42 1 Mathematical Physics
1.21 Consider the Fourier integral theorem
f(x)=
2
π
∫∞
0
cosaxda
∫∞
0
e−ucosaudu
Putf(x)=e−x. Now the definite integral
∫∞
0
e−bucos(au)du=
b
b^2 +a^2
Here
∫∞
0
e−ucosaudu=
1
1 +a^2
∴
2
π
∫∞
0
cosax
1 +a^2
dx=f(x)or
∫∞
0
cosax
1 +a^2
=
π
2
e−x
1.22 The Gaussian distribution is centered ont =0 and has root mean square
deviationτ.
̃f(ω)=√^1
2 π
∫∞
−∞
f(t)e−iωtdt
=
1
√
2 π
∫∞
−∞
1
τ
√
2 π
e−t
(^2) / 2 τ 2
e−iωtdt
=
1
√
2 π
∫∞
−∞
1
τ
√
2 π
e−[t
(^2) + 2 τ (^2) iωt+(τ (^2) iω) (^2) −(τ (^2) iω) (^2) ]/ 2 τ 2
dt
=
1
√
2 π
e−
τ^22 ω^2
{
1
τ
√
2 π
∫∞
−∞
e
−(t+iτ^2 ω^2 )^2
2 τ^2 dt
}
The expression in the Curl bracket is equal to 1 as it is the integral for a
normalized Gaussian distribution.
∴ ̃f(ω)=
1
√
2 π
e−
τ^2 ω^2
2
which is another Gaussian distribution centered on zero and with a root mean
square deviation 1/τ.
1.3.3 Gamma and Beta Functions
1.23Γ(z+1)=limT→∞
∫T
0 e
−xxzdx
Integrating by parts
Γ(z+1)= lim
T→∞
[−xze−x|T 0 +z
∫T
0
e−xxz−^1 dx]
=zlimT→∞
∫T
0
e−xxz−^1 dx=zΓ(z)
becauseTze−T→0asT→∞
Also, sinceΓ(1)=
∫∞
0 e
−xdx= 1
Ifzis a positive integern,
Γ(n+1)=n!