Prove the following improper integral converges

Question

So I have the following integral:

$$\int_{0}^{1} \sin(x+\frac{1}{x})\quad dx$$

And I have to prove this converges. The way I was taught to do it was to solve the integral and take the limit as $a \to 0$. However this function does not have an antiderivative, so this method wouldn't work.

I got the following hint: Perform the substitution $y=\frac{1}{x}$, then we'd have:

$$\int_{0}^{1} \frac{\sin(y+\frac{1}{y})}{y} \quad dy$$ (although I think I should change the integration limits to $f(a)$and $f(b)$ but I don't know how would you evaluate them).

Now since $|\sin(x)|\leq 1$, then we have that $\frac{\sin(y+\frac{1}{y})}{y} \leq \frac{1}{y}$ and since:

$$\int_{0}^{1}\frac{1}{y} dy$$ converges, so does $$\int_{0}^{1} \frac{\sin(y+\frac{1}{y})}{y} \quad dy$$ Is that correct, or where did I go wrong?

Answer 1

The function integrand $f $ is bounded at $(0,1 ]$, so the integral is convergent.

let $\epsilon>0$. $f $ is continuous at $[\frac {\epsilon}{4},1] $ thus it is integrable. there exist a subdivision $\sigma $ such that $$U (f,\sigma)-L (f,\sigma)<\frac {\epsilon}{2} $$

put $\sigma'=\sigma \cup \{0\} $, then $$U (f,\sigma')-L (f,\sigma')\leq$$ $$U(f,\sigma)-L (f,\sigma)+2\frac{\epsilon}{4}<\epsilon $$

$f $ is integrable at $[0,1] $. we can put $f (0)=1$ for example.

We can also observe that $$|\sin (x+\frac {1}{x})|=$$ $$| \sin (x)\cos (\frac {1}{x})+\cos (x) \sin (\frac {1}{x} )|\leq$$ $$\sin (x)+\cos (x) $$ so your integral is absolutely convergent.

Answer 2

$g(x)=x+\frac{1}{x}$ is a decreasing function on $(0,1)$, hence $$ \int_{0}^{1}\sin\left(x+\frac{1}{x}\right)\,dx = \int_{2}^{+\infty}\frac{2\sin(t)}{\sqrt{t^2-4}\left(t+\sqrt{t^2-4}\right)}\,dt $$ and the RHS is convergent by Dirichlet's test: $\sin(t)$ has a bounded primitive and $\frac{1}{\sqrt{t^2-4}\left(t+\sqrt{t^2-4}\right)}$ is decreasing towards zero on $(2,+\infty)$.