Prove using the Fourier integral of a gate function
$f(x) = \left\{ \begin{array}{ll} 0 & T <|x|\lt \infty \ \\ \frac{1}{2} & |x| =T \\ 1 & -T < x < T \\ \end{array} \right. $ that $\int_0^\infty\frac{\sin u}{u}du = \frac{\pi}{2}$.
I started developing the function into the integral, resulting in: $$f(x) = \frac{2}{\pi}\int_0^\infty\frac{\sin\lambda T\cos\lambda x}{ \lambda}d\lambda$$
Great. I assumed I needed to evaluate it at $x = 0$ since that would just leave me with: $$f(0) = 1 =\frac{2}{\pi}\int_0^\infty \frac{\sin(\lambda T)}{\lambda}d\lambda$$ However, now I'm at a loss. I asked around and I got told (by a guy who doesn't really want to understand the math) that it's simply possible to rewrite it like:
$$\frac{2}{\pi}\int_0^\infty \frac{\sin(\lambda T)}{\lambda}d\lambda = \frac{2}{\pi}\int_0^\infty \frac{\sin(\lambda T)}{\lambda T}d(\lambda T)$$
From there on I get how to finish the proof, however I don't understand how it's possible to make such a substitution, and he just learned it without understanding and couldn't show me how he got there, just "it's the way it is". So, how can the left integral be turned into the right one?
