   Chapter 15.3, Problem 40E

Chapter
Section
Textbook Problem

(a) We define the improper integral (over the entire plane ℝ2) I = ∬ R 2 e − ( x 2 + y 2 )   d A =   ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 )   d y   d x =   lim a → ∞ ∬ D a e − ( x 2 + y 2 )   d A where Da is the disk with radius a and center the origin. Show that   ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 )   d A   =   π (b) An equivalent definition of the improper integral in part (a) is ∬ R 2 e − ( x 2 + y 2 )   d A   =   lim a → ∞   ∬ S a e − ( x 2 + y 2 )   d A where Sa is the square with vertices (±a, ±a). Use this to show that ∫ − ∞ ∞ e − x 2 d x   ∫ − ∞ ∞ e − y 2 d y   =   π (c) Deduce that ∫ − ∞ ∞ e − x 2 d x   =   π (d) By making the change of variable t = 2 x , show that ∫ − ∞ ∞ e − x 2 / 2 d x   =   2 π (This is a fundamental result for probability and statistics.)

(a)

To determine

To show: e(x2+y2)dA=π.

Explanation

Given:

The improper integral,

Here, Da is the disk centered at origin and radius a.

Formula used:

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π, then, Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

From the given region D, it is observed that r varies from 0 to a and θ varies from 0 to 2π. Convert the given function completely into polar coordinates including the limit points by substituting x=rcosθ and y=rsinθ.

f(x,y)=e(x2+y2)=e(r2)

Therefore, by the equation (1) the given integral becomes,

e(x2+y2)dA=lima02π0ae(r2)(r)drdθ=lima02π0are(r2)drdθ

Integrate the function with respect to r and θ by using the equation (2).

02π0are(r2)drdθ=0are(r2)dr02πdθ

Let t=r2

(b)

To determine

To show: ex2dxey2dy=π.

(c)

To determine

To show: ex2dx=π.

(d)

To determine

To show: ex22dx=2π.

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