Behaviour of a double integral

Question

Question: How do I go on proving the finiteness of $$ \int_0^\infty \int_0^\infty \frac{e^{-(x+y)}}{x+y} dx dy $$

To be clear, just a hint or two will suffice.

Answer 1

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Given the pair of positive real numbers $\left(\epsilon,R\right)\in\mathbb{R}_{>0}^{2}$ such that $0<\epsilon<R$, let $T_{\epsilon,R}$ denote the following trapezoidal region of the plane:

$$T_{\epsilon,R}:=\{\left(x,y\right)\in\mathbb{R}^{2}\mid0\le x\land0\le y\land\epsilon\le x+y\le R\}.$$

Note that in the double limit as $R\to+\infty$ and $\epsilon\to0$ from the right, the region $T_{\epsilon,R}$ becomes the entire first quadrant of the plane. Thus, you can express your doubly improper double integral as the limit,

$$\begin{align} \mathcal{I} &=\iint_{\left[0,\infty\right)\times\left[0,\infty\right)}\mathrm{d}x\,\mathrm{d}y\,\frac{e^{-x-y}}{x+y}\\ &=\lim_{\epsilon\to0^{+}}\lim_{R\to\infty}\iint_{T_{\epsilon,R}}\mathrm{d}x\,\mathrm{d}y\,\frac{e^{-x-y}}{x+y},\\ \end{align}$$

Then,

$$\begin{align} \mathcal{I} &=\lim_{\epsilon\to0^{+}}\lim_{R\to\infty}\iint_{T_{\epsilon,R}}\mathrm{d}x\,\mathrm{d}y\,\frac{e^{-x-y}}{x+y}\\ &=\lim_{\epsilon\to0^{+}}\lim_{R\to\infty}\left[\iint_{\{\left(x,y\right)\in T_{\epsilon,R}\mid0\le x\le\epsilon\}}\mathrm{d}x\,\mathrm{d}y\,\frac{e^{-x-y}}{x+y}+\iint_{\{\left(x,y\right)\in T_{\epsilon,R}\mid\epsilon\le x\le R\}}\mathrm{d}x\,\mathrm{d}y\,\frac{e^{-x-y}}{x+y}\right]\\ &=\lim_{\epsilon\to0^{+}}\lim_{R\to\infty}\left[\int_{0}^{\epsilon}\mathrm{d}x\int_{\epsilon-x}^{R-x}\mathrm{d}y\,\frac{e^{-x-y}}{x+y}+\int_{\epsilon}^{R}\mathrm{d}x\int_{0}^{R-x}\mathrm{d}y\,\frac{e^{-x-y}}{x+y}\right]\\ &=\lim_{\epsilon\to0^{+}}\lim_{R\to\infty}\left[\int_{0}^{\epsilon}\mathrm{d}x\int_{\epsilon}^{R}\mathrm{d}t\,\frac{e^{-t}}{t}+\int_{\epsilon}^{R}\mathrm{d}x\int_{x}^{R}\mathrm{d}t\,\frac{e^{-t}}{t}\right];~~~\small{\left[y=t-x\right]}\\ &=\lim_{\epsilon\to0^{+}}\lim_{R\to\infty}\left[\epsilon\int_{\epsilon}^{R}\mathrm{d}t\,\frac{e^{-t}}{t}+\int_{\epsilon}^{R}\mathrm{d}t\int_{\epsilon}^{t}\mathrm{d}x\,\frac{e^{-t}}{t}\right],\\ \end{align}$$

...And from there things get pretty obvious, so I leave the rest to our fearless reader. Cheers :)

Answer 2

Since $f(s):={e^{-s}\over s}$ $(s>0)$ is positive it is sufficient to prove that the integrals over finite subregions of the first quadrant are bounded.

For given $a>0$ consider the trapezoid $T_a:=\bigl\{(x,y)\bigm|x\geq 0,\ y\geq 0,\ a\leq x+y\leq 2a\bigr\}$. One has $$\int_{T_a} f(x,y)\>{\rm d}(x,y)\leq {e^{-a}\over a}\cdot {\rm area}(T_a)={3a\over2}e^{-a}\ .$$ It follows that $$\sum_{k=0}^N \int_{T_{2^k}} f(x,y)\>{\rm d}(x,y)\leq{3\over 2}\sum_{k=0}^N 2^k\>e^{-2^k}<{3\over2}\sum_{k=0}^\infty 2^k\>e^{-k}={3e\over2(e-2)}$$ and $$\sum_{k=1}^N \int_{T_{2^{-k}}} f(x,y)\>{\rm d}(x,y)\leq{3\over2}\sum_{k=1}^N 2^{-k}<{3\over2}\ .$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}\int_{0}^{\infty}{\expo{-\pars{x + y}} \over x + y} \,\dd x\,\dd y & = \int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}} \bracks{\int_{0}^{\infty}\expo{-\pars{x + y}t}\,\dd t} \,\dd x\,\dd y \\[5mm] & = \int_{0}^{\infty}\bracks{\int_{0}^{\infty}\expo{-\pars{1 + t}x}\dd x}^{2}\dd t = \int_{0}^{\infty}{\dd t \over \pars{1 + t}^{2}} = \bbx{1} \end{align}