0 is not positive, the manual states

next thing the limit $b\rightarrow\infty$ (or $\mu\rightarrow\infty$) requests the integral $\int_\infty^\infty ... dx$ why this should evaluate to 1?

Sorry

One of the MarcumQ properties is that Limit[MarcumQ[1, 3, \[Mu]],\[Mu] -> Infinity] = 1

the picture does not suggest that

and if one is willing to accept $8$ as a good approximation for $\infty$ than the graphical result agrees with the Fundamental theorem of calculus.

If you had have the $\mu$ in the first argument of MarcumQ the limit would be meaningful

and now you have to ask yourself what $\lgroup\frac{x}{a}\rgroup^{\mu-1} I_{\mu-1}(a x)$ for $\mu\rightarrow\infty$ is. For this approximations may exist.

Hi, I disagree with most of what you said. But thanks anyways for your participation.

It appears that Mathematica does not know this information about MarcumQ. It also seems not to be able to compute a power series expansion about mu=0, oddly just returning the MarcumQ function it self:

In[1]:= Series[MarcumQ[1, 3, \[Mu]], {\[Mu], 0, 2}]

Out[1]= MarcumQ[1, 3, \[Mu]]

This is so, even though Mathematica knows how to compute derivatives, e.g.:

In[2]:= D[MarcumQ[1, 3, \[Mu]], {\[Mu], 3}]

Out[2]= 
3 E^(1/2 (-9 - \[Mu]^2)) \[Mu] Hypergeometric0F1Regularized[1, (
   9 \[Mu]^2)/4] - 
 E^(1/2 (-9 - \[Mu]^2)) \[Mu]^3 Hypergeometric0F1Regularized[1, (
   9 \[Mu]^2)/4] - 
 27/2 E^(1/2 (-9 - \[Mu]^2)) \[Mu] Hypergeometric0F1Regularized[2, (
   9 \[Mu]^2)/4] + 
 9 E^(1/2 (-9 - \[Mu]^2)) \[Mu]^3 Hypergeometric0F1Regularized[2, (
   9 \[Mu]^2)/4] - 
 81/4 E^(1/2 (-9 - \[Mu]^2)) \[Mu]^3 Hypergeometric0F1Regularized[3, (
   9 \[Mu]^2)/4]