The value of the following double integration

Question

$$ \mbox{I need to the value of the following double integration:} \int_{0}^{\infty}\int_{0}^{\infty} \exp\left[1 - \left(x - y - 1\right)^{\, 2} \over 2\left(x - y - 1\right)\right] \,\mathrm{d}y\,\mathrm{d}x. $$ Certain solution is not to my mind

Answer 1

Let $A:=\{(x,y): 0< x-2 \leq y\leq x-1\}$ then for any $(x,y)\in A$, $$\frac{1-(x-y-1)^{2}}{2(x-y-1)}\geq 0$$ which implies that $$\exp\left[\frac{1-(x-y-1)^{2}}{2(x-y-1)}\right]\geq 1.$$ Since exponential is positive and $A\subset \mathbb{R^+}\times \mathbb{R^+}$, it follows that $$\int_{0}^{\infty}\int_{0}^{\infty}\exp\left[\frac{1-(x-y-1)^{2}}{2(x-y-1)}\right]dydx\geq \iint_{A}1 dydx=|A|\\=\int_{x=2}^{+\infty}\int_{y=x-2}^{y=x-1}1 dy dx=\int_{x=2}^{+\infty}1 dx=+\infty.$$

Answer 2

Since the integrand is positive everywhere then $$\int_0^\infty \int_{x}^\infty \exp\left[\frac{1-(x-y-1)^2}{2(x-y-1)}\right] dydx < \int_0^\infty \int_0^\infty \exp\left[\frac{1-(x-y-1)^2}{2(x-y-1)}\right] dydx.$$

The double integral on the left hand side has $y$ running from $x$ to $\infty$ so $x \leq y < \infty$. Thus, $x-y-1 \leq -1$. Let $\alpha = x-y-1$, then $\alpha \leq -1$. Notice that $\frac{1-\alpha^2}{2\alpha} = \frac{(1-\alpha)(1+\alpha)}{2\alpha} \geq 0$ since $\alpha \leq -1$. Since $\frac{1-\alpha^2}{2\alpha} \geq 0$, then $\exp[ \frac{1-\alpha^2}{2\alpha}] \geq 1$. Thus, $$\int_0^\infty \int_0^\infty \exp\left[\frac{1-(x-y-1)^2}{2(x-y-1)}\right] dydx \geq \int_0^\infty \int_{x}^\infty\exp\left[\frac{1-(x-y-1)^2}{2(x-y-1)}\right] dydx \geq \int_0^\infty \int_{x}^\infty 1 dydx = \infty.$$

So the original integral diverges to $\infty$.