1000 Solved Problems in Modern Physics

68 1 Mathematical Physics

=

λAλBNA^0

(λB−λA)

[

1

λA

{

1

s

−

1

s+λA

}

−

1

λB

{

1

s

−

1

s+λB

}]

=N 10

[

1

s

−

λB

(λB−λA)

1

(s+λA)

+

λA

(λB−λA)

1

(s+λB)

]

∴Nc=NA^0

[

1 +

1

λB−λA

(λAexp(−λBt)−λBexp(−λAt))

]

1.74 (a)L{eax}=

∫∞

0

e−sxeaxdx=

∫∞

0

e−(s−a)xdx

=

1

s−a

,ifs>a

(b) and (c). From part (a),L(eax)=s−^1 aReplaceabyai

L(eiax)=L{cosax+isinax}

=L{cosax}+iL{sinax}

=

1

s−ai

=

s+ai

s^2 +a^2

=

s

s^2 +a^2

+

ia

s^2 +a^2

Equating real and imaginary parts:

L{cosax}=

s

s^2 +a^2

;L{sinax}=

a

x^2 +a^2

1.3.10 Special Functions

1.75 ExpressHnin terms of a generating functionT(ξ,s).

T(ξ,s)=exp[ξ^2 −(s−ξ)^2 ]=exp[−s^2 +^2 sξ]

=

∑∞

n= 0

Hn(ξ)sn

n!

(1)

Differentiate (1) first with respect toξand then with respect to s.

∂T

∂ξ

= 2 sexp(−s^2 + 2 sξ)=

∑

n

2 sn+^1 Hn(ξ)

n!

=

∑

n

snHn′(ξ)

n!

(2)

Equating equal powers of s

Hn′=^2 nHn− 1 (3)

∂T

∂s

=ξ(− 2 s+ 2 ξ)exp(−s^2 + 2 sξ)=

∑

n

(− 2 s+ 2 ξ)snHn(ξ)=

∑

n

sn−^1 Hn(ξ)

(n−1)!

(4)

Equating equal powers of s in the sums of equations

Hn+ 1 = 2 ξHn− 2 nHn− 1 (5)

It is seen that (5) satisfies the Hermite’s equation