We can compute an improper integral as a limit over regions of a different shape: SS S₁₂e-²-³ dady = lim 11.² e a-00 Sa -2²-y² dxdy where Sa is the square [-a, a] x [-a, a]. Use this definition to conclude that: (1 e ² dz) (edy) = ₁ -2-² dx T
We can compute an improper integral as a limit over regions of a different shape: SS S₁₂e-²-³ dady = lim 11.² e a-00 Sa -2²-y² dxdy where Sa is the square [-a, a] x [-a, a]. Use this definition to conclude that: (1 e ² dz) (edy) = ₁ -2-² dx T
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Use the definition for how to compute an improper
![We can compute an improper integral as a limit over regions of a different shape:
\[
\iint_{\mathbb{R}^2} e^{-x^2-y^2} \, dx \, dy = \lim_{a \to \infty} \iint_{S_a} e^{-x^2-y^2} \, dx \, dy
\]
where \( S_a \) is the square \([-a, a] \times [-a, a]\). Use this definition to conclude that:
\[
\left( \int_{-\infty}^{\infty} e^{-x^2} \, dx \right) \left( \int_{-\infty}^{\infty} e^{-y^2} \, dy \right) = \pi
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8de5cb45-7800-4644-b643-4cdb8453c3e2%2F0ea323b4-5fe1-46e3-90d4-b5dbecf0f2d6%2Fekwf2q_processed.png&w=3840&q=75)
Transcribed Image Text:We can compute an improper integral as a limit over regions of a different shape:
\[
\iint_{\mathbb{R}^2} e^{-x^2-y^2} \, dx \, dy = \lim_{a \to \infty} \iint_{S_a} e^{-x^2-y^2} \, dx \, dy
\]
where \( S_a \) is the square \([-a, a] \times [-a, a]\). Use this definition to conclude that:
\[
\left( \int_{-\infty}^{\infty} e^{-x^2} \, dx \right) \left( \int_{-\infty}^{\infty} e^{-y^2} \, dy \right) = \pi
\]
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