Fresnel Integral

Question

Does someone know how they go from the left side to the right side?

$$\left(\int_{0}^{R}e^{-x^2/2}\,dx\right)^2 = \iint_{[0,R]^2} e^{-(x^2+y^2)/2}\,dx\,dy $$

Thanks.

I guess $G$ is $[0,R]^2$, and the diagram is quite irrelevant. — Jack D'Aurizio, Apr 22 '17 at 16:17

score 1 · Answer 1 · answered Apr 22 '17 at 15:52

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Note that we can write

$$\begin{align} \int_0^R e^{-x^2/2}\,dx&=\sqrt{\int_0^R e^{-x^2/2}\,dx\,\int_0^R e^{-y^2/2}\,dy}\\\\ &=\sqrt{\int_0^R\int_0^R e^{-x^2/2}e^{-y^2/2}\,dx\,dy}\\\\ &=\sqrt{\int_0^R\int_0^R e^{-(x^2+y^2)/2}\,dx\,dy} \end{align}$$

answered Apr 22 '17 at 15:52

Mark Viola

179,405

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Why the square roots? – Bernard Apr 22 '17 at 15:53
@Bernard The left-hand side has not been squared. – Mark Viola Apr 22 '17 at 15:53
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I mean, in the O.P.'s question, it is squared. – Bernard Apr 22 '17 at 15:55
@Bernard I'm starting with the integral that I presume the OP is interested in evaluating. That's all really. – Mark Viola Apr 22 '17 at 15:56

score 1 · Answer 2 · answered Apr 22 '17 at 15:53

Similar to the Gaussian integral calculation $$ \left(\int_0^Re^{-\frac{x^2}{2}}\,dx\right)^2= \int_0^Re^{-\frac{x^2}{2}}\,dx\cdot \int_0^Re^{-\frac{y^2}{2}}\,dy= \int_0^R\int_0^Re^{-\frac{x^2+y^2}{2}}\,dx\,dy=\iint_{[0,R]\times[0,R]}e^{-\frac{x^2+y^2}{2}}\,dxdy. $$

Fresnel Integral

2 Answers2