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For the function defined by $$F(x)=\begin{cases}\displaystyle\int_x^{2x}\sin t^2\,\mathrm dt,&x\neq0\\0,&x=0\end{cases}$$ analyze continuity and derivability at the origin. Is $F$ derivable at point $x_0=\sqrt{\pi/2}$? Justify the answer, and if possible, calculate $F'(x_0)$.


I have been told that I must use the Fundamental Theorem of Integral Calculus but I do not know how to apply it to this case.

For the function to be continuous at the origin, it must happen that $F(0)=\lim_{x\to0}F(x)$. We know that $F(0)=0$, and $$\lim_{x\to0}F(x)=\lim_{x\to0}\int_x^{2x}\sin t^2\,\mathrm dt\;{\bf\color{red}=}\int_0^{2\cdot0}\sin t^2\,\mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $\bf\color{red}=$.

To find the derivative at $x_0=0$ I tried the differentiate directly $F(x)$ but it is wrong, so I have been told that I must use the definition. So we have to find $$F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}{x-0}=\lim_{x\to0}\frac{\int_x^{2x}\sin t^2\,\mathrm dt}x.$$ Why we have to bound $\left|\sin t^2\right|\leq t^2$? How can we do that?

Finally, I do not know how to use the aforementioned theorem to justify that the function is derivable in $\sqrt{\pi/2}$. Using the definition again:

\begin{align*} F'\left(\sqrt{\frac\pi2}\right)&=\lim_{x\to\sqrt{\frac\pi2}}\frac{F(x)-F\left(\sqrt{\frac\pi2}\right)}{x-\sqrt{\frac\pi2}}\\ &=\lim_{x\to\sqrt{\frac\pi2}}\frac{\int_x^{2x}\sin t^2\,\mathrm dt-\int_{\sqrt{\pi/2}}^{2\sqrt{\pi/2}}\sin t^2\,\mathrm dt}{x-\sqrt{\frac\pi2}}\\ &\leq\lim_{x\to\sqrt{\frac\pi2}}\frac{\int_x^{2x}t^2\,\mathrm dt-\int_{\sqrt{\pi/2}}^{2\sqrt{\pi/2}}t^2\,\mathrm dt}{x-\sqrt{\frac\pi2}}\\ &\underbrace=_{A=\sqrt{\pi/2}}\lim_{x\to A}\frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\\ &=\frac73\lim_{x\to A}\frac{x^3-A^3}{x-A}\\ &=\frac73\lim_{x\to A}\frac{(x-A)(x^2+Ax+A^2)}{x-A}\\ &=\frac73(A^2+A^2+A^2)\\ &=7A^2\\ &=\frac{7\pi}2, \end{align*}

but it is wrong.

How can we solve the statement?

Thanks!

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    $\begingroup$ Would it help if you write $\int_x^{2x}=\int_0^{2x}-\int_0^x$? $\endgroup$ – A.Γ. Jan 3 at 23:59
  • $\begingroup$ @A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous? $\endgroup$ – manooooh Jan 4 at 0:01
  • $\begingroup$ It is because the integral is additive. Or if you wish $\int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $\int_0^y$. $\endgroup$ – A.Γ. Jan 4 at 0:05
  • $\begingroup$ @A.Γ. how do you know that $0\in[x,2x]$ for all $x\neq0$? Because for example if $x\in(0,\infty)$ then $0\not\in[x,2x]$. $\endgroup$ – manooooh Jan 4 at 0:29
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    $\begingroup$ One does not need $0\in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$\int_{a} ^{b} f(x) \, dx+\int_{b} ^{c} f(x) \, dx=\int_{a} ^{c} f(x) \, dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $\int_{a} ^{a}f(x)\,dx=0$ and $\int_{b} ^{a} f(x) \, dx=-\int_{a} ^{b} f(x) \, dx$. $\endgroup$ – Paramanand Singh Jan 4 at 7:23
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You just need to use the fundamental theorem of calculus. Since the integrand $\sin(t^2)$ is continuous everywhere we can write $$F(x) =\int_{0}^{2x}\sin t^2\,dt-\int_{0}^{x}\sin t^2\,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2\int_{0}^{x}\sin (4z^2)\,dz-\int_{0}^{x}\sin t^2\,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2\sin (4x^2)-\sin x^2$$ for all $x\in\mathbb {R} $.


For reference I mention FTC explicitly :

Fundamental Theorem of Calculus Part 1: Let the function $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $. Then the function $F:[a, b] \to\mathbb {R} $ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ is continuous on $[a, b] $ and if $f$ is continuous at some point $c\in[a, b] $ then $F$ is differentiable at $c$ with derivative $F'(c) =f(c) $.

Using the above theorem it can be proved that if a function $f:\mathbb {R} \to\mathbb {R} $ is Riemann integrable on every bounded and closed interval then the function $F:\mathbb {R} \to\mathbb {R} $ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ for some $a\in\mathbb {R} $ is continuous everywhere and if $f$ is continuous at some point $c\in \mathbb {R} $ then $F$ is differentiable at $c$ with $F'(c) =f(c) $.

