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:  $${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}\mathrm{d}x$$Firstly, let's define a function $I(b)$ as follows:$$I(b)={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}{\mathrm{e}}^{-bx}\mathrm{d}x\text{where }b>=0$$Now let's differentiate both sides with respect to $b$, not $x$:$$I^{\prime }(b)=-{\int }_0^{\mathrm{\infty }}\sin x{\mathrm{e}}^{-bx}\mathrm{d}x$$Using integration by parts, $I(b)$ can be expressed as follows: $$I^{\prime }(b)={\frac{{\mathrm{e}}^{-bx}(\cos x+b\sin x)}{b^2+1}|}_{x=0}^{x=\mathrm{\infty }}=-\frac{1}{b^2+1}$$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:$$-\arctan b+\mathrm{C}={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}{\mathrm{e}}^{-bx}\mathrm{d}x$$Now as $b\to \mathrm{\infty }$, the equation reduces to: $$-\frac{\pi }{2}+\mathrm{C}=0\text{ and so }\mathrm{C}=\frac{\pi }{2}$$Finally, we can write:$$I(b)=-\arctan (b)+\frac{\pi }{2}={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}{\mathrm{e}}^{-bx}\mathrm{d}x$$Now setting $b$=0, we achieve our desired result, namely:$$I(0)={\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi }{2}$$A video demonstration of this technique (known as Feynman's Technique, presumably after the famous physicist) can be found here: