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: