I’m trying to find a connection between Marcum-Q function, which is defined as: $$Q_M(a,b)=a^{1-M}e^{-\frac{a^2}{2}}\int_{b}^{\infty} x^M \exp^{-\frac{x^2}{2}} \mathrm I_{M-1}(a x)\mathrm dx$$ where $\mathrm I_M(\cdot)$ - is the modified Bessel function and Gauss-Q function, which is defined as: $$Q (b)= \frac{1}{\sqrt{2\pi}}\int_{b}^{\infty} e^{-\frac{x^2}{2}} \mathrm dx$$
I know that
$$\lim_{a\to\infty}\sqrt{a}e^{-a}\mathrm I_{M-1}(a)= \frac{1}{\sqrt{2\pi}}$$ and since $x$ is real and positive I’d like to use this identity.
So I deduce that I need to cancel out an “extra” $ a^{1-M}e^{-\frac{a^2}{2}}$ factor. To do this I multiply by its inverse. Moreover I need to get rid of $x^M$ so I can set $M=0$.
Then I get:
$$
\begin{eqnarray}
\lim_{a\to\infty} Q_0(a,b)\sqrt{a}e^{-a}a^{-1} e^{\frac{a^2}{2}}&=&\lim_{a\to\infty} \frac{ Q_0(a,b)}{ \sqrt{a e}}e^{\frac{(a-1)^2}{2}}=\\
&=&\lim_{a\to\infty}\int_{b}^{\infty} e^{-\frac{x^2}{2}}\sqrt{a}e^{-a} \mathrm I_{-1}(a x)\mathrm dx=\\
&=&\int_{b}^{\infty} e^{-\frac{x^2}{2}}\lim_{a\to\infty}\left(\sqrt{a}e^{-a} \mathrm I_{-1}(a x) \right) \mathrm dx=\\
&=& \frac{1}{\sqrt{2\pi}}\int_{b}^{\infty} e^{-\frac{x^2}{2}} \mathrm dx=Q (b)
\end{eqnarray}
$$
But after I perform numeric comparison in Mathematica I deduce that the obtained result is wrong. But where is the mistake?