I want to show the existence of $\int_0^{\infty}\frac{\sin(x)}{x}dx$.
And my questions is: It does not help when I show that $|\int_{0}^{\infty}\frac{\sin(x)}{x}dx|<\infty$, right? Because the sequence does also attain negative values.
What else can I do?
The only proof I know that $\int_{0}^{\infty} \frac{\sin x}{x} \, dx$ exists is as follows; I apologize if it seems unmotivated.
Consider the iterated integral $$\int_{0}^{\infty} \left(\int_{0}^{\infty} ye^{-xy}(1- \cos x) \, dy \right)dx. \, \, \, \, (\diamondsuit)$$ Evaluating this integral in the order given we have $$\int_{0}^{\infty} \left(\int_{0}^{\infty} ye^{-xy}(1- \cos x) \, dy \right)dx \, = \, \int_{0}^{\infty} (1 - \cos x) \left(\int_{0}^{\infty} ye^{-xy} \, dy \right)dx \, \, \, \, (\star).$$ We evaluate the inner integral by parts with $u = y$ and $dv = e^{-xy} \, dy$ to give $$\int_{0}^{\infty} ye^{-xy} \, dy = \frac{-y}{x}e^{-xy} \bigg|_{0}^{\infty} - \int_{0}^{\infty} \frac{-1}{x}e^{-xy} \, dy \, = \frac{1}{x^2}.$$ (Note that for $$\frac{-y}{x}e^{-xy} \bigg|_{0}^{\infty}$$ to vanish, we are using the fact that exp grows faster at infinity than any polynomial (including the identity function).) We substitute this result into $(\star)$ and integrate by parts with $u = (1 - \cos x)$ and $dv = x^{-2}dx$ to give $$\int_{0}^{\infty} \left(\int_{0}^{\infty} ye^{-xy}(1- \cos x) \, dy \right)dx = \int_{0}^{\infty} \frac{(1 - \cos x)}{x^2} \, dx = \int_{0}^{\infty} \frac{\sin x}{x} \, dx \text{,} \, \, \, \, (*)$$ the integral whose existence you want to prove.
$$ $$ Meanwhile, by Tonelli’s Theorem, since $$F(x,y) := ye^{-xy}(1-\cos x)$$ is a nonnegative measurable function on $[0, \infty) \times [0, \infty)$, we may reverse the order of integration in $(\diamondsuit)$ to give $$\int_{0}^{\infty} \left( \int_{0}^{\infty} ye^{-xy} (1-\cos x)\,dy \right) dx \, = \, \int_{0}^{\infty} y\left( \int_{0}^{\infty}e^{-xy}(1-\cos x) \, dx \right) dy.$$ We integrate by parts with $u = 1 - \cos x$ and $dv = e^{-xy} \, dy$ to give $$\int_{0}^{\infty} \left( \int_{0}^{\infty} ye^{-xy} (1-\cos x)\,dy \right) dx \, = \, \int_{0}^{\infty}y \left(\int_{0}^{\infty} \frac{1}{y} e^{-xy} \sin{x} \, dx \right)dy \, = \, \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-xy} \sin x \, dx \right) dy.$$ Finally, as our appended work shows, $$ \int_{0}^{\infty} e^{-xy} \sin x \, dx \, = \, \frac{1}{1+y^2}. \, \, \, \, (\dagger)$$ Hence, $$\int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-xy} \sin x \, dx \right) dy \, = \, \int_{0}^{\infty} \frac{1}{1 + y^2} \, dy \, = \, \frac{\pi}{2}.$$ Combining this result with $(*)$ gives $$ \int_{0}^{\infty} \frac{\sin x}{x} \, dx \, = \, \frac{\pi}{2}.$$
$$ $$ $$ $$ $$ $$ Work for $(\dagger)$: Let $u = \sin x$, $dv = e^{-xy} \, dx$. Then integrating by parts gives $$\int_{0}^{\infty} e^{-xy} \sin x \, dx \,= \, -\frac{1}{y} e^{-xy} \, \sin x \bigg|_{x=0}^{\infty} - \int_{0}^{\infty} -\frac{1}{y} e^{-xy} \cos x \, dx \, = \, \int_{0}^{\infty} \frac{1}{y}\,e^{-xy} \, \cos x \, dx.$$ Integrating by parts again (this time with $u = \cos x$ and $dv = \frac{1}{y}e^{-xy} \, dy$), we have $$\int_{0}^{\infty}e^{-xy} \sin x \, dx \, = \, - \frac{1}{y^2} \, e^{-xy} \, \cos x \bigg| _{0}^{\infty} \, - \, \frac{1}{y^2} \int_{0}^{\infty} e^{-xy} \, \sin x \, dx = 1 - \frac{1}{y^2} \int_{0}^{\infty} e^{-xy} \, \sin x \, dx.$$ Hence, $$\int_{0}^{\infty}e^{-xy} \sin x \, dx \, = \, \frac{1}{1+y^2}.$$