Is the function $\hat{i}_0(x) = e^{-|x|} \sqrt{\frac{\pi}{2x}} I_{\frac{1}{2}}(x)$ positive or negative for negative $x$?
$I_{\alpha}(x)$ above is a modified Bessel function.
Here are my arguments. Considering that $I_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \sinh(x)$, $\hat{i}_0(x)$ can be represented as follows:
$ \hat{i}_0(x) = e^{-|x|} \sqrt{\frac{\pi}{2x}} \sqrt{\frac{2}{\pi x}} \sinh(x)$
$ \hat{i}_0(x) = \frac{e^{-|x|} \sinh(x)}{(\sqrt{x})^2}$
Using the convention that $\sqrt{x} = i\sqrt{-x}$ for negative $x$, we get
$ \hat{i}_0(x) = \frac{e^{-|x|} \sinh(x)}{x}$
which is positive for negative $x$:
However, using the original formula, Wolfram Alpha says that the function is negative for negative $x$:
Am I missing something?