572 Real Analysis
(Gh. Bucur, E. Câmpu, S. G ̆aina, ̆ Culegere de Probleme de Calcul Diferen ̧tial ̧si
Integral(Collection of Problems in Differential and Integral Calculus), Editura Tehnic ̆a,
Bucharest, 1967)
518.The method is similar to that for computing the Fresnel integrals, only simpler. If
we denote the integral byI, then
I^2 =∫∞
−∞e−x
2
dx∫∞
−∞e−y
2
dy=∫∞
−∞∫∞
−∞e−(x(^2) +y (^2) )
dxdy.
Switching to polar coordinates, we obtain
I^2 =
∫ 2 π0∫∞
0e−r
2
rdrdθ=∫ 2 π0(
−
1
2
)
e−r
2∣∣
∣∣
∞0dθ=∫ 2 π01
2
dθ=π.Hence the desired formulaI=
√
π.519.Call the integralI. By symmetry, we may compute it over the domain{(u, v, w)∈
R^3 | 0 ≤v≤u≤ 1 }, then double the result. We substituteu=rcosθ,v=rsinθ,w=
tanφ, taking into account that the limits of integration become 0 ≤ θ,φ≤ π 4 , and
0 ≤r≤secθ. We have
I= 2
∫π 40∫ π 40∫secθ0rsec^2 φ
( 1 +r^2 cos^2 θ+r^2 sin^2 θ+tan^2 φ)^2drdθdφ= 2
∫π 40∫ π 40∫secθ0rsec^2 φ
(r^2 +sec^2 φ)^2
drdθdφ= 2
∫π 40∫ π 40sec^2 φ− 1
2 (r^2 +sec^2 φ)∣
∣
∣∣
r=secθr= 0dθdφ=−
∫ π 40∫ π 40sec^2 φ
sec^2 θ+sec^2 φ
dθdφ+(π
4) 2
.
But notice that this is the same as
∫ π 4
0∫ π 40(
1 −
sec^2 φ
sec^2 θ+sec^2 φ)
dθdφ=∫ π 40∫ π 40sec^2 θ
sec^2 θ+sec^2 φdθdφ.If we exchange the roles ofθandφin this last integral we see that
−
∫ π 40∫ π 40sec^2 φ
sec^2 θ+sec^2 φdθdφ+(π
4) 2
=
∫ π 40∫ π 40sec^2 φ
sec^2 θ+sec^2 φdθdφ.Hence
∫ π 4
0∫ π 40sec^2 φ
sec^2 θ+sec^2 φ
dθdφ=π^2
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