I am wondering about the limiting behavior as $k\rightarrow\infty$ of the following integral:
$$I(k)=\frac{2^{-k/2}}{\Gamma(k/2)}\int_{f(k)}^\infty t^{k/2-1}e^{-t/2}\left(\frac{\gamma\left(\frac{k}{2},\frac{t-f(k)}{2}\right)}{\Gamma(k/2)}\right)^{g(k)-1}dt$$ where $\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}dt$ is the Gamma function, $\gamma(s,x)=\int_0^x t^{s-1}e^{-t}dt$ is the lower incomplete Gamma function, function $g(x)$ is asymptotically greater than $k$, i.e. $g(k)=\omega(k)$ (for example, $g(k)$ could be $k^2$), and $f(k)$ is function that is asymptotically greater than $\sqrt{k/\log(g(k))}$ (that is, $f(k)=\omega(\sqrt{k/\log(g(k))})$). Specifically, I am wondering if $\lim_{k\rightarrow\infty}I(k)=1$ or $\lim_{k\rightarrow\infty}I(k)=0$ or if it converges to some other number.
Where this comes from
Suppose that I have a sequence of $g(k)$ i.i.d. random variables $X_1,X_2,\ldots,X_{g(k)}$ on $[0,\infty)$ and that I order their realizations in a decreasing order: $X^{(1)}\geq X^{(2)}\geq\ldots\geq X^{(g(x))}$. The probability that the difference between the maximum $X^{(1)}$ in this sequence and the second largest $X^{(2)}$ exceeds $f(k)$ can be written down as follows:
$$P\left[X^{(1)}-X^{(2)}\geq f(k)\right]=\int_{f(k)}^\infty f_X(t)[F_X(t-f(k))]^{g(k)-1}dt$$
where $f_X(t)$ and $F_X(t)$ are respectively the probability density and cumulative distribution functions of $X$.
I am interested in learning the complimentary cumulative distribution function for the difference between largest and second-largest elements in a sequence of i.i.d. chi-squared random variables with $k$ degrees of freedom.