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I've been trying to figure out how to evaluate $$ \frac{d}{dx}\int_{0}^{x} \sin(1/t) dt $$ at $x = 0$. I know that the integrand is undefined at $x = 0,$ but is there any way to "extend" the derivative to the point? Or is it not differntiable there - and if so, why?

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    So you have $f(x) = \int_0^x \sin (1/t) \text{d}t$ which you want to differentiate with respect to $t$? Or with respect to $x$? – 2017-02-03
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    It should be with respect to $x$. I fixed it. – 2017-02-03
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    @Hetebrij it's not a function of t – 2017-02-03
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    Exactly, that is why I asked if it was intentionally with respect to $t$ instead of $x$, since then the question would be really easy. – 2017-02-03
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    @Hetebrij fair enough – 2017-02-03

3 Answers 3

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Hint:

$$\frac{1}{x}\int_0^x \sin(1/t)\,dt = \frac{1}{x}\int_{1/x}^\infty \frac{\sin u}{u^2}\,du.$$

Integrate by parts.

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    now it's clear :) – 2017-02-03
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    Could you explain? I'm not sure how you did that. – 2017-02-03
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    They are computing the derivative at $x=0$, the only troublesome point – 2017-02-03
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    @MathQ let $F(x) = \int_0^x \sin(1/t)\, dt.$ Then $F(0)= 0.$ We want to find $\lim_{x\to 0} (F(x)-F(0))/(x-0).$ That's why there's a $1/x$ in front. – 2017-02-03
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    No the derivative exists and equals $0.$ – 2017-02-05
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    @zhw. I'm sorry, I've spent several hours trying to finish the problem but have been unable to. Could you please give me a bit more of the solution? – 2017-02-05
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    Let me walk you through it. Are we OK with my displayed equation and that we need to show the limit of this expression exists? – 2017-02-05
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    Yes. I got that part, but I'm not sure where to go next. – 2017-02-05
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    Did you integrate by parts? – 2017-02-05
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    Yes. I got $\frac{-\sin u}{u} + \int_{1/x}^{\infty} \frac{\cos u}{u} du.$ – 2017-02-05
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    No, you want this: $\int (\sin u)(u^{-2})\,du = (-\cos u)(u^{-2}) - \int (-\cos u)(-2)u^{-3}\, du.$ – 2017-02-05
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    Alright. I understand the first part of the sum (it goes to $0$), but how do you evaluate the second integral? – 2017-02-05
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    You don't evaluate it. You estimate it. How big can $\int_{1/x}^\infty u^{-3}\, du$ be? – 2017-02-05
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    It can be as large as $.5 x^2,$ which tends to $0.$ Thanks! – 2017-02-05
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Use the function $g(x) =x^{2}\cos(1/x),g(0)=0$ so that $g$ is differentiable with $$g'(x) =2x\cos(1/x)+\sin(1/x),g'(0)=0$$ and hence upon integrating we get $$\frac{1}{x}\int_{0}^{x}\sin(1/t)\,dt=\frac{g(x)}{x}-\frac{2}{x}\int_{0}^{x}t\cos(1/t)\,dt$$ Taking limits as $x\to 0$ we can see that the RHS tends to $0$ so the desired derivative is $0$.

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Use the identity

$$ \frac{{\rm d}}{{\rm d}x} \int \limits_{a(x)}^{b(x)} f(t)\,{\rm d}t = b'(x) f\left( b(x) \right)-a'(x) f\left(a(x)\right)$$

to get

$$\frac{{\rm d}}{{\rm d}x} \int \limits_{0}^{x} \sin \left( \frac{1}{t} \right)\,{\rm d}t = 1\, \lim_{t \rightarrow x} \sin \left( \frac{1}{t} \right) - 0 = \sin \left( \frac{1}{x} \right)$$

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    Which works for $x>0$ , not for $x=0$ since $f(b(0))$ not defined. – 2017-02-03