Convergence of $\int_0^\infty\frac{\sin x}{x}dx$

I'm working on proving the convergence of the below integral, the value of which I know to be $\displaystyle \frac{\pi}{2}$. $$ \int_0^\infty\frac{\sin x}{x}dx $$ I'm also well aware that this question has been answer several times on this site. Yet, I'm writing this as a question because I've tried the below method which looks (to me) like it has some hope, yet I'm not able to see it.

Note: I'm trying to find a proof with the following constraints:

• No using of vector calculus or multiple integrals
• No using of techniques of infinite sequence and series
• I've just been introduced to improper integrals and hence could only use the direct comparison or limit comparison tests or something elementary like that.

My Proof:
Consider $$ \begin{aligned} I &= \int_a^\infty\frac{\sin x}{x}dx \\ &= \lim_{b\rightarrow\infty}\int_a^b\frac{\sin x}{x}dx\\ &= -\lim_{b\rightarrow\infty}\left[\frac{\cos x}{x}\right]_a^b-\lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx \\ &= \frac{\cos a}{a}-\lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx \\ \end{aligned} $$ Now $$ \lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx \leq \lim_{b\rightarrow\infty}\int_a^b\frac{dx}{x^2}=\frac{1}{a} $$ Hence we have $$ \lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx =\frac{1}{a}-h \text{ | }h\geq0 $$ Substituting this result back to $I$, we get $$ I=\frac{\cos a}{a}-\left[\frac{1}{a}-h\right]=\frac{\cos a-1}{a}+h $$

Now $$ \begin{aligned} \int_0^\infty\frac{\sin x}{x}dx &= \lim_{a\rightarrow0}\int_a^\infty\frac{\sin x}{x}dx \\ &= \lim_{a\rightarrow0}\frac{\cos a-1}{a}+h \\ &= h \end{aligned} $$

This looks neat, but since $h$ is not bound on the other end, I don't see how I could proceed.

Solution 1:

The integral $\int^\infty_0\frac{\sin x}xdx$ converges in the same fashion that the sum $\sum_{n=1}^\infty\frac{(-1)^n }n$ converges. This can be seen if we brake the infinte interval $[0,\infty]$ into successive intervals over half periods of the sine function - then the integrand over each half period roughly decreases as $\frac 1x$ and has alternating sign due to the sine function. This idea can be used to show that the integral converges. Firstly we note that the integral of a continous function ($\sin x/x$ is such at $x=0$ due to removability of the singulatrity) over a finite interval always exist. So we only need to show that the integral $\int_{2\pi}^{m\pi}\frac{\sin x}xdx$ exists as $m\to \infty$ for integers $m$. Indeed: $$ I(k)=\int_{2\pi}^{2\pi k }\frac{\sin x}xdx=\frac12\bigg\{\int_{2\pi}^{2\pi k }\frac{\sin x}xdx+\int_\pi^{2\pi k-\pi}\frac{\sin (x+\pi)}{x+\pi}dx\bigg\} $$ where the second term is obtained from the first by change of veariable $x'=x-\pi$. Taking the limit $k\to \infty$ and subtracting and adding finite integrals (over $[2\pi k,2\pi k-\pi]$ and $[2\pi,\pi]$, to the second integral on the right (calling the result now $I_1(k)$) we find: $$ I_1(\infty)=\frac12\lim_{k\to\infty}\int_{2\pi}^{2\pi k}\sin x\bigg[\frac1x-\frac1{x+\pi}\bigg]dx\le\frac{\pi}2\int_{2\pi}^\infty\frac{dx} {x^2} $$ and the last integral is known to exist.