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Can anyone give me a counter example of the statement

If $\int_0^\infty f(x) $ exists and $f(x)$ is differentiable then $\lim _{x \to \infty} f'(x)$ exists.

My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.

1) $1/x^2$

2) $-1/x^2$

3) $x= 1$.

Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.

I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.enter image description here

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Yep, this works perfectly! You can rigorize this sort of idea by defining some function like

$$\frac{\sin\left(x^{10}\right)}{x^2}$$

(where the exponent of $10$ is simply to make sure our function oscillates fast enough).

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    $\begingroup$ Actually simply $\sin (x^2)$ suffices. $\endgroup$ – Szeto Nov 22 '18 at 5:24
  • $\begingroup$ @Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-) $\endgroup$ – Carl Schildkraut Nov 22 '18 at 5:47
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    $\begingroup$ Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :) $\endgroup$ – Szeto Nov 22 '18 at 5:49

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