# Fundamental Theorem of Calculus for $\lim\limits_{x\to 0}\frac{\int_0^x(x-t)\sin t^2\ dt}{x\sin^3x}$

How to integrate this?

Evaluate $$\lim_{x\to0}\frac{\displaystyle\int_0^x(x-t)\sin t^2\ dt}{x\sin^3x}.$$

I've had difficulty in using L'Hopital rule. At the same time failed to really understand how to differentiate or evaluate the limit of the $x\sin^3x$ and $\int_0^x(x-t)\sin t^2\ dt$

Would like to appreciate your help

• l'Hospital....? – DonAntonio Oct 20 '16 at 11:58
• I've already did that integrated it three times. but i still kept getting zero. And it was getting more complicated for the denominator which kind of contradicts the use of L'Hospital's Rule. – sallyg Oct 20 '16 at 12:01
• How would it contradict the use of l'Hopital? – Sean Roberson Oct 20 '16 at 12:02
• I've been taught that when using L'Hopital rule, the differentiated should not make the whole thing complicated. However, should I just proceed in still doing L'Hopital rule and differentiate it the fourth time? A friend and I have differentiated xsin^3x but kept getting 0 to no avail. – sallyg Oct 20 '16 at 12:16
• @sallyg Is it ok now? – hamam_Abdallah Oct 20 '16 at 12:28

Use repeatedly $\sin u=u\,{\rm sinc}(u)$, whereby $\lim_{u\to0}{\rm sinc}(u)={\rm sinc}(0)=1$. We have $$\int_0^x (x-t)\sin(t^2)\>dt=x^4\int_0^1(1-\tau)\,\tau^2{\rm sinc}(x^2\tau^2)\>d\tau$$ and $$\>x\sin^3 x=x^4\>{\rm sinc}^3(x)\ ,$$ so that $${\int_0^x (x-t)\sin(t^2)\>dt \over x\sin^3 x}={\int_0^1(1-\tau)\,\tau^2\bigl(1+r(x,\tau)\bigr)\>d\tau\over 1+\bar r(x)}\ ,$$ whereby $\lim_{x\to0}r(x,\tau)=0$ uniformly in $\tau$, and $\lim_{x\to0}\bar r(x)=0$ as well. It follows that the limit in question is $$\int_0^1(1-\tau)\,\tau^2\>d\tau={1\over12}\ .$$

• Thank you! It helped me a lot. i forgot the Lemma Rule – sallyg Oct 25 '16 at 6:43

Write the numerator as $$f(x)=x\int_0^x\sin t^2\,dt-\int_0^x t\sin t^2\,dt$$ Then the FTC and the product rule tell you that $$f'(x)=\int_0^x\sin t^2\,dt+x\sin x^2-x\sin x^2=\int_0^x\sin t^2\,dt$$ Therefore $$f''(x)=\sin x^2$$ Thus, applying l'Hôpital twice, $$\lim_{x\to0}\frac{f(x)}{x^4}= \lim_{x\to0}\frac{f'(x)}{4x^3}= \lim_{x\to0}\frac{f''(x)}{12x^2}=\frac{1}{12}$$ Hence $$\lim_{x\to0}\frac{f(x)}{x\sin^3x}= \lim_{x\to0}\frac{f(x)}{x^4}\frac{x^3}{\sin^3x}=\frac{1}{12}$$

Let

$f(x)=x\int_0^x\sin(t^2)dt-\int_0^xt\sin(t^2)dt$

and

$g(x)=x\sin^3(x)$

$g(x)$ is equivalent to $h(x)=x^4$ and we will have the same limit if we replace

$g(x)$by $h(x)$.

we will compute

$\lim_{x\to0}\frac{f(x)}{h(x)}\frac{h(x)}{g(x)}$

and use

$\lim_{x\to 0}\frac{h(x)}{g(x)}=1$.

$f'(x)=\int_0^x\sin(t^2)dt+x\sin(x^2)-x\sin(x^2)=x\int_0^x\sin(t^2)dt$

thus your limit is $\color{red}{\frac{1}{12}}$ after a second l' Hopital rule.

