}\right)\sim_0-\frac12\frac{(x-\frac{x^3}{6})^2-x^2}{x^4}\sim_0\frac16$$ and now we prove that the two integrals have the same limit by: $$0\le\int_{\sin x}^x\left(\frac{1}{t^{3}}-\frac{1}{t^{3}(1+\sqrt t)}\right)dt=\int_{\sin x}^x\frac{\sqrt t}{t^3(1+\sqrt t)}dt\\\le\int_{\sin x}^x t^{-5/2}dt=-\frac25t^{-3/2}\Bigg|_{\sin x}^x\\=-\frac25\left(\frac1{x^{3/2}}-\frac1{\sin^{3/2}x}\right)\sim_0\frac25\frac{x^{7/2}}{4x^3}\xrightarrow{x\to0}0$$

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This is not much different from Sami Ben Romdhane's answer. Since the function $g(t)=\frac{1}{t^3(1+\sqrt{t})}$ is decreasing over $\mathbb{R}^+$, by the mean value theorem: $$(x-\sin x) \frac{1}{\sin^3 x\,(1+\sqrt{\sin x})}\leq\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}\leq (x-\sin x)\frac{1}{x^3(1+\sqrt{x})},$$ but when $x$ approaches $0$ both the LHS and the RHS approaches $\frac{1}{6}$.

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When you substitute a new variable in, you must not forget to change the limits of your integral.

So,

instead of $$x, \sin x$$

We have

$$\sqrt{x}, \sqrt{\sin x}$$

Everything else looks correct.

EDIT

To solve the limit:

$$\lim_{x\to 0} \frac{2}{\sqrt{x}} + \frac{2}{\sqrt{\sin x}}$$

$$= \lim_{x\to 0} \frac{2(\sqrt{\sin{x}} - \sqrt{x})}{\sqrt{x}\sqrt{\sin x}}$$

Use l'hopsitals

Can you get it from here?

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Yeah, it's just a typo though, sorry I'll edit. –  Jules Aug 2 '14 at 11:32
@Jules well then everything else seems to be correct. –  Varun Iyer Aug 2 '14 at 11:32
Ok, but what about $\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}$ or $\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}$? I don't know how to show it is $0$ when $x$ approaches $0$. –  Jules Aug 2 '14 at 11:41
Combine the two fractions. Then use l'hospital's rule –  Varun Iyer Aug 2 '14 at 11:45
I did that and it just made it look more complicated. Yeah, but I'll try again, thanks. –  Jules Aug 2 '14 at 11:47