Novel Integration Technique

How does one integrate the integral shown below? Well, the standard approaches don't work here and so a little trickery is called for. Here's the integral in question:   $$\int_0^\infty \! \frac{\sin x}{x} \mathrm{d}x$$ Firstly, let's define a function $I(b)$ as follows: $$I(b)=\int_0^\infty \! \frac{\sin x}{x}\mathrm{e}^{-bx} \mathrm{d}x \; \text{where } b>=0$$ Now let's differentiate both sides with respect to $b$, not $x$: $$I'(b)=-\int_0^\infty  \! \sin x \, \mathrm{e}^{-bx} \mathrm{d}x \;$$ Using integration by parts, $I(b)$ can be expressed as follows: $$I'(b)=\left. \frac{\mathrm{e}^{-bx}(\cos x + b \sin x)}{b^2+1} \right| _{x=0} ^{x=\infty} =-\frac{1}{b^2+1}$$ The demonstration of this integration by parts can be found here. Now integrating both sides with respect to b, we get: $$I(b)=-\int \!\frac{1}{b^2+1} \mathrm{d}b =-\arctan b+\mathrm{C}$$ Comparing this result for $I(b)$ with our earlier result, we can write: $$-\arctan b +\mathrm{C}=\int_0^\infty \! \frac{\sin x}{x}\mathrm{e}^{-bx} \mathrm{d}x$$ Now as $b \rightarrow \infty$, the equation reduces to: $$-\frac{\pi}{2} + \mathrm{C}=0 \text{ and so } \mathrm{C}=\frac{\pi}{2}$$ Finally, we can write: $$I(b)=-\arctan(b)+\frac{\pi}{2}=\int_0^\infty \! \frac{\sin x}{x}\mathrm{e}^{-bx} \mathrm{d}x$$ Now setting $b$=0, we achieve our desired result, namely: $$I(0)=\int_0^\infty \! \frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$$ A video demonstration of this technique (known as Feynman's Technique, presumably after the famous physicist) can be found here: