?The double improper integral \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}

Chapter 5, Problem 180

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The double improper integral \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}+y^{2} / 2} d y d x\) may be defined as the limit value of the double integrals \(\iint_{D_{a}} e^{-x^{2}+y^{2} / 2} d A\) over disks \(D_{a}\) of radii a centered at the origin, as a increases without bound; that is, \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}+y^{2} / 2} d y d x=\lim _{a \rightarrow \infty} \iint_{D_{a}} e^{-x^{2}+y^{2} / 2} d A\) .

a. Use polar coordinates to show that \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}+y^{2} / 2} d y d x=2 \pi\) .

b. Show that \(\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}\) , by using the relation \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}+y^{2} / 2} d y d x=\left(\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2} / 2} d y\right)\).

Text Transcription:

int_-infty^infty int_-infty^infty e^-x^2 + y^2 / 2 dy dx

iint_D_a e^-x^2 + y^2 / 2 dA

D_a

int_-infty^infty int_-infty^infty e^-x^2 + y^2 / 2 dy dx = lim_a rightarrow infty iint_D_a e^-x^2 + y^2 / 2 dA

int_-infty^infty int_-infty^infty e^-x^2 + y^2 / 2 dy dx = 2 pi

int_-infty^infty e^-x^2 / 2 dx = sqrt 2 pi

int_-infty^infty int_-infty^infty e^-x^2 + y^2 / 2 dy dx = (int_-infty^infty e^-x^2 / 2 dx)(int_-infty^infty e^-y^2 / 2 dy)