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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!

Thanks in advance!

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    $\begingroup$ What if you make the change of variables $t=1/x$? $\endgroup$
    – Gary
    May 2, 2022 at 11:09
  • $\begingroup$ 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 $\endgroup$
    – Aishgadol
    May 2, 2022 at 11:43

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With $t=1/x$ and by parts, you get $$\int_0^1\frac{\cos(1/x)}{x}dx=\int_1^\infty\frac{\cos t}{t}dt=\left[\frac{\sin t}{t}\right]_1^\infty+\int_1^\infty\frac{\sin t}{t^2}dt.$$

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  • $\begingroup$ 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 $\dfrac{1}{x}$ ? $\endgroup$
    – Aishgadol
    May 2, 2022 at 11:45
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    $\begingroup$ @Aishgadol More details: $$ \int_0^1 {\frac{{\cos (1/x)}}{x}dx} = \int_{ + \infty }^1 {\frac{{\cos t}}{{1/t}}\frac{{dx}}{{dt}}dt} = \int_{ + \infty }^1 {\frac{{\cos t}}{{1/t}}\frac{{d(1/t)}}{{dt}}dt} = - \int_{ + \infty }^1 {\frac{{\cos t}}{t}dt} = \int_1^{ + \infty } {\frac{{\cos t}}{t}dt} . $$ $\endgroup$
    – Gary
    May 2, 2022 at 11:53

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