Q:

Expectation of a Non-Central Chi-Squared distribution

I am trying to find following  expectation
\begin{equation}
\mathbb{E} [\frac{1}{1+X}]
\end{equation}
where $X$ is a non central Chi Squared distribution with $K$ degree of freedom and $K$ non-centrality parameter. I know it must be related to Marqum Q function however I can't find something like this in integral tables and I don't have Mathematica.

A:

To define notation, let $X \sim \text{NoncentralChisquared}(k,k)$ with pdf $f(x)$:

Symbolic solution to $\mathbb{E} [\frac{1}{1+X}]$
A closed form solution to $\mathbb{E} [\frac{1}{1+X}]$ does not seem readily obtainable, whether by transforming the pdf to $Y = 1/(1+X)$ or otherwise. 
The neatest form I have obtained is first by deriving the mgf $M_X(t) = \mathbb{E}[e^{t X}]$ as:

where I am using the Expect function from mathStatica/Mathematica to automate the nitty gritties (I should note I am one of the authors of the function). 
The mgf can be used to derive this negative moment (for non-negative random variables) via:
$$\mathbb{E} [\frac{1}{1+ aX}] \quad = \quad \int_0^{\infty } e^{-t} M(-a t) \;dt$$
When $a = 1$, this is:  $$\mathbb{E}[\frac{1}{1+X}] \quad = \quad \int_0^{\infty } \exp\big(-\frac{k t}{1+2t}-t\big) (1+2t)^{-k/2} \;dt$$
which can be solved for odd-valued integers. 
For example, when $k = 1$, the exact solution is:  $$\frac{e^{-i} \sqrt{\pi } \left(\text{erf}\left((-1)^{3/4}\right)+e^{2 i} \text{erfc}\left(\sqrt[4]{-1}\right)+1\right)}{2 \sqrt{2}}  \quad \approx \quad  0.520862$$ 
Approximation
The above seems far too messy. 
Fortunately, an exceptional and simple approximation is: $$\mathbb{E} [\frac{1}{1+X}] \; \; \approx  \; \frac{1}{2(k-1)}  \quad \quad \text{for} \quad k> 7$$ 
The following figure compares:

$\mathbb{E} [\frac{1}{1+X}]$ (calculated by numerical integration) - BLUE curve
... to the simple $\frac{1}{2(k-1)}$ approximation  - ORANGE dashed curve

For $k> 10$, the approximation appears to be accurate to at least 3 decimal places.

