TITLE: Calculate $\int_{\mathbb{R}} {e^{-x^2}\, d\lambda (x) } $ using Fubini's theorem
QUESTION [0 upvotes]: Calculate $\int_{\mathbb{R}} {e^{-x^2}\, d\lambda (x) } $ using Fubini's theorem.

A hint is added to the task, suggesting we first calculate: $$\int_{\mathbb{R}^2} {e^{-(x^2+y^2)}\,d(\lambda \times \lambda)}.$$
I calculated the integral using polar coordinates and using Tonelli's theorem:   $$\int_{\mathbb{R}^2} {e^{-(x^2+y^2)}\,d(\lambda \times \lambda)} = \pi $$ but I don't know how this hint relates to the integral of the prompt.

REPLY [2 votes]: Consider
$$
\int_\Bbb{R\times R}e^{-(x^2+y^2)}d(x\times y)\stackrel{\rm Fubini}{=}\int_\Bbb{R}\int_\Bbb{R}e^{-(x^2+y^2)}d\lambda(x)\,d\lambda(y)=\int_\Bbb{R} \int_\Bbb{R}e^{-x^2}e^{-y^2}d\lambda(x)\,d\lambda(y)=\int_\Bbb{R}e^{-x^2}d\lambda(x) \int_\Bbb{R}e^{-y^2}d\lambda(y)=\left(\int_\Bbb{R}e^{-x^2}d\lambda(x)\right)^2.
$$
Thus:
$$
\Leftrightarrow \int_\Bbb{R}e^{-x^2}d\lambda(x)=\sqrt{\int_\Bbb{R\times R}e^{-(x^2+y^2)}d(x\times y)}=\sqrt{\pi}.
$$
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