# Convergence of improper integral, $\cos(1/x)$

I'm trying to deduce weather this improper integral is convergent or not: $$\int_{0}^{1}\dfrac{\cos(\frac{1}{x})}{x}dx.$$ I've tried using Dirichlet's test for convergence, yet I cant seem to properly 'place' the functions under the needed terms.

I was hinted by a colleague that substitution can be applied here yet I see no way of utilizing that method.

I'm not sure how to move forward at this point, hints are happily accepted!

• What if you make the change of variables $t=1/x$?
– Gary
May 2, 2022 at 11:09
• I've had this thought, though the same colleague that recommended substitution stated that the limits of the integral change aswell, as I'm not sure how these change accordingly to the substitution, I did not walk down that path May 2, 2022 at 11:43

With $$t=1/x$$ and by parts, you get $$\int_0^1\frac{\cos(1/x)}{x}dx=\int_1^\infty\frac{\cos t}{t}dt=\left[\frac{\sin t}{t}\right]_1^\infty+\int_1^\infty\frac{\sin t}{t^2}dt.$$
• Interesting, except for the replacements for $\dfrac{1}{x} \rightarrow t$, you simply placed $1,x \rightarrow 0$ at the limits and switched them for $\dfrac{1}{x}$ ? May 2, 2022 at 11:45
• @Aishgadol More details: $$\int_0^1 {\frac{{\cos (1/x)}}{x}dx} = \int_{ + \infty }^1 {\frac{{\cos t}}{{1/t}}\frac{{dx}}{{dt}}dt} = \int_{ + \infty }^1 {\frac{{\cos t}}{{1/t}}\frac{{d(1/t)}}{{dt}}dt} = - \int_{ + \infty }^1 {\frac{{\cos t}}{t}dt} = \int_1^{ + \infty } {\frac{{\cos t}}{t}dt} .$$