Let $(\mathbb{R}^{2},B(\mathbb{R}^{2}),\lambda_2)\; $ , where $\lambda_2$ is the Lebesgue measure for $\mathbb{R}^{2}$
We define :
$$f(x,y) = ye^{-y^{2}(1+x^{2})}1_{(x,y)\,\in\,\mathbb{R}_+\times \,\mathbb{R}_+}$$
We introduce
$$I = \int_{0}^{+\infty}{e^{-{\frac{x^{2}}{2}}}}dx$$
How can we calculate the value of $I$ using $\int fd\lambda _2$ ?
Thanks in advance for your help
I am not sure how it can only be solved by Lebesgue measure but I guess we can somehow use improper Riemann integral and Fubini's (or Tonelli's) theorem.
$\displaystyle \int_{\mathbb{R}^2} f \, d\lambda_2 = \int_{\mathbb{R}_+ \times \mathbb{R}_+} f \, d\lambda_2 = \iint_{\mathbb{R}_+ \times \mathbb{R}_+} ye^{-y^2(1+x^2)} \, dA =: J$
The last improper double integral $J$ can be calculated as iterated integrals in 2 ways by Tonelli-Fubini Theorem:
$\displaystyle J = \int_{0}^{\infty} \! \! \! \int_{0}^{\infty} ye^{-y^2(1+x^2)} \, dydx = \int_{0}^{\infty} \! -\frac{e^{-y^2(1+x^2)}}{2(1+x^2)} \Bigg \vert_{y=0}^{y=\infty} \, dx = \int_{0}^{\infty} \! \frac{1}{2(1+x^2)} \, dx = \frac{1}{2} \arctan(x) \bigg\vert_{0}^{\infty} = \frac{\pi}{4}$
On the other hand
$\displaystyle J = \int_{0}^{\infty} \! \! \! \int_{0}^{\infty} ye^{-y^2(1+x^2)} \, dxdy = \int_{0}^{\infty} \! \! \! \int_{0}^{\infty} ye^{-y^2}e^{-(yx)^2} \, dxdy = \int_{0}^{\infty} \! \! \! \int_{0}^{\infty} \frac{1}{\sqrt{2}} e^{-y^2}e^{-u^2/2} \, dudy$
$\displaystyle = \frac{1}{\sqrt{2}} \int_{0}^{\infty} e^{-y^2} \left( \int_{0}^{\infty} e^{-u^2/2} \, du \right) dy = \frac{1}{\sqrt{2}} \int_{0}^{\infty} e^{-y^2} I \, dy = \frac{I}{\sqrt{2}} \int_{0}^{\infty} e^{-y^2} \, dy = \frac{I}{\sqrt{2}} \int_{0}^{\infty} e^{-v^2/2} \, \frac{dv}{\sqrt{2}}$
$\displaystyle = \frac{I}{2} \int_{0}^{\infty} e^{-v^2/2} \, dv = \frac{I}{2} I = \frac{I^2}{2}$
Thus, $\displaystyle \frac{I^2}{2} = \frac{\pi}{4} \Rightarrow I = \sqrt{\frac{\pi}{2}}$
(We use $u=\frac{yx}{\sqrt{2}}$ and $y=\frac{v}{\sqrt{2}} $ change of variables to obtain integral $I$)
Note: Please feel free to edit (to enrich) this answer.