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Here is my attempt:

$\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx=\int_0^1 \frac{1}{\arctan(t)}dt$

We know that for every $x \ge 0$ we have $0 \le \arctan(x) \le x$, so $\frac{1}{x} \le \frac{1}{\arctan(x)}$, and we know that $\int_0^1 \frac{1}{x}dx=\infty$,

So by the test, also $\int_0^1 \frac{1}{\arctan(t)}dt=\infty$, so $\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx=\infty$.

Is that correct?

Thanks!

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    $\begingroup$ How did you get $\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx=\int_0^1 \frac{1}{\arctan(t)}dt$? It almost seems as if you were “simplifying” $\frac{\arcsin(x)}{\arccos(x)} $ to $ \arctan(x)$. $\endgroup$
    – Martin R
    Apr 26 '21 at 14:27
  • $\begingroup$ Ohh I mean $\arcsin(x)$ you are right. $\endgroup$
    – user853637
    Apr 26 '21 at 14:32
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    $\begingroup$ The equality of the integrals you started with is not correct. $\endgroup$
    – user
    Apr 26 '21 at 14:32
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No, this is not correct. You didn't state which substitution $t=f(x)$ gave you $\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx=\int_0^1 \frac{1}{\arctan(t)}dt$.

One possible correct solution is: $$\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx\ge\frac12\int_0^\frac\pi3\frac{dx}{x}=\infty$$

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  • $\begingroup$ Can you explain how you changed to interval? what is $t$ in this case? $\endgroup$
    – user853637
    Apr 26 '21 at 15:06
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    $\begingroup$ @Math4me In general, you can always shorten the interval when the integrand is positive. That is, $\int_0^{\pi/2}\cos x/x dx\ge\int_0^{\pi/3}\cos x/x dx$. $\endgroup$
    – Kenta S
    Apr 26 '21 at 15:25
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Simplest way that comes to mind is a power-series expansion of $\cos x$

$$\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx = \int_0^{\frac{\pi}{2}} \frac{1}{x}(1-\frac{x^2}{2!}+\frac{x^4}{4!}...)dx$$

$$=\int_0^{\frac{\pi}{2}}(\frac{1}{x}-\frac{x}{2!}+\frac{x^3}{4!}...)dx$$

$$=(\ln x - \frac{x^2}{4} + \frac{x^4}{96}...)\rvert^{x=4}_{x=0}$$

One can see that

$$\lim_{n\to0} \ln n = \infty$$

And therefore this integral is going to diverge.

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    $\begingroup$ Splitting up the integrand into its power series is not always justified. You have to show this using Fubini, DCT, etc. $\endgroup$
    – K.defaoite
    Apr 26 '21 at 15:10
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$\cos$ is differentiable at $0$, so, by definition, there exists a function $r$ with $\lim_{x \to 0} \frac{r(x)}{x}=0$, so that $$\cos(x)=\cos(0)+\cos'(0)x+r(x)=1+r(x)$$ Note that $r$ is continuous because $\cos-1$ is continuous. This means that: $$ \frac{\cos(x)}{x}=\frac{1}{x}-\frac{r(x)}{x}$$ The second part is not problematic, as it can be extended to a continuous function, but the integral of the first term's divergent, hence the whole integral is divergent.

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