where $Q(s,x)=\frac{\Gamma(s,x)}{\Gamma(s)}$ is the upper incomplete gamma function $\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$ regularized by gamma function $\Gamma(s)$, $Q^{-1}(s,z)$ is the solution for $x$ in $z=Q(s,x)$, $a>0$ is a constant, $\frac{1}{2}<b<1$ is also a constant, and $g(x)=\omega(x)$ is a function that is asymptotically greater than $x$.

I am interested in the behavior of the limit $\lim_{x\rightarrow\infty}L(x)$ for varying asymptotics of $f(x)$. Specifically, numerical evaluations seem to indicate that when $f(x)=\omega(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=0$ and when $f(x)=\mathcal{O}(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=1$. However, I am having trouble proving this analytically (my usual Taylor expansion tricks don't seem to work here). Any help?


1 Answer 1


Below I give a sketch of a proof.

First an approximation for the inverse incomplete gamma function, $Q^{-1}$, is needed. Henceforth I assume that

$$ g(x)\sim \gamma x^s $$

for large $x$ with $s>0$. Then ($0<b<1$)

$$ 1-b^{1/g(x)}\sim (\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right), $$

which is small for large $x$.

Now we use a formula due to Tricomi, which can be found here

$$ Q(\alpha+1,\alpha+\sqrt{2 \alpha} y) = \frac{1}{2} \text{erfc}(y), $$

for $\alpha\rightarrow \infty$.

Setting $\alpha=x/2$ one gets for large $x$ (dropping constant and lower terms) the following approximation

$$ Q\left(\frac{x}{2},\frac{x}{2}\left(1+\frac{2 y}{\sqrt{x}}\right)\right) \sim \frac{1}{2} \text{erfc}(y) . $$

Now we choose $y$ such that the right side is approximately equal to $(\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right)$. With the help of the expansion of the inverse erfc, $\text{inverfc}$ (see e.g. here .)

$$ \text{inverfc} (y) \sim \frac{1}{\sqrt{2}} \sqrt{-\ln\left(\pi y^2 \ln \frac{1}{y}\right)} . $$

Again using the fact that $x$ is large, we find $y\sim \sqrt{s} \sqrt{\ln x}$, keeping only the highest order in $\ln x$.

This gives

$$ Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)} \right) \sim \frac{x}{2}\left(1+2\sqrt{ s}\sqrt{ \frac{\ln x}{x}}\right), $$

independent of $\gamma$ and $b$.

Insertion and once more dropping terms of lower order in $x$ (assuming $f(x)/\sqrt{x}\rightarrow 0$ for $x\rightarrow \infty$) $$ L(x) \sim Q\left(\frac{x}{2},\frac{x}{2}\left[1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right)\right]\right). $$

Another asymptotic formula for the incomplete gamma functions comes into play (see, e.g., here):

$$ \Gamma(z,\lambda z) \sim (\lambda z)^z e^{-\lambda z}\frac{1}{z(\lambda - 1)}, $$

For this formula to be valid $\lambda > 1$.

Using the above and Stirling's formula for the $\Gamma(z)$ with $z=x/2$ and setting

$$ \lambda = \lambda(x) := 1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right). $$

after a lot of cancellations the result is

$$ L(x) \sim (2 \sqrt{s \ln x}- f(x)/a)^{-1}\rightarrow 0 $$

for $x \rightarrow \infty$.

Note that

$$ \lambda(x)^{x/2} \sim e^{x(\lambda(x)-1)/2}, $$

which cancels exactly with the exponential term in the approximation, thus leaving only the $\sqrt{x}/(\lambda(x)-1)$.

If we have $\lambda = 1$, i.e. $f(x) = 2 a \sqrt{s \ln x}$, another formula might be used (see, e.g. here)

$$ \Gamma(z,z) \sim \sqrt{\frac{\pi}{2}} z^{z-\frac{1}{2}}e^{-z} . $$

With this I find

$$ L(x)\sim Q \left(\frac{x}{2},\frac{x}{2}\right)\sim \frac{1}{2}, $$

which is in disagreement with the numerical results. One has instead to take into account next order terms.

  • $\begingroup$ Thanks for the derivation! I haven't gotten a chance to completely go through it yet, however, there are two things that I noticed right away in the beginning: 1) I am fine with the assumption of a $g(x)$ being a power function of $x$, but I think you meant $s>1$ (you have $s>0$ which I think is just a typo); 2) Are you sure that $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx\frac{1}{2}\operatorname{erfc}(y)$? The DLMF link you posted states that $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx\frac{1}{2}\operatorname{erfc}(-y)$... Does that impact the result? $\endgroup$
    – Bullmoose
    Aug 22, 2013 at 22:31
  • $\begingroup$ Back to going through your proof and noticed that I made a mistake in my comment: indeed $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx \frac{1}{2}\operatorname{erfc}(y)$ since Tricomi's formula in DLMF is for the complement $P(\alpha+1,\alpha+\sqrt{2\alpha}y)=1-Q(\alpha+1,\alpha+\sqrt{2\alpha}y)$... my mistake, though point #1 in my previous comment still stands. :) Continuing reading the proof... $\endgroup$
    – Bullmoose
    Aug 23, 2013 at 2:03
  • $\begingroup$ I finished reading the proof -- I think it's correct (that was quite a tour de force through DLMF). I think there is one more typo in the final approximation of $Q^{-1}$ (after "This gives"): I think you want the $\ln x$ under the square root sign (it's there for the remainder of the proof). Regarding your last statement, my numerical evaluations weren't perfect and indeed $L(x)$ might be going to $\frac{1}{2}$ instead of 1. Still, it looks like that there is indeed a threshold on the asymptotics of $f(x)$. Thanks! $\endgroup$
    – Bullmoose
    Aug 23, 2013 at 2:29
  • $\begingroup$ Thank you Bullmoose for going through the proof ! I had indeed to flip a lot of DLMF pages. I corrected one typo (the ln is now under the square root). Thank you for that. . $s>0$ is no typo. The proof works perfect for all $s>0$, I guess. One just needs that $g(x)$ grows infinitely as $x \rightarrow \infty$. $\endgroup$ Aug 23, 2013 at 7:08
  • $\begingroup$ Thank you for the proof! I really like how you inverted the $Q$ function -- that's a neat trick that I'll use in the future. While the proof works as is, after a brief search I found this paper by Temme. I haven't had a chance to read it in detail, but it seems to apply to a very general range of scenarios, so I am now wondering whether your derivation of $Q^{-1}$ matches the form given in that paper (I'll try to read it carefully this coming week.) $\endgroup$
    – Bullmoose
    Aug 25, 2013 at 5:01

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