A proof of an equation containing improper gamma distribution

Tomonari MASADA @ Nagasaki University

January 21, 2013

We’d like to prove the following equation:

exp

{−∫ ∞

0

a(1 − e−tz)z−1e−zdz

}= (1 + t)−a . (1)

This equation appears on page 92 of Poisson Processes by J. F. C. Kingman, where we replace µ(A) by a.By taking the logarithm of both sides, we obtain∫ ∞

0

(1 − e−tz)z−1e−zdz = ln(1 + t) (2)

and would like to prove this.

e−tz can be expanded as follows:

e−tz = 1 − t

1!z +

t2

2!z2 − t3

3!z3 +

t4

4!z4 − · · · . (3)

Therefore, we can rewrite the integral∫∞

0(1 − e−tz)z−1e−zdz as follows:∫ ∞

0

(1 − e−tz)z−1e−zdz

=

∫ ∞0

(t

1!z − t2

2!z2 +

t3

3!z3 − t4

4!z4 + · · ·

)z−1e−zdz

=t

1!

∫ ∞0

z1−1e−zdz − t2

2!

∫ ∞0

z2−1e−zdz +t3

3!

∫ ∞0

z3−1e−zdz − t4

4!

∫ ∞0

z4−1e−zdz + · · ·

=t

1!Γ(1) − t2

2!Γ(2) +

t3

3!Γ(3) − t4

4!Γ(4) + · · ·

=t

1− t2

2+t3

3− t4

4+ · · · , (4)

where we use the gamma integral formula, i.e.,∫∞

0zb−1e−azdz = Γ(b)

ab .

On the other hand, ln(1 + t) can be expanded as follows:

ln(1 + t) = 0 +1

1 + t

∣∣∣∣0

· t1!

− 1!

(1 + t)2

∣∣∣∣0

· t2

2!+

2!

(1 + t)3

∣∣∣∣0

· t3

3!− 3!

(1 + t)4

∣∣∣∣0

· t4

4!+ · · ·

=t

1− t2

2+t3

3− t4

4+ · · · . (5)

Therefore, ∫ ∞0

(1 − e−tz)z−1e−zdz = ln(1 + t) . (6)

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