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    $\begingroup$ @manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment. $\endgroup$ – Paramanand Singh Jan 5 at 1:34
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    $\begingroup$ @manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =\int_{a} ^{g(x)} f(t) \, dt$$ then $F'(x) =f(g(x)) g'(x) $. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x) $. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral. $\endgroup$ – Paramanand Singh Jan 5 at 1:39
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    $\begingroup$ @manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2\,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$\int_{0}^{x}\sin t^2\,dt=2\int_{0}^{x/2}\sin (4z^2)\,dz$$ $\endgroup$ – Paramanand Singh Jan 5 at 4:32
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    $\begingroup$ @manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $\int_{0}^{x}\sin t^2\,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$). $\endgroup$ – Paramanand Singh Jan 5 at 4:37
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    $\begingroup$ @manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question. $\endgroup$ – Paramanand Singh Jan 5 at 4:40
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If you want to evaluate the limit:

$$\displaystyle\lim_{x\to 0}F(x)=\lim_{x\to 0}\int_{x}^{2x}\sin(t^2)dt$$

you can observe that $\forall x>0$ (the case $x<0$ is the same), $f(t)=\sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $\xi_{x}\in (x,2x)$ such that

$$\int_{x}^{2x}\sin(t^2)dt=\sin(\xi_{x}^2)(2x-x)\implies F(x)=\sin(\xi_{x}^2)x$$

Now $\xi_{x}\to 0$ for $x\to 0^{+}$ so:

$$\lim_{x\to 0^{+}}F(x)=\lim_{x\to 0}\sin(\xi_{x}^2)x=[\sin(0)\cdot 0]=0$$

Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:

$$\lim_{x\to 0^{+}}\frac{F(x)-F(0)}{x-0}=\lim_{x\to 0}\frac{\sin(\xi_{x}^2)x}{x}=\lim_{x\to 0}\sin(\xi_{x}^2)=0$$

For $x_0=\sqrt{\frac{\pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.

$F'(x)=2\sin(4x^2)-\sin(x^2)\implies F'\left(\sqrt{\frac{\pi}{2}}\right)=2\sin(4\cdot\frac{\pi}{2})-\sin(4\cdot\frac{\pi}{2})=-1$

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  • $\begingroup$ Thanks for the answer! How do you know that $F(x)=f(\xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $\xi_{x}\in (x,2x)$ such that $F(x)=\frac{F(2x)-F(x)}{2x-x}$. $\endgroup$ – manooooh Jan 4 at 0:49
  • $\begingroup$ I applied the mean value theorem for the function $f(t)=\sin(t^2)$ over the intervall $[x,2x]$. So exists $\xi_x\in(x,2x)$ such that $\int_{x}^{2x}f(t)dt=f(\xi_x)(2x-x)$. $\endgroup$ – Ixion Jan 4 at 0:52
  • $\begingroup$ I do not know how do you get $f(\xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $\xi_x\in(x,2x)$ such that $\require{cancel}f'(\xi_x)=\dfrac{f(2x)-f(x)}{2x-x}=\dfrac{\sin(4x^2)-\sin(2x)}x\cancel\implies f(\xi_x)(2x-x)$. $\endgroup$ – manooooh Jan 4 at 5:31
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    $\begingroup$ The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $\xi\in (a,b)$ such that: $$\int_{a}^{b}f(t)dt=f(\xi)(b-a)$$ In this case $f(t)=\sin(t^2), \ a=x$ and $b=2x.$ $\endgroup$ – Ixion Jan 4 at 21:06
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Since $\frac {\sin\, x} x\to1 $as $x \to 0$ we can find $\delta >0$ such that $\frac 1 2 t^{2} \leq\sin(t^{2})\leq 2t^{2}$ for $|t| <\delta$. This gives $\frac 7 6 x^{3} \leq F(x) \leq \frac {14} 3x^{3}$ for $0<x<\sqrt {\delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=\int_0^{2x}\sin(t^{2})\, dt -\int_0^{x}\sin(t^{2})\, dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2\sin(4x^{2})-\sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.

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  • $\begingroup$ Thanks for the answer! Why do you use "Since $\frac {\sin\, x} x\to1$ as $x\to0$ (...)"? $\endgroup$ – manooooh Jan 4 at 6:35
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    $\begingroup$ @manooooh it makes it easy to see that $\frac {F(x)} x \to 0$ as $x\to 0$. I can use just the definition of derivative instead of using MVT, etc. $\endgroup$ – Kavi Rama Murthy Jan 4 at 6:40
  • $\begingroup$ Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}{x-0}=\lim_{x\to0}\frac{\int_x^{2x}\sin t^2\,\mathrm dt}x$$ I should use the fact that $\left|\sin t^2\right|\leq t^2$ so now the limit becomes $\lim_{x\to0}\left|\frac{\int_x^{2x}t^2\,\mathrm dt}x\right|$? $\endgroup$ – manooooh Jan 4 at 6:44
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    $\begingroup$ @manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right? $\endgroup$ – Kavi Rama Murthy Jan 4 at 7:17
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    $\begingroup$ @manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy! $\endgroup$ – Kavi Rama Murthy Jan 4 at 7:30

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