# check convergence of $\int _1^{\infty }\:\frac{\sin\left(x-1\right)}{x}dx$

I"m trying to check if $$\int _1^{\infty }\:\frac{\sin\left(x-1\right)}{x}dx$$ is convergence or not .

I tried that $\sin(x-1)\le 1 , x\in [1,\infty)$ but then i get that $\frac{\sin\left(x-1\right)}{x}\le\frac{1}{x}$ but this don't help alot .

I also tried to set $f(x)=\frac{\sin\left(x-1\right)}{x}$ and $g(x) =\frac{1}{x}$ and to find $\lim _{x\to \infty }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)$

any idea how i can solve it ?

thanks

• Have you tried to integrate by parts on a compact [1,a] ? – Atmos Dec 27 '17 at 12:05
• @Atmos no but i"ll try it right now – J.Doe Dec 27 '17 at 12:10

Hint $$\sin(x-1) = \sin(x) \cos(1) -\sin(1) \cos(x).$$

and we know that $$\int _1^{\infty }\:\frac{\sin\left(x\right)}{x}dx$$

converges and $$\int _1^{\infty }\:\frac{\cos\left(x\right)}{x}dx$$ converges too by Dirichlet test.

• Why does the second integral diverge ? – Atmos Dec 27 '17 at 12:17
• @Atmos sorry I think it converges – Guy Fsone Dec 27 '17 at 12:21
• Not really ^^ but i thought -Ci(1) existed. For me, your argument is absolutely good but it proves that this integral converges. – Atmos Dec 27 '17 at 12:23
• Well, if we could use Dirichlet test, the conclusion would directly follow from it without expanding $\sin(x-1)$. – Cave Johnson Dec 27 '17 at 12:41

Hint. Let $M>1$. Then integrating by parts, $$\int_1^M\frac{\sin\left(x-1\right)}{x}dx=\left[\frac{-\cos\left(x-1\right)}{x}\right]_1^M-\int_1^M\frac{\cos\left(x-1\right)}{x^2}dx$$ By letting $M \to \infty$, one sees that the given integral is convergent since the latter integral is convergent.

• what i did , Let $f(x)=sin(x-1) , g(x)=\frac{1}{x}$ and then we get that $|\int _1^Msin\left(x-1\right)\:|=|-\cos \left(1-x\right)|^M_0\le2$ ... then we get by Dirichlet test that the integral is convergent ( $g(x)$monotonic goes to 0 and $\int _{\:}^{\:}f\left(x\right)$ is bounded for $M\in(1,\infty)$) – J.Doe Dec 27 '17 at 12:42
• @J.Doe What you did is correct too :) (In your comment above you mean $f$ is bounded, not $\int f$ is bounded). – Olivier Oloa Dec 27 '17 at 13:50