1000 Solved Problems in Modern Physics

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!