• I don't see it clearly: first, that "$\;g(x)\;$ is equivalent..." part is not clear at all: why is it true? You can say that $\;|\sin x|<x\;$ for small values (in abs. val.) of $\;x\;$ , yet that the values of both expressions in the limits, "when doing l'Hospital", will be the same is, I believe, something that must be proved. Also, how did you get the equality $$\int_0^x\sin(t^2)dt+x\sin(x^2)-x\sin(x^2)=x\int_0^x\sin(t^2)dt\;\;?$$ Where does the $\;x\;$ factor in the rightmost side come from? – DonAntonio Oct 20 '16 at 12:35
• @Ab No, it isn't still clear: in the middle expression of $\;f'(x)\;$ , the second and third summand cancel with each other, and you're then left only with the integral $\;\int_0^x \sin t^2\,dt\;$ . Why do you multiply this by $\;x\;$ on the right side? – DonAntonio Oct 20 '16 at 12:43
• Your $f'(x)$ has a factor of $x$ too much. It's probably just a typo, since the end result is correct. – Daniel Fischer Oct 20 '16 at 14:18
• The problem with that factor $\;x\;$ there is that it has a very big importance for the limit, specially because of the claim that the denominator "is like $\;x^4\;$ in the limit", which seems to be a rather important thing to state without a proof or, at least, without a hint why it works that way in this case. – DonAntonio Oct 20 '16 at 15:01

Let $$f(x) = \int_{0}^{x}(x - t)\sin t^{2}\,dt$$ then it can be easily proved (using integration by parts) that $$f''(x) = \sin x^{2}, f(0) = f'(0) = 0$$ and therefore we can use L'Hospital's Rule twice to get the answer easily.

We have \begin{align} L &= \lim_{x \to 0}\frac{f(x)}{x\sin^{3}x}\notag\\ &= \lim_{x \to 0}\frac{f(x)}{x^{4}}\cdot\frac{x^{3}}{\sin^{3}x}\notag\\ &= \lim_{x \to 0}\frac{f(x)}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{f'(x)}{4x^{3}}\text{ (via L'Hospital's Rule)}\notag\\ &= \lim_{x \to 0}\frac{f''(x)}{12x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{12}\lim_{x \to 0}\frac{\sin x^{2}}{x^{2}}\notag\\ &= \frac{1}{12} \end{align}

More generally if $$F(x) = \frac{1}{(n - 1)!}\int_{a}^{x}(x - t)^{n - 1}f(t)\,dt$$ then $$F^{(n)}(x) = f(x), F(a) = F'(a) = F''(a) = F^{(n - 1)}(a) = 0$$ Further note that it is never a good idea to jump to L'Hospital Rule directly (unless the problem is too simple) and it is better to transform the expression into a form where application of L'Hospital's Rule is easy. Thus in the solution provided above we have effectively replaced the expression $x\sin^{3}x$ in denominator with $x^{4}$ and then applied L'Hospital's Rule. This considerably simplifies the process of L'Hospital's Rule.

An idea with Taylor expansions:

$$(x-t)\sin t^2=(x-t)\left(t^2-\frac{t^6}6+\ldots\right)=xt^2-\frac{xt^6}6-t^3+\frac{t^7}6+\ldots\implies$$

$$\int_0^x(x-t)\sin t^2\,dt=\left.\left(\frac x3t^3-\frac14t^4+\ldots\right)\right|_0^x=\frac{x^4}3-\frac{x^4}4+\ldots=\frac{x^4}{12}+\ldots$$

Observe that the $\;\ldots\;$ above is an expression which, upon substituting $\;t\to x\;$, has an exponent higher than $\;4\;$ . Because of the following this won't contribute anything in the limit.

We also have

$$x\sin^3x=x\left(x-\frac{x^3}6+\ldots\right)^3=x(x^3+\ldots)=x^4+\ldots$$

Thus our limit is

$$\lim_{x\to0}\frac{\frac{x^4}{12}}{x^4}=\frac1{12}$$

• No, in your $f(t)$ there is an $(x-t)\sin(t^2)$. – hamam_Abdallah Oct 20 '16 at 12:46
• What is the derivative of $\int_0^x xdt$. – hamam_Abdallah Oct 20 '16 at 12:51
• @AbdallahHammam $2x$ – MathematicsStudent1122 Oct 20 '16 at 12:51
• You have the wrong $F'$ (and hence a wrong end result). $F'(x) = \int_0^x \sin (t^2)\,dt$. – Daniel Fischer Oct 20 '16 at 13:49
• That is indeed surprising. For it is $\frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ and not $\frac{1}{3} + \frac{1}{4}$. – Daniel Fischer Oct 20 '16 at 14:56