Leibniz’ Rule
d Z v(x)
dv
du Z v(x) ∂f (x, t)
f (x, t)dt = f (x, v(x)) − f (x, u(x)) +
dt
dx u(x)
dx
dx
∂x
u(x)
1
d Z 2/x sin xt
dt
dx t=1/x t
=
Z 2/x
2
1
sin 2
sin 1
·− 2 −
·− 2 +
cos xt dt
2/x
x
1/x
x
t=1/x
=−
sin xt
sin 2 sin 1
+
+
x
x
x
2/x
t=1/x
sin 2 sin 1
sin 2 sin 1
=−
+
+
−
=0
x
x
x
x
2
s=
Z v
u
1 − e−t
dt
t
∂s
1 − ev l0 H −ev
=
→ −1
∂v
v
→
1
∂s
1 − eu l0 H +eu
=−
→ +1
∂u
u
→
1
3
Evaluate the integral
F (α) =
Z 1
0
xα − 1
dx (α ≥ 0)
ln x
by differentiating under the integral sign
Differentiate both sides with respect to α:
F 0 (α) =
Z 1
0
=
Z 1
0
∂ xα − 1
∂α
ln x
dx
Z 1
1 α
xα+1
x ln x dx =
xα dx =
ln x
α+1
0
"
#1
=
0
1
1+α
Integrating now with respect to α we obtain F (α) = ln(1 + α) + C.
Since F (0) = 0, C = 0 ⇒ F (α) = ln(1 + α)
4
Evaluate the integral
F (α) =
Z ∞
e−x ·
0
Differentiate both sides with respect to α:
5
sin αx
dx
x
Using
R ∞ −x2
e
0
√
=
π
,
2
show that I =
R ∞ −x2
e
0
√
cos αx dx =
π −α2 /4
e
2
Differentiate both sides with respect to α:
Z ∞
dI
2
=
e−x · (−x sin αx) dx
dα
0
2
Integrate “by parts” with u = sin αx, dv = −xe−x dx ⇒ du = α cos αx dx, v =
2
−e−x /2:
∞
1
1 Z ∞ −x2
2
= − e−x sin αx +
e α cos αx dx
2
2 0
0
The first term approaches zero at both limits and the integral is the original integral
I multiplied by α:
α
dI
= I
dα
2
We might recognize this differential equation in the form
ln y =√14 x2 + C ⇒ y = Cex
C = π/2 so:
2 /4
. Thus I = Ce
Z ∞
−α2
4
e
cos αx dx =
0
6
=
xy
2
⇒
dy
y
− 21 x dx ⇒
and at α = 0 the given value yields
√
−x2
dy
dx
π −α2 /4
e
2