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?
What you're missing is that $\sqrt{1/x}$ is not the same as $1/\sqrt{x}$ when $x$ is negative. Indeed, $\sqrt{1/(-1)} = \sqrt{-1} = i$ but $1/\sqrt{-1} = 1/i = -i$. You might try asking Wolfram Alpha for $e^{-|x|} \dfrac{\sqrt{\pi}}{\sqrt{2x}} I_{1/2}(x)$.