Whilst trying to count certain types of bipartite graphs, I'm lead to try to bound the following quantity
$$
I:=\int_0^\infty \int_0^\infty (x+y)^m e^{-\frac{x^2}{2i} – \frac{y^2}{2j}} dx\,dy
$$
where $i,j$ and $m$ are integers, and I'm interested in the asymptotics for large $i$ and $j$ and potentially $m$ (although it would suffice to have a good upper bound when $i \approx j$ and $m=o(i)$).
One can derive an exact expression for the integral by multiplying out the terms and using known identities for the quantities $\int_0^\infty x^k e^{-\frac{x^2}{2i}} dx$, however the asymptotics of this sum is unclear to me.
It would seem more natural to use a type of `saddle-point' method here, approximate the logarithm of the function around its maximum at $(x_0,y_0) = \left(i \sqrt{\frac{m}{i+j}},j \sqrt{\frac{m}{i+j}} \right)$ using the first two terms of the Taylor series, and so evaluate the integral in this region as a standard Gaussian, and then show that the contribution from outside this region is negligible.
This would lead to the following bound, which I would guess is in fact the correct asymptotic order
$$
I \approx \exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\pi\sqrt{2ij}.
$$
However, I can't get the regions in which the approximation is correct and the region in which the integral is negligible to overlap.
I suspect that this integral will have been considered somewhere in the literature, or at the very least will be susceptible to standard techniques in a field I'm not familiar with.
Best Answer
A upper bound
(With the help of Maple)
With the substitution $u = x+y, v = y$, we have \begin{align} I &= \int_0^\infty \int_0^u u^m \mathrm{e}^{-(u-v)^2/(2i) - v^2/(2j)} \mathrm{d} v \mathrm{d}u\\ &= \int_0^\infty \sqrt{\frac{\pi ij}{2i+2j}}\, u^m \mathrm{e}^{-\frac{u^2}{2i+2j}} \left[\mathrm{erf}\Big(\tfrac{u}{i}\sqrt{\tfrac{ij}{2i+2j}}\Big) + \mathrm{erf}\Big(\tfrac{u}{j}\sqrt{\tfrac{ij}{2i+2j}}\Big) \right] \mathrm{d}u\\ &= \int_0^\infty \sqrt{\frac{\pi ij}{2i+2j}}\, u^m \mathrm{e}^{-\frac{u^2}{2i+2j}} \mathrm{erf}\Big(\tfrac{u}{i}\sqrt{\tfrac{ij}{2i+2j}}\Big)\mathrm{d}u \\ &\qquad + \int_0^\infty \sqrt{\frac{\pi ij}{2i+2j}}\, u^m \mathrm{e}^{-\frac{u^2}{2i+2j}} \mathrm{erf}\Big(\tfrac{u}{j}\sqrt{\tfrac{ij}{2i+2j}}\Big) \mathrm{d}u\\ &= \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}i^{m+1}\int_0^\infty w^m \mathrm{erf}(w)\mathrm{e}^{-w^2i/j} \mathrm{d} w\\ &\qquad + \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}j^{m+1}\int_0^\infty w^m \mathrm{erf}(w)\mathrm{e}^{-w^2j/i}\mathrm{d} w\\ &= \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}i^{m+1} \Big(\int_0^\infty w^m \mathrm{e}^{-w^2i/j} \mathrm{d} w - \int_0^\infty w^m (1 - \mathrm{erf}(w))\mathrm{e}^{-w^2i/j} \mathrm{d} w\Big)\\ &\qquad + \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}j^{m+1} \Big(\int_0^\infty w^m \mathrm{e}^{-w^2j/i}\mathrm{d} w - \int_0^\infty w^m (1-\mathrm{erf}(w))\mathrm{e}^{-w^2j/i}\mathrm{d} w\Big)\\ &= 2\sqrt{\pi}2^{m/2-1}(i+j)^{m/2}\sqrt{ij}\, \Gamma(\tfrac{m+1}{2})\\ &\qquad - \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}i^{m+1} \int_0^\infty w^m (1 - \mathrm{erf}(w))\mathrm{e}^{-w^2i/j} \mathrm{d} w\\ &\qquad - \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}j^{m+1} \int_0^\infty w^m (1-\mathrm{erf}(w))\mathrm{e}^{-w^2j/i}\mathrm{d} w\\ &\le 2\sqrt{\pi}2^{m/2-1}(i+j)^{m/2}\sqrt{ij}\, \Gamma(\tfrac{m+1}{2})\\ &\qquad - \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}i^{m+1} \int_0^\infty w^m \Big(\sqrt{\frac{2\mathrm{e}}{\pi}}\frac{\sqrt{\beta-1}}{\beta}\mathrm{e}^{-\beta w^2}\Big)\mathrm{e}^{-w^2i/j} \mathrm{d} w\\ &\qquad - \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}j^{m+1} \int_0^\infty w^m \Big(\sqrt{\frac{2\mathrm{e}}{\pi}}\frac{\sqrt{\beta-1}}{\beta}\mathrm{e}^{-\beta w^2}\Big)\mathrm{e}^{-w^2j/i} \mathrm{d} w\\ &= 2\sqrt{\pi}2^{m/2-1}(i+j)^{m/2}\sqrt{ij}\, \Gamma(\tfrac{m+1}{2})\\ &\qquad - \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}i^{m+1} \sqrt{\frac{\mathrm{e}}{2\pi}}\frac{\sqrt{\beta-1}}{\beta} (\beta +\tfrac{i}{j})^{-(m+1)/2}\Gamma(\frac{m+1}{2})\\ &\qquad - \sqrt{\pi}(\tfrac{2i+2j}{ij})^{m/2}j^{m+1} \sqrt{\frac{\mathrm{e}}{2\pi}}\frac{\sqrt{\beta-1}}{\beta} (\beta +\tfrac{j}{i})^{-(m+1)/2}\Gamma(\frac{m+1}{2}) \end{align} where $\mathrm{erf}(w) = \frac{2}{\sqrt{\pi}}\int_0^w \mathrm{e}^{-t^2}\mathrm{d} t$ is the error function, and we have used $1 - \mathrm{erf}(w) \ge \sqrt{\frac{2\mathrm{e}}{\pi}}\frac{\sqrt{\beta-1}}{\beta}\mathrm{e}^{-\beta w^2}$ (for $w\ge 0$, $\beta > 1$; see https://en.wikipedia.org/wiki/Error_function). We may choose $\beta = \frac{5}{4}$.