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\end{equation}  Such a result is very curious... Another alternative is that only the modulus of the 'regular derivatiive' is taken and the derivative of $\sqrt{\pi}$ is constant, or that this result is a conincidence... We can attempt a new integral and postulate the result... Try \begin{equation}  \frac{d}{d\mathbb{I}}\int_{-\infty}^{\infty} e^{-x^4} \;dx =\frac{d}{d\mathbb{I}} 2\Gamma(\frac{5}{4})  \end{equation}  I would have no idea where to plop an $a$ on the left hand side if we brought one into the exponent, it may well be on the bottom but I don't know at this stage. However using our theorem that we can replace $x$ times the regular derivative on the left hand side and flip the sign of the constant right we would have  \begin{equation}  \int_{-\infty}^{\infty} x^4e^{-x^4} \;dx =\frac{1}{2}\Gamma(\frac{5}{4})  \end{equation}  Our postulated result.