Proof of Laplace Operator for Harmonic Function

i)
I =
∫ 1
0
∫ 1
0
x − y
(x + y)3 dxdy
Using Wolfram Alpha Calculator, we get
∫ 1
0
∫ 1
0
x − y
(x + y)3 dxdy = −1
2
ii)
I =
∫ 1
0
∫ 1
0
x − y
(x + y)3 dydx
Using Wolfram Alpha Calculator, we get
∫ 1
0
∫ 1
0
x − y
(x + y)3 dydx = 1
2
Even though both integrals are not equal, Fubini’s Theorem is not violated.
This is because, (B, Hass, Heil and Weir, 2018) “Fubini’s Theorem states if,
f is continuous in the region R : [a, b] × [c, d] and ”
∫ ∫
R
|f (x, y)| d(x, y) ≤ ∞
then ∫ b
a
(∫ d
c
f (x, y) dx
)
dy =
∫ d
c
(∫ b
a
f (x, y) dy
)
dx
1

For the given integral(s)
∫ ∫
R
∣
∣
∣
∣
x − y
(x + y)3
∣
∣
∣
∣ d(x, y) ≥
∫ 1
0
(∫ x
0
x − y
(x + y)3 dx
)
dy
≥
∫ 1
0
− −y + x
2 (y + x)2 − 1
2y + x
2y (x + y) dy
[the term,
∫ 1
0
1
2y dy = divergent]
≥ ∞
Thus, Fubini’s Theorem criteria is not met for this particular integral and it
cannot be applied to equate the results of i) and ii).
Answer
According to B, Hass, Heil and Weir (2018) “Fubini’s Theorem states that
the order of integration of a function of two variables does not affect the value
of the double integral. Clairaut’s theorem states that the order of differen-
tiation of such a function does not affect the value of the second derivative.
Both of the theorems require continuity (although Fubini’s allows a finite
number of smooth curves to contain discontinuities).”
Now,
gxy = ∂
∂x
∂
∂y g(x, y)
= ∂
∂x
∂
∂y
∫ x
a
∫ y
c
f (s, t) dtds
2

= ∂
∂x
∫ x
a
(∫ y
c
∂
∂y f (s, t) dt
)
ds
[Using Fundamental Theorem:
∫ y
c
∂
∂y f (s, t) dt = f (s, y) − f (s, c)]
[f (s, c) is outside the region of interest since, c < y < d]
[∴ f (s, y) − f (s, c) = f (s, y)]
= ∂
∂x
(∫ x
a
f (s, y) ds
)
= f (x, y)
By Clairaut’s theorem, we have
gxy = gyx
Therefore,
gxy = gyx = f (x, y)
3

Answer
a)
If f is continuous on [0,1] then
∫ 1
0
∫ 1
0
f (x)f (y) dxdy =
[∫ 1
0
f (x) dx
]2
We use Fubini’s Theorem corollary that if f = g(x)h(y) and f is continuous
in R : [a, b] × [c, d], then
∫ b
a
∫ d
c
f dxdy =
∫ b
a
g(x) dx
∫ d
c
h(y) dy
For the given problem, g = f and h = f and [a, b] = [c, d] = [0, 1]. Therefore,
by Fubini’s Theorem corollary, the given statement is True.
4