I'm interested in the asympotic behaviour (assuming an exact solution is intractable) of $$I(m,d)=\int_0^{\infty} \left[ Q(m,x)\right]^d dx$$

for fixed $d \in \mathbb{N}$ (in particular, for $d=3$) and $m\to +\infty$.

Here $Q(m,x) = \frac{\Gamma(m,x)}{\Gamma(m)} $ is the upper regularized gamma function.

Empirically, it seems that $I(m,3) = m - a \sqrt{m} +O(1)$ with $a \approx 0.835$



The answer for $d=3$ is $a=\frac{3}{2\sqrt{\pi}}=0.846283 \dots$.


The integral is splitted in two at $x = m$. $$ \int_{0}^{m} \ Q(m,x)^d \ dx + \int_{m}^{\infty} \ Q(m,x)^d \ dx $$ For the asymptotic calculation of the integrals two approximations of the reguralized gamma function, $Q(m,x)$, for large $m$ are needed:

  1. for $0\le x \le m$ (see, e.g., http://dlmf.nist.gov/8.12.E4 ): $$ Q(m, x)\sim \frac{1}{2} \text{erfc} \left(-\sqrt{m}\ \sqrt{\frac{x}{m}-1- \ln \frac{x}{m}}\right), $$
  2. for $m \le x$ (see, e.g., http://dlmf.nist.gov/8.12.E18 ) $$ Q(m, x)\sim \frac{x^{m-\frac{1}{2}} e^{-x}}{\Gamma(m)} \ e^{\frac{(x-m)^2}{2\ x}} \text{erfc} \left(\frac{x-m}{\sqrt{2\ x}}\right). $$

ad 1.

First we derive the following asymptotic integral for $m \rightarrow \infty$ ($d \in \mathbb{N}$) using an integral representation of the error function (see, e.g., http://dlmf.nist.gov/7.7.E1 ). Please, note that there is no minus sign in the argument. $$ \int_{0}^{1} dy \ \text{erfc}(\sqrt{m}\ \sqrt{y-1-\ln y})^{n} $$

$$ = \left( \frac{2}{\pi}\right)^{n} \int_{0}^{1} dy \ e^{-n\ m (y-1-\ln\ y)} \ \prod_{i=1}^{n} \left( \ \int_{0}^{\infty} dt_{i}\ (1 + t_{i}^{2} )^{-1} \ e^{- m \ t_{i}^{2} (y-1-\ln y) }\right) $$

$$ = \left( \frac{2}{\pi} \right)^{n} \left( \prod_{i=1}^{n}\int_{0}^{\infty} dt_{i} \right)\ \left( \prod_{i=1}^{n} (1 + t_{i}^{2})^{-1} \right) \int_{0}^{1} dy \ e^{-m\ (n + \sum_{i}t_{i}^{2})\ (y-1-\ln y)}. $$ where the summation in the exponent is understood from $i=1$ to $n$. We assume, that all integrals converge to justify the exchange of integrations.

We use Laplace' method for asymptotic expansion of the $y$ integral: The exponent as a function of $y$ is expanded around its minimum $y=1$. With $\eta= y-1$ $$ y-1-\ln y \sim \frac{1}{2}\eta^2. $$ After changing the integral variable to $\eta$ and extending the integration limit neglecting an exponentially small error, we get for the inner integral $$ \int_{0}^{1} dy \ e^{-m\ (d+\sum_{i}t_{i}^{2})\ (y-1-\ln y)} $$

$$ \sim \int_{0}^{\infty} d\eta \ e^{-m\ (d+\sum_{i}t_{i}^{2})\frac{\eta^2}{2}\ } $$

$$ = \sqrt{\frac{\pi}{2\ m}} \left(d+\sum_{i=1}^{d}t^{2}_{i}\right)^{-1/2}. $$

Together $$ \int_{0}^{1} dy \ \text{erfc}(\sqrt{m}\ \sqrt{y-1-\ln y})^{n} \sim \left(\frac{2}{\pi}\right)^{n-\frac{1}{2}} \ m^{-1/2}\ I_{n}, $$

with $$ I_{n}:= \left(\prod_{i=1}^{n}\ \int_{0}^{\infty}dt_{i}\right)\ \left(n + \sum_{i=1}^{n} t_{i}^{2}\right)^{-1/2}\prod_{i=1}^{n}\ \left(1 + t^{2}_{i} \right)^{-1} $$

Mathematica 11 gives the following results for $I_{n}$ $$ I_{1} = 1, \ I_{2} = \pi\left(1-\frac{1}{\sqrt{2}}\right), \ I_{3} = 1.0356625\dots, $$

$$ I_{4} = 1.273085\dots, \ I_{5} = 1.6458\dots, $$

(I did not invest much effort to calculate the integrals symbolically. This might be worth a question on MO ?)

