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As the title says, I need to show that $$\int_0^{1} \sin\left(x + \frac{1}{x}\right)\, dx$$ exists.

After performing the substitution $x = 1/u, dx = -1/u^2 du$, the integral becomes $$\int_1^{\infty}\frac{\sin(u + 1/u )}{u^2}\, du.$$ The integrand oscillates between being positive and negative infinitely many times in $[1, \infty).$ If we define a function $f$ to be equal to the integrand when it is non-negative, and $0$ when the integrand is negative, then $\int_1^{\infty}f$ clearly exists, because $f(u) \leq 1/u^2$ (since $|\sin| \leq 1$). If we define a function $g$ in the opposite way as we defined $f$, so that $g$ is equal to the integrand when it is negative, and $0$ when it is non-negative, then we should also have that $\int_1^{\infty}g$ exists, because $g(u) \geq -1/u^2.$

We should therefore have that $$\int_1^{\infty}f(u)+g(u)\, du = \int_1^{\infty}\frac{\sin(u + 1/u)}{u^2}\, du,$$ so the latter integral exists, and therefore so does $\int_0^1 \sin(x + 1/x)\, dx$.

Does this proof seem correct? The only solution I've seen to this problem is quite different.

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  • $\begingroup$ I'm not going to have time to look over this entire thing, but if I were you, I would be very cautious about that substitution you made. Furthermore, you're missing a negative sign on $\text{d}x$. $\endgroup$ – Clarinetist Jun 26 '14 at 17:06
  • $\begingroup$ Thanks for pointing out the missing negative sign. I see nothing wrong with the substitution though. $\endgroup$ – James Pirlman Jun 26 '14 at 17:09
  • $\begingroup$ $\Large\tt M$ yields $\large\approx 0.3761995686$. $\endgroup$ – Felix Marin Jun 26 '14 at 17:11
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    $\begingroup$ The integrand is both continuous and bounded on $(0,1]$. $\endgroup$ – Random Variable Jun 26 '14 at 17:42
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    $\begingroup$ Seems basically fine. One might explicitly show that $\int_1^B f(u)\,du$ has a limit as $B\to\infty$. Since our function is continuous bounded in $(0,1]$ there was really no need of the substitution, but that $u^2$ in the denominator does have rhetorical power. $\endgroup$ – André Nicolas Jun 26 '14 at 17:44
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I think the argument is correct, but actually there is no need to split the function into positive and negative part and we to do need neither the substitution. The integral is (absolutely) convergent because the absolute value of the $\sin$ is bounded by $1$ on and we integrate over a bounded interval.

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  • $\begingroup$ Why is this downvoted? $\endgroup$ – Cave Johnson Jan 2 '18 at 9:42

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