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\end{equation}  Which is making the results look suspiciously like $x$ times the 'regular derivative'. Just going by the rule that we times the factor $(1-\delta)$ to any occurence of $x$ in the function would leave the derivative of a constant as $0$ still, (which would indeed be $x$ times the regular derivative). However, the first integral shown, the right hand side had the derivative taken, this scenario needs to be protected or contradication will occur! We now wish to perform \begin{equation}  \frac{d}{d\mathbb{I}}\int_{-\infty}^{\infty} e^{-x^2} \;dx =\frac{d}{d\mathbb{I}} \sqrt{\pi}  \end{equation}  Now instead of using an $a$ parameter, we may replace the derivative with the 'regular derivative' times $x$  \begin{equation}  \int_{-\infty}^{\infty} x(-2xe^{-x^2}) \;dx =\frac{d}{d\mathbb{I}} \sqrt{\pi}  \end{equation}  We see the factor of $2$ has presented itself via a new method! Rather than the square root index it has now been generated by the exponential function, however, to keep consistency, if this is true it must be the case that \begin{equation}  \frac{d}{d\mathbb{I}} \sqrt{\pi}=-\sqrt{\pi}  \end{equation}  Such a result is very curious...