42
1 Mathematical Physics
1.21 Consider the Fourier integral theorem
f (x) =
2
π
∞
0
cos ax da
∞
0
e
−u cos au du
Put f (x) = e
−x . Now the definite integral
∞
0
e
−bu cos(au) du =
b
b 2 + a 2
Here
∞
0
e
−u cos au du =
1
1 + a 2
∴
2
π
∞
0
cos ax
1 + a 2 dx = f (x) or
∞
0
cos ax
1 + a 2 =
π
2
e
−x
1.22 The Gaussian distribution is centered on t = 0 and has root mean square
deviation τ.
˜ f (ω) =
1
√
2π
∞
−∞
f (t)e
−iωt dt
=
1
√
2π
∞
−∞
1
τ
√
2π
e
−t
2 /2τ
2 e
−iωt dt
=
1
√
2π
∞
−∞
1
τ
√
2π
e
−[t
2 +2τ
2 iωt+(τ
2 iω)
2 −(τ
2 iω)
2 ]/2τ
2 dt
=
1
√
2π
e
−
τ 2 ω 2
2
1
τ
√
2π
∞
−∞
e
−(t+iτ 2 ω 2 ) 2
2τ 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
√
2π
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
1.23 Γ(z + 1) = lim T →∞
T
0 e
−x x
z dx
Integrating by parts
Γ(z + 1) = lim
T →∞
[−x
z e
−x
|
T
0 + z
T
0
e
−x x
z−1 dx]
= z lim T →∞
T
0
e
−x x
z−1 dx = zΓ (z)
because T
z e
−T
→ 0 as T → ∞
Also, since Γ(1) =
∞
0 e
−x dx = 1
If z is a positive integer n,
Γ(n + 1) = n!