$\begingroup$

I am trying to show that the following integral is convergent but not absolutely.

$$\int_0^\infty\frac{\sin x}{x}dx.$$

My attempt:

I first obtained the taylor series of $\int_0^x\frac{sin x}{x}dx$ which is as follows: $$x-\frac{x^3}{3 \times 3!}+\frac{x^5}{5\times5!}-\frac{x^7}{7 \times 7!}+\cdots = \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) \times (2n+1)!} $$

Now $\int_0^\infty\frac{\sin x}{x}dx=\lim_{x\to \infty} \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) \times (2n+1)!}$

and I got stuck here! What is the next step?

Answers