3Partial differentiation
IA Differential Equations
3.4
Differentiation of an integral wrt parameter in the
integrand
Consider a family of functions
f
(
x, c
). Define
I
(
b, c
) =
R
b
0
f
(
x, c
)d
x
. Then by
the fundamental theorem of calculus, we have
∂I
∂b
=
f
(
b, c
). On the other hand,
we have
∂I
∂c
= lim
δc→0
1
δc
"
Z
b
0
f(x, c + δc)dx −
Z
b
0
f(x, c) dx
#
= lim
δc→0
Z
b
0
f(x, c + δc) − f(x, c)
δc
dx
=
Z
b
0
lim
δc→0
f(x, c + δc) − f(x, c)
δc
dx
=
Z
b
0
∂f
∂c
dx
In general, if I(b(x), c(x)) =
R
b(x)
0
f(y, c(x))dy, then by the chain rule, we have
dI
dx
=
∂I
∂b
db
dx
+
∂I
∂c
dc
dx
= f(b, c)b
0
(x) + c
0
(x)
Z
b
0
∂f
∂c
dy.
So we obtain
Theorem (Differentiation under the integral sign).
d
dx
Z
b(x)
0
f(y, c(x)) dy = f(b, c)b
0
(x) + c
0
(x)
Z
b
0
∂f
∂c
dy
This is sometimes a useful technique that allows us to perform certain
otherwise difficult integrals, as you will see in the example sheets. However, here
we will only have time for a boring example.
Example. Let I =
R
1
0
e
−λx
2
dx. Then
dI
dλ
=
Z
1
0
−x
2
e
−λx
2
dx.
If I =
R
λ
0
e
−λx
2
dx. Then
dI
dλ
= e
−λ
3
+
Z
λ
0
−x
2
e
−λx
2
dx.