Question: Show that 
$$ 
\int\_0^{\infty} \frac{\sin \pi \omega \sin x\omega}{1-\omega^2}d\omega=
\begin{cases}
\frac{\pi}{2}\sin x,&\mbox{ if } 0\leq x\leq\pi\\
\quad\\
0,&\mbox{ if } x\geq\pi
\end{cases}
$$

Here is what I did 
$$
f(x)=
\begin{cases}
\sin x,&\mbox{ if } 0\leq x\leq\pi\\
\quad\\
0,&\mbox{ if } x\geq\pi
\end{cases}
$$
Then I tried to compute the integral

$$
\int\_0^{\infty}[A(\omega)\cos(\omega x)+B(\omega)\sin(\omega x)]d\omega
$$

But I never reach my target. What is wrong?

Answer: 
If you agree that

$$\lim\_{x\to0^+}{1\over x}=\lim\_{x\to0}{1\over |x|}$$

then, using an appropriate "multiplication" theorem for limits of the form $\lim\_{x\to a}(f(x)g(x))=\lim\_{x\to a}(f(x))\lim\_{x\to a}(g(x))$, we have

$$\lim\_{x\to0}{1\over x^2}=\lim\_{x\to0}{1\over|x|^2}=\left(\lim\_{x\to0}{1\over|x|} \right)^2=\left(\lim\_{x\to0^+}{1\over x} \right)^2=\infty$$

**Added later** (at OP's request): All you really need here is a theorem that says 

$$\lim\_{x\to a}f(x)=\lim\_{x\to a}g(x)=\infty\implies \lim\_{x\to a}(f(x)g(x))=\infty$$

But this follows easily from the definition of $\infty$ as a limit: $\lim\_{x\to a}f(x)=\infty$ if for all $M\gt0$ (no matter how large) there exists a $\delta\gt0$ such that $|x-a|\lt\delta\implies f(x)\gt M$. Here's the proof: Let $M\gt0$, and let $M=M\_1M\_2$ with $M\_1,M\_2\gt0$. Since $\lim\_{x\to a}f(x)=\infty$, there is a $\delta\_1\gt0$ such that $|x-a|\lt\delta\_1\implies f(x)\gt M\_1$. Likewise, since $\lim\_{x\to a}g(x)=\infty$, there is a $\delta\_2\gt0$ such that $|x-a|\lt\delta\_2\implies g(x)\gt M\_2$. Now let $\delta=\min(\delta\_1,\delta\_2)$. Then $|x-a|\lt\delta$ implies $f(x)\gt M\_1$ and $g(x)\gt M\_2$, hence $f(x)g(x)\gt M\_1M\_2=M$, and thus $\lim\_{x\to a}(f(x)g(x))=\infty$.

Note: These basic "arithmetic" theorems for limits *do* require the formal (epsilon-delta) definitions for their proofs. But once you have the theorems at your disposal, you can cut way back on the need for the formal definitions. You just have to make sure you're applying the theorems correctly.
You could use a general result:

> 
> If there exists $\epsilon>0$ such that $f(x)\geq g(x)$ for all $0<x<\epsilon$, and $\lim\_{x\to 0^+} g(x)=+\infty$, then $\lim\_{x\to 0^+} f(x)=+\infty$.
> 
> 
> 

Proving that lemma will require some $\epsilon-\delta$ result, but at least it separates it out into a lemma that is obviously true. It is essentially the squeeze theorem at infinity.

Alternatively, if you could use a lemma:

> 
> If $g$ is a function such that $\lim\_{y\to+\infty} g(y)=+\infty$, and $f$ such that $\lim\_{x\to 0^+} f(x)=+\infty$, then $\lim\_{x\to 0^+} g(f(x))=+\infty$.
> 
> 
> 

Basically, the claim that $\lim\_{y\to+\infty} g(y) = +\infty$ is roughly stating that $g$ is continuous at $+\infty$, if $g(+\infty)$ is defined as $+\infty.$

Yet another possible lemma: 

> 
> $$\lim\_{x\to 0^+} f(x)=+\infty \iff \lim\_{x\to 0^+} \arctan f(x) = \frac{\pi}{2}$$
> 
> 
> 

This explains why my first lemma really is the squeeze theorem, since $\arctan$ is strictly increasing.

There are other functions with the same properties as $\arctan$, like $\frac{x}{\sqrt{1+x^2}}$, namely, it is strictly increasing, and:

> 
> $$\lim\_{x\to 0^+} f(x)=+\infty \iff \lim\_{x\to 0^+} \frac{f(x)}{\sqrt{1+f(x)^2}}= 1$$
> 
> 
>