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Section 15.3: Integration in Polar Coordinates

Exercise 40

  1. We define the improper integral (over the entire plane \(\mathbb{R}^2\)) \(I = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)} dA = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dy dx = \lim_{a \to \infty} \iint_{D_a} e^{-(x^2+y^2)} dA\) where \(D_a\) is the disk with radius \(a\) and center the origin. Show that \(\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dA = \pi\).
  2. An equivalent definition of the improper integral in part (a) is \(\iint_{\mathbb{R}^2} e^{-(x^2+y^2)} dA = \lim_{a \to \infty} \iint_{S_a} e^{-(x^2+y^2)} dA\) where \(S_a\) is the square with vertices \((\pm a, \pm a)\). Use this to show that \(\int_{-\infty}^\infty e^{-x^2} dx \int_{-\infty}^\infty e^{-y^2} dy = \pi\).
  3. Deduce that \[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}. \]
  4. By making the change of variable \(t = \sqrt{2}x\), show that \(\int_{-\infty}^\infty e^{-x^2/2} dx = \sqrt{2\pi}\). (This is a fundamental result for probability and statistics.)