find $\frac{d}{dx} \int_{\sqrt{x}}^{2\sqrt{x}} x \sin(t^2) dt$

Question

Find the derivative of $g(x)=\int_{\sqrt{x}}^{2\sqrt{x}} x \sin(t^2)\,dt$. I tried using taking out the $x$ and using the product rule $g'(x) = \int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\,dt + x \frac{d}{dx}\int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\,dt$ and the second term can be solved with the fundamental theorem of calculus but I am stuck with integrating the first term which seems impossible to integrate. Is there another approach?

Answer 1

Hint: by leibniz-rule-derivation

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(x,t) dt=f(x,b(x)) \frac{d}{dx}b(x) -f(x,a(x)) \frac{d}{dx}a(x)+\int_{a(x)}^{b(x)} \frac{d}{dx} f(x,t) dt$$

so

$$\frac{d}{dx} \int_{\sqrt{x}}^{2\sqrt{x}} x \sin(t^2) dt=f(x,2\sqrt{x}) \frac{d}{dx}2\sqrt{x} -f(x,\sqrt{x}) \frac{d}{dx}\sqrt{x}+ \int_{\sqrt{x}}^{2\sqrt{x}} \frac{d}{dx} x \sin(t^2) dt $$

$f(x,t)=x\sin(t^2)$, $f(x,2\sqrt{x})=x\sin(4|x|)$ and $f(x,\sqrt{x})=x\sin(|x|)$

to calculate $$\int_{\sqrt{x}}^{2\sqrt{x}} \sin(t^2) dt$$

see what-is-the-int-sin-x2-dx

Answer 2

I see it's just x(F(4x)-F(x)), you can use the multiplication rule, and get the result pretty fast.

$F(x) = \int_{\delta}^{x} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu$, where $\delta$ is any positive number < x.

$[x(F(4x)-F(x))]'=(F(4x)-F(x))+x(4F'(4x)-F'(x))$, $F'(x) =\frac{\sin(x)}{2\sqrt{x}}$.

Done.

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I see your question is about the first term.

BTW, though I am not sure I can answer your question, perhaps it is not easy to avoid that complicated integral by changing the approach to the differentiation.

(Though some approach of differentiation similar to recursive use of integration by parts could possibly work?)

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Original answer:

Let $\mu=t^2$, then

$\int_{\sqrt{x}}^{2\sqrt{x}} x \sin(t^2)\,dt =x\int_{{x}}^{4{x}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu$.

According to the definition of differential,

$$\frac{d}{dx}\int_{\sqrt{x}}^{2\sqrt{x}} x \sin(t^2)\,dt =\lim_{h\to0}\frac{(x+h)\int_{{x+h}}^{4{(x+h)}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu-x\int_{{x}}^{4{x}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu}{h} =\lim_{h\to0}\frac{h\int_{{x+h}}^{4{(x+h)}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu+x\int_{{4x}}^{4{(x+h)}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu-x\int_{{x}}^{x+h} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu}{h}$$

$\lim_{h\to0}\frac{x\int_{{x}}^{x+h} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu}{h}=x\lim_{h\to0}\frac{(F(x+h)-F(x))}{h}$, (where F(x) is the integral function of $\frac{\sin(\mu)}{2\sqrt{\mu}}$), according either to the fundamental theorem of calculus or to MVT, it equals $x\frac{\sin(x)}{2\sqrt{x}}$. Similarly $\lim_{h\to0}\frac{x\int_{{4x}}^{4(x+h)} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu}{h}=4\lim_{h\to0}\frac{x\int_{{4x}}^{4(x+h)} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu}{4h}=4x\frac{\sin(4x)}{4\sqrt{x}}$.

Besides, $\lim_{h\to0}\frac{h\int_{{x+h}}^{4{(x+h)}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu}{h} =\int_{{x}}^{4{x}} \frac{\sin(\mu)}{2\sqrt{\mu}}\,d\mu=\int_{{\sqrt{x}}}^{2\sqrt{x}} {\sin(t^2)}\,dt$, say it's $\eta$, which I don't know how to calculate, but you can refer to the link.

Therefore, $$\frac{d}{dx}\int_{\sqrt{x}}^{2\sqrt{x}} x \sin(t^2)\,dt=\eta+\sqrt{x}\left({\sin(4x)}-\frac{\sin(x)}{2}\right).$$

I assume here all variables are real, and $x\neq 0$ (i.e. x>0).

Answer 3

We can start by factoring the $x$. $$g'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(x\int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\mathrm{d}t\right)$$ Now we use the product rule. $$g'(x)=\int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\mathrm{d}t+x\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\mathrm{d}t\right)$$ Let's state the fundamental theorem of calculus: $$\frac{\mathrm{d}}{\mathrm{d}x}\int_{g_1(x)}^{g_2(x)}f(t)\mathrm{d}t=f(g_2(x)){g_2}'(x)-f(g_1(x)){g_1}'(x)$$ Thus, $$g'(x)=\int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\mathrm{d}t+x\left(\sin\left((2\sqrt{x})^2\right)\frac{1}{\sqrt{x}}-\sin\left((\sqrt{x})^2\right)\frac{1}{2\sqrt{x}}\right)$$ $$g'(x)=\int_{\sqrt{x}}^{2\sqrt{x}}\sin(t^2)\mathrm{d}t+\sqrt{x}\left(\sin(4x)-\frac{\sin(x)}{2}\right)$$ Can't really simplify it much from here without working out the integral. This integral is non-elementary, but can be stated in terms of Fresnel integrals.