For a function defined by parts study continuity, and differentiability at two points

Question

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!

Answer 1

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) $.

Answer 2

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.

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

Answer 3

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 $-1$.