# Prove that $\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx=\infty$

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!

• 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 $\arctan(x)$. Apr 26 '21 at 14:27
• Ohh I mean $\arcsin(x)$ you are right.
– user853637
Apr 26 '21 at 14:32
• The equality of the integrals you started with is not correct.
– user
Apr 26 '21 at 14:32

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: $$\int_0^{\frac{\pi}{2}} \frac{\cos x}{x}dx\ge\frac12\int_0^\frac\pi3\frac{dx}{x}=\infty$$

• Can you explain how you changed to interval? what is $t$ in this case?
– user853637
Apr 26 '21 at 15:06
• @Math4me In general, you can always shorten the interval when the integrand is positive. That is, $\int_0^{\pi/2}\cos x/x dx\ge\int_0^{\pi/3}\cos x/x dx$. Apr 26 '21 at 15:25

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.

• Splitting up the integrand into its power series is not always justified. You have to show this using Fubini, DCT, etc. Apr 26 '21 at 15:10

$$\cos$$ 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.