Convergence test for series $\sum_{k\geq 0}\int_{k\pi}^{(k+1)\pi}\frac{\sin{x}}{x}dx=\frac{\pi}{2}$

Question

I didn't understand the following statement made by the author in the accompanying image of a page.

"By a simple estimate $a_k\leq\frac{1}{k\pi}$ for $k \geq 1$, so $a_k \rightarrow 0$. Thus $\displaystyle\int_0^{\infty}\frac{\sin{x}}{x}dx=\displaystyle\sum_{k\geq0}(-1)^k a_k$ converges."

Now if $a_1=\displaystyle\int_{\pi}^{2\pi}\frac{|\sin{x}|}{x}dx=0.43378547585$ radians. and it is greater than $\frac{1}{k\pi}$ where k=1. So this is in contradiction to the statement mentioned above.

Am i wrong in the interpretation of the above statement?

I have added second page to this question. I think the statements made by the author on this page are not correct. I want to know whether i am correct or wrong?

If any member knows the correct answer to this question may reply with correct answer.

Answer 1

The inequality $a_k\leq\frac1{k\pi}$ is false for all $k\geq1$. I suspect that they meant to write $a_k\leq\frac1{k}$.

In fact, we can estimate $$ a_k=\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}xdx >\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{(k+1)\pi}dx=\frac{2}{(k+1)\pi}. $$ If $a_k\leq\frac1{k\pi}$ was true, we would have $$ \frac{2}{(k+1)\pi}<\frac1{k\pi}, $$ which is false for all $k\geq1$.

Answer 2

The integral $\int_{0}^{\infty}\frac{\sin x}{x}dx$ converges in sense of improper Riemann integral. But, $\int_{0}^{\infty}\big|\frac{\sin x}{x}\big| dx$ does not converge as improper Riemann integral.

And this an example of a function which is Riemann integrable( improper ) but not, Lebesgue integrable.