Altogether and by using the identity for the error function, $$ \text{erfc} (-z)=2-\text{erfc}(z), $$

one gets for the first part of the integral (a change to variable $y=x/m$ is included) $$ \int_{0}^{m}dx \ Q(m,x)^d \sim 2^{-d} \int_{0}^{m}dx \ \text{erfc} \left(-\sqrt{m}\ \sqrt{\frac{x}{m}-1-\ln \frac{x}{m}}\right)^{d} $$

$$ = m \int_{0}^{1}dy\ \left[1-\frac{1}{2}\text{erfc} (\sqrt{m}\ \sqrt{y-1-\ln y})\right]^{d} $$

$$ = m - m \sum_{i=1}^{d}(-2)^{-i} {d \choose i} \int_{0}^{1}dy\ \text{erfc}(\sqrt{m}\ \sqrt{y-1-\ln y})^{i} $$

$$ = m - m^{-1/2} \sqrt{\frac{\pi}{2}}\sum_{i=1}^{d}(-\pi)^{-i} {d \choose i}\ I_{i} $$

ad 2.

For the second part transform integration variable to $y=x/m$, use the same integral representation of the error function (see, e.g., http://dlmf.nist.gov/7.7.E1 ) and interchange integrations $$ \int_{m}^{\infty}dx \ Q(m,x)^{d} \sim \int_{m}^{\infty}dx\ \left(\sqrt{\frac{\pi}{2}} \frac{x^{m-\frac{1}{2}} \ e^{-x}}{\Gamma(m)} e^{-\frac{(x-m)^2}{2x}}\ \text{erfc}\left(\frac{x-m}{\sqrt{2x}}\right)\right)^{d} $$

$$ = \frac{m^{d(m-\frac{1}{2})+1}}{\Gamma(m)^d} \left(\frac{\pi}{2}\right)^{d/2} \int^{\infty}_{1} dy\ e^{-d m \left(y+\frac{(y-1)^2}{2y}\right)}\ y^{d \left(m-\frac{1}{2}\right)} \ \text{erfc} \left(\sqrt{m}\frac{y-1}{\sqrt{2y}}\right)^{d} $$

$$ = \frac{m^{d \left(m-\frac{1}{2}\right) + 1}}{\Gamma(m)^d} \left(\frac{2}{\pi}\right)^{d/2} \left( \prod_{i=1}^{d}\int_{0}^{\infty}dt_{i} \ (1+t_{i}^{2})^{-1} \right) \int_{1}^{\infty} dy \ e^{-m \left(d+\sum_{i}t_{i}^{2}\right) \frac{(y-1)^{2}}{2y} }\ y^{d \left( m - \frac{1}{2} \right)} $$

The $y$ integral is approximated by Laplace' method. The integrand is written as $$ \exp \left[ - m \left( \left( d + \sum_{i} t^{2}_{i} \right) \frac{ (y-1)^{2}}{2y} + d \ln y \right) \right] y^{-d/2}. $$

The term in the square brackets is extremal at the lower integration limit $y=1$. Expanding to second order in $\eta=y-1$ gives $$ -m d - m \ \eta^2 \ \frac{\sum_{i}t_{i}^{2}+d}{2}. $$

Integrating from $\eta=0$ to infinity, again neglecting errors by exponentially small contributions, gives for the asymptotic approximation of the second part of the integral, $I(m,d)$, $$ \int_{m}^{\infty}dx \ Q(m,x)^d \sim \frac{m^{d \left(m-\frac{1}{2}\right) + 1}}{\Gamma(m)^d} \left(\frac{2}{\pi}\right)^{(d-1)/2} e^{- d m} m^{-1/2} I_{d} $$

The term $ y^{-d/2}$ in the integrand is not dependent on $m$ and can be set to 1 in this approximation.

After approximating the $\Gamma$ function by Stirling this gives for the second part of the integral, $I(d,m)$, $$ \int_{m}^{\infty}dx \ Q(m,x)^d \sim \frac{\pi^{-d+\frac{1}{2}}}{\sqrt{2}} m^{-1/2} I_{d} $$

Taken all together we have asymptotically for large $m$ $$ I(m,d) \sim m - m^{-1/2} \sqrt{\frac{\pi}{2}}\left[ \sum_{i=1}^{d-1}\left( \pi^{-i} (-1)^{i+1} {d \choose i}\ I_{i}\right) - (1 + (-1)^{d})\ \pi^{-d} I_{d} \right] $$

and finally $$ I(m,3) \sim m - m^{1/2}\frac{3}{2\sqrt{\pi}} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.