Proving $\int_0^{1} \sin\left(x + \frac{1}{x}\right)\, dx$ Exists

Question

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.

Answer 1

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.

Answer 2

For any $x \in \left(0, 1\right],$ $$ \begin{aligned} 1\geq\sin \left(x+\frac{1}{x}\right) \geq -1 \end{aligned} $$ Therefore $\sin \left(x+\frac{1}{x}\right) $ is bounded on $(0,1]$ and continuous on $(0,1]$ and hence the integral exists.