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I want to separate the variables $y$ and $u$ in this double integral:

$$\int_{0}^{t} \int_{\mathbb{R}} \text{exp}\left(-\frac{y^{2}}{2u}\right)dy du,\qquad t>0.$$

Can we do this change of variable

$$x=\frac{y}{\sqrt{2u}}$$ giving

$$\int_{0}^{t} \int_{\mathbb{R}} \sqrt{2u}\: \text{exp}\left(-x^2\right)dxdu\,?$$

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  • $\begingroup$ I edited your title to reflect what you most likely meant to ask. I hope you do not mind! $\endgroup$ Commented Jan 6, 2018 at 5:47

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Yes, your change of variable is correct. You may then write $$ \int_{0}^{t} \int_{\mathbb{R}} \sqrt{2u}\: \text{exp}(-x^{2})dxdu=\left(\int_{0}^{t} \sqrt{2u}\:du\right)\left( \int_{\mathbb{R}}\text{exp}(-x^{2})dx\right) $$ and conclude with the gaussian integral.

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