Showing existence of an improper integral by estimating the absolute value 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?
 A: 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.
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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
$$ \int_{0}^{\infty} \frac{\sin x}{x} \, dx \, = \, \frac{\pi}{2}.$$
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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,
$$\int_{0}^{\infty}e^{-xy} \sin x \, dx \, = \, \frac{1}{1+y^2}.$$
