276 CHAPTER 11. FUNDAMENTAL TRANSFORMS

Then using the dominated convergence theorem to differentiate,

H ′ (s) =∫R

(−e−ct2

)t sin(st)dt

=

(e−ct2

2csin(st) |∞−∞−

s2c

∫R

e−ct2cos(st)dt

)=− s

2cH (s) .

Also H (0) =∫R e−ct2

dt. Thus H (0) =∫R e−cx2

dx≡ I and so

I2 =∫R2

e−c(x2+y2)dxdy =∫

∞

0

∫ 2π

0e−cr2

rdθdr =π

c.

For another proof of this which does not use change of variables and polar coordinates, seeProblems 4, 5 below. Hence

H ′ (s)+s

2cH (s) = 0, H (0) =

√π

c.

It follows that H (s) = e−s24c

√π√c . Hence 1√

2π

∫R e−ct2

e−istdt =√

π

c1√2π

e−s24c =

( 12c

)1/2e−

s24c .

This proves the formula in the case of one dimension. The case of the inverse Fouriertransform is similar. The n dimensional formula follows from Fubini’s theorem. ■

With these formulas, it is easy to verify F,F−1 map G to G and F ◦F−1 = F−1 ◦F = id.

Theorem 11.5.8 Each of F and F−1 map G to G . Also for ψ ∈ G , F−1 ◦F (ψ) =ψ

and F ◦F−1 (ψ) = ψ .

Proof: To make the notation simpler,∫

will symbolize 1(2π)n/2

∫Rn . Also, fb (x) ≡

e−b|x|2 . Then from the above, F fb = (2b)−n/2 f(4b)−1 The first claim will be shown if it is

shown that Fψ ∈ G for ψ (x)≡ xα e−b|x|2 because an arbitrary function of G is a finite sumof scalar multiples of functions such as ψ . Using Lemma 11.5.7,

Fψ (t) ≡∫

e−it·xxα e−b|x|2dx

= (−i)−|α|Dαt

(∫e−it·xe−b|x|2dx

), (Differentiating under integral)

= (−i)−|α|Dαt

(e−|t|24b

(√π√b

)n)by Lemma 11.5.7

and this is clearly in G because it equals a polynomial times e−|t|24b . Similarly, F−1 : G → G .

Now consider F−1 ◦F (ψ)(s) where ψ = xα e−b|x|2 was just used. From the above, andintegrating by parts,

F−1 ◦F (ψ)(s) = (−i)−|α|∫

eis·tDαt

(∫e−it·xe−b|x|2dx

)dt

= (−i)−|α| (−i)|α| sα

∫eis·t(∫

e−it·xe−b|x|2dx)

dt

= sα F−1 (F ( fb))(s)