limit of a quotient of integrals involving $e^{-(x+y)^2}$ and $e^{-x^2-y^2}$

Question

I am interested in computing the following limit: $$ \lim_{r\ \to\ {+}0}\,\,\frac{\displaystyle\int_{-r}^{r} \int_{-r}^{r}{\rm e}^{-\left(x\ +\ y\right)^{\,\,2}}\,\, {\rm d}x\,{\rm d}y}{\displaystyle\int_{-r}^{r}\int_{-r}^{r} {\rm e}^{-x^{2}\ -\ y^{2}}\,\,\,{\rm d}x\,{\rm d}y} $$ My first idea was to use polar coordinates and then Fubini's theorem. However since we are not integrating over a "circular" region this didn't work out so well. Then, I remembered a corollary of the Lebesgue differentiation theorem which tells us that if $f \in L_{1}(\mathbb{R})$ then $$ f(b)=\displaystyle\lim_{t \to{+}0}{} \displaystyle \frac{1}{2t}\int_{b-t}^{b+t} f $$ for every $b$ in which $f$ is continuous. We can try to apply this result to the denominator integral writing $$ \int_{-r}^{r} \int_{-r}^{r} e^{-x^2-y^2} dx dy= \left( \int_{-r}^{r} e^{-x^2} dx \right)\left( \int_{-r}^{r} e^{-y^2} dy \right) $$ But it won't work for the other integral since we can't "separate" as in this case.

Any help?

In advance thank you very much.

Answer 1

After some consideration I've found that this limit is trivial! You just have to use a similar corollary of the one you wrote: $$f(0,0)=\lim_{r\to 0^+}\frac{1}{4r^2}\int_{[-r,r]\times [-r,r]}f(x,y)d(x,y),$$ when $f$ is continuous at $(0,0)$.

So the limit is $1$.

Answer 2

Too long for comments.

Using plain algebra and calculus, all antiderivatives and integrals involved are simple since we face a series of error functions.

By the end, we have $$I=\frac{\displaystyle\int_{-r}^{r} \int_{-r}^{r} e^{-(x+y)^2} dx dy}{\int_{-r}^{r} \int_{-r}^{r} e^{-x^2-y^2} dx dy}=\frac{2 \sqrt{\pi } r\, \text{erf}(2 r)+e^{-4 r^2}-1}{\pi \, \text{erf}(r)^2}$$

Using series expansion $$I=1+\frac{2 r^4}{9}\Bigg[1+\frac 1{15}\sum_{n=1}^\infty (-1)^n\, \frac{a_n}{n!}\,r^{2n} \Bigg]$$ where the $a_n$ form the (unknown ?) sequence $$\left\{8,\frac{254}{35},\frac{928}{105},\frac{1508656}{121275},\frac{59 34464}{315315},\frac{140460752}{4729725},\frac{2326232576}{48243195},\cdots \right\}$$

Using the above terms only, this gives a very good approximation up to $r=1$

Answer 3

In fact the problem can be solved simply applying L'Hopital's rule. Let $$F(r)=\int_{-r}^{r}\int_{-r}^{r}e^{-(x+y)^2}dxdy=\int_{-r}^{r}(\int_{-r}^{r}e^{-(x+y)^2}dy)dx$$ $$G(r)=\int_{-r}^{r}\int_{-r}^{r}e^{-x^2-y^2}dxdy=(\int_{-r}^{r}e^{-x^2}dx)^2$$(These two transformations apply the Fubini theorem.)

We recall that if $$\Phi(r)=\int_{f(r)}^{h(r)}\psi(r,x)dx$$,then $$\Phi'(r)=\psi(r,g(r))g'(r)-\psi(r,f(r))f'(r)+\int_{f(r)}^{g(r)}\psi_r(r,x)dx$$

And this formula fits the condition for determining $F'(r)$, where $$\psi(x,r)=\int_{-r}^{r}e^{-(x+y)^2}dy$$

After boring but straightforward calculation we have $$F'(r)=2\int_{-r}^{r}e^{-(r+y)^2}dy+2\int_{-r}^{r}e^{-(r-y)^2}dy\longrightarrow 0$$ $$(r\longrightarrow 0)$$ $$F''(r)=4+4e^{-4r^2}\longrightarrow 8$$ $$(r\longrightarrow0)$$

$$G'(r)=4e^{-r^2}\int_{-r}^{r}e^{-x^2}dx\longrightarrow 0$$ $$(r\longrightarrow 0)$$ $$G''(r)=4(2e^{-2r^2}+e^{-r^2}2r\int_{-r}^{r}e^{-x^2}dx)\longrightarrow 8$$ $$(r\longrightarrow 0)$$

Then the L'Hopital's rule claims that the answer is $\frac{8}{8}=1$