# Show that $\int_0^{\infty} x^{-a} \cos \left( \frac{1}{x}\right) dx$ exists for $a\in(1,2)$

Show that $\displaystyle \int_0^{\infty} x^{-a} \cos \left(\frac{1}{x}\right) dx$ exists for $a\in(1,2)$

I have already shown that $\displaystyle \int_0^{\infty} x^{-a} \sin \left(\frac{1}{x}\right) dx$ exists for $a \in (0,1)$.

My thoughts were to integrate by parts, which gives $$\displaystyle \int_0^{\infty} x^{-a} \cos \left(\frac{1}{x}\right) dx = \left[-\frac{1}{a}x^{1-a}\cos x \right]_0^{\infty} - \frac{1}{a} \int_0^{\infty}\left(x^{a-1} \right)\sin\frac{1}{x}dx$$The second term seems exactly as I want, I have convergence for $(a-1) \in (0,1) \Rightarrow a \in (1,2)$ by my previous workings, as I need.

However, it is the first term, I cannot evaluate. I don't think the limit exists for $a\in(1,2)$.

Help is appreciated.

• Your integration by parts isn't right. With $dv=x^{-a}dx$ and $u=cos(1/x)$ we have $v=\frac{1-a}x^{1-a}$ and $du=-x^{-2}\sin(1/x)$. Nov 13 '17 at 11:23
• try to prove that $\int_0^c f$ and $\int_c^\infty f$ for some $c>0$ exists separately. Nov 13 '17 at 11:31

Let $f(x):=x^{-a}\cos(x^{-1})$, then we can see that
$$\left|\int_c^\infty f(x)\,\mathrm dx\right|\le\int_c^\infty |x^{-a}\cos(x^{-1})|\,\mathrm dx\le\int_c^\infty x^{-a}\,\mathrm dx=\frac{c^{1-a}}{a-1}<\infty,\quad\forall c>0$$
Then it remains to show that $\int_0^c f$ converges. But with the change of variable $y:=x^{-1}$ we get
$$\int_0^c f(x)\,\mathrm dx=\int_{c^{-1}}^\infty y^{-b}\cos y\,\mathrm dy,\quad b\in(0,1)$$