
Differential Entropies for Probability Distributions 151
Using integration by parts with:
u = x − µ dv = (x − µ)e
−(x−µ)
2
/2σ
2
dx
du = dx v = −σ
2
e
−(x−µ)
2
/2σ
2
yields
h
X
= ln
Zσ
√
2π
+
1
2Zσ
3
√
2π
"
−σ
2
(x − µ)e
−(x−µ)
2
/2σ
2
|
b
a
+ σ
2
Z
b
a
e
−(x−µ)
2
/2σ
2
dx
#
= ln
Zσ
√
2π
+
1
2Z
√
2π
a − µ
σ
e
−(a−µ)
2
/2σ
2
−
b − µ
σ
e
−(b−µ)
2
/2σ
2
+
1
2
Z
b
a
1
Zσ
√
2π
e
−(x−µ)
2
/2σ
2
dx
= ln
Zσ
√
2π
+
1
2Z
a − µ
σ
φ
a − µ
σ
−
b − µ
σ
φ
b − µ
σ
+
1
2
= ln
σZ
√
2πee
1
2Z
[(
a−µ
σ
)
φ
(
a−µ
σ
)
−
(
b−µ
σ
)
φ
(
b−µ
σ
)]
In bits, the differential entropy becomes
h
X
= log
2
√
2πeσZe
1
2Z
[(
a−µ
σ
)
φ
(
a−µ
σ
)
−
(
b−µ
σ
)
φ
(
b−µ
σ
)]