Section 16.4, Problem 36
Part (a): If Da is the disk of radius a centered at the origin
ZZ
Z 2π Z a
2
2
2
e−r r dr dθ
e−(x +y ) dA =
(1)
0
0
Da
where we have written the integral in polar coordinates using x2 + y 2 = r2 and dA = r dr dθ. This
integral is straightforward to evaluate. Making the substitution u = −r2 , du = −2r dr, we have
obtain
Z 2π Z −a2
Z 2π
Z 2π
2
du
1 u −a2
1 −a2
1 −a2
eu
dθ =
[e ]0 dθ =
(e
(e
− 1) dθ = 2π
− 1) = π(1 − e−a )
−2
−2
−2
−2
0
0
0
0
(2)
RR
2
2
2
Thus Da e−(x +y ) dA = π(1 − e−a ). To obtain the improper integral over the entire plane, we
take the limit as a → ∞.
ZZ
ZZ
2
2
2
−(x2 +y 2 )
e−(x +y ) dA = lim π(1 − e−a ) = π
I=
e
dA = lim
(3)
a→∞
R2
a→∞
Da
Part (b): The equivalent definition of the improper integral is as the limit as a → ∞ of integrals
over squares Sa = {(x, y) | −a ≤ x ≤ a, −a ≤ y ≤ a}. Clearly,
Z aZ a
ZZ
Z aZ a
2
2
2
2
2
2
e−(x +y ) dA =
e−(x +y ) dx dy =
e−x e−y dx dy
(4)
−a
Sa
−a
−a
−a
2
Since e−y is constant with respect to x, we can pull it out of the inner integral:
Z a Z a
2
−x2
e
dx e−y dy
−a
(5)
−a
Since the expression in square brackets is constant with respect to y, we can pull it out of the
dy-integral:
Z a
Z a
2
2
e−x dx
e−y dy
(6)
−a
−a
Thus
ZZ
I = lim
a→∞
Sa
2
e−(x
+y 2 )
a
Z
a→∞
2
e−x dx
dA = lim
Z
−a
a
2
e−y dy =
−a
Since in part (a) we found I = π, we get
Z ∞
Z
2
e−x dx
−∞
∞
Z
∞
−∞
2
e−x dx
Z
∞
−∞
(7)
−∞
2
e−y dy = π
(8)
−∞
Part (c): Now we observe that both integrals in the equation above are actually equal:
R∞
2
e−y dy
R∞
−∞
2
e−x dx =
2
e−y dy, since x and y are just dummy variables. Thus
Z
∞
e
−x2
−∞
1
2
dx = π
(9)
R∞
√
2
Hence the integral in question is ± π. It’s clear that −∞ e−x dx is positive, since it is the integral
√
of a positive function. Thus it equals π:
Z ∞
√
2
e−x dx = π
(10)
−∞
√
√
2 x. Then x2 = t2 /2, and dt = 2 dx, and when we substitute:
Z ∞
Z ∞
Z ∞
√
2
2
2
dt
1
e−t /2 √ = √
e−x dx =
π=
e−t /2 dt
(11)
2
2
−∞
−∞
−∞
√
Moving the factor of 2 over to the other side and changing the dummy variable from t back to x
gives
Z ∞
√
2
e−x /2 dx = 2π
(12)
Part (d): Let t =
−∞
√
Note: The function f (x) = (e−x /2 )/ 2π is called Rthe Gaussian or normal distribution, which
∞
is important in probability theory. This function has −∞ f (x) dx = 1, which is required for any
√
probability distribution. This requirement explains the importance of the factor 2π.
2
2