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Let $U:\{(x,y)\in \mathbb R^2;x\gt 0,y\gt 0\}$ and $f:U\to \mathbb R$ be defined as $f(x,y)=e^{-(x^2+y^2)}$.

Let's define the elements of exhaustion $U=\cup L_k$ as

$$L_k=\{re^{i\theta};1/k\le r\le k,1/k\le \theta\le \pi/2-1/k\}$$

I want to prove the integral $\int_Uf(x,y)dxdy$ is convergent and its value is $\pi/4$.

My attempt

Using polar coordinates we have:

$$\int_Uf(x,y)dxdy=\lim_{k\to\infty}\int_{L_k}f(x,y)dxdy=\int_{1/k}^k\int_{1/k}^{\pi/2-1/k}e^{-r^2}rdrd\theta$$

Therefore we have to prove the following:

$$\lim_{k\to\infty}\int_{1/k}^k\int_{1/k}^{\pi/2-1/k}e^{-r^2}rdrd\theta=\pi/4$$

Thus developing the integral inside the limit we have

$\int_{1/k}^k\int_{1/k}^{\pi/2-1/k}e^{-r^2}rdrd\theta$

$=-\frac{1}{2}\int_{1/k}^{\pi/2-1/k}[e^{-r^2}]_{1/k}^kd\theta\ldots$

Am I right so far?

I need help

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    The integrand function is trivially integrable over $(0,+\infty)^2$ since it is positive and bounded by the integrable function $\frac{1}{1+\left(\frac{x^2+y^2}{2}\right)^2}$ (or just because functions in a Schwartz space are obviously integrable.) By symmetry the integral equals one fourth of the pretty well-known integral $$\iint_{\mathbb{R}^2}e^{-x^2-y^2}\,dx\,dy$$ that equals $\pi$. – 2017-01-29
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    @JackD'Aurizio is this integral well-known? where can I read more about this integral? thank you – 2017-01-29
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    Using the polar coordinates $(\int_{-\infty}^\infty e^{-x^2}dx)^2 = \iint_{\mathbb{R}^2}e^{-x^2-y^2}\,dx\,dy = \int_0^\infty 2\pi r e^{-r^2}dr = 2\pi \frac{-e^{r^2}}{2}|_0^\infty = \pi$ – 2017-01-30
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    @user1952009 one of those $-$ signs should be in the exponent – 2017-01-30
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    @user1952009 why does these equalities hold? thank you for your comment – 2017-01-30
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    The second one is a polar change of variable – 2017-01-30
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    @user1952009 I'm having troubles to understand the first equality. Thank you again – 2017-01-30
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    ?? $(\int_{-\infty}^\infty e^{-x^2}dx)^2 = \int_{-\infty}^\infty e^{-x^2}dx \int_{-\infty}^\infty e^{-y^2}dy = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2} e^{-y^2}dxdy$ and it is not needed for what you asked – 2017-01-30
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    Do you know the name of the theorem you're using in the second equality of your previous comment. Thank you! – 2017-01-30
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    For the operation performed here: $$\int_Uf(x,y)dxdy=\lim_{k\to\infty}\int_{L_k}f(x,y)dxdy=\int_{1/k}^k\int_{1/k}^{\pi/2-1/k}e^{-r^2}rdrd\theta$$ How would do this but for $f(x,y)$ be defined in $\mathbb{C}$ ? – 2018-01-24

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