Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5 pdf

Thus the solution of the integral equation is
φ(x) = x −
λ
λ
2
+ 5λ + 60

5(λ − 24)
6
x
2
+
λ + 60
4

.
2. For x < 1 the integral equation reduces to
φ(x) = x.
For x ≥ 1 the integral equation becomes,
φ(x) = x + λ

1
0
sin(xy)φ(y) dy.
We could solve this problem by writing down the Neumann series. Instead we will use an eigenfunction expansion.
Let {λ
n
} and {φ
n
} be the eigenvalues and orthonormal eigenfunctions of

φ(x) = λ

1
0
sin(xy)φ(y) dy.
We expand φ(x) and x in terms of the eigenfunctions.
φ(x) =
∞

n=1
a
n
φ
n
(x)
x =
∞

n=1
b
n
φ
n
(x), b
n
= x, φ
n
(x)
We determine the coefficients a
n

by substituting the series expansions into the Fredholm equation and equating
2134
coefficients of the eigenfunctions.
φ(x) = x + λ

1
0
sin(xy)φ(y) dy
∞

n=1
a
n
φ
n
(x) =
∞

n=1
b
n
φ
n
(x) + λ

1
0
sin(xy)
∞


n=1
a
n
φ
n
(y) dy
∞

n=1
a
n
φ
n
(x) =
∞

n=1
b
n
φ
n
(x) + λ
∞

n=1
a
n
1
λ
n

φ
n
(x)
a
n

1 −
λ
λ
n

= b
n
If λ is not an eigenvalue then we can solve for the a
n
to obtain the unique solution.
a
n
=
b
n
1 − λ/λ
n
=
λ
n
b
n
λ
n

− λ
= b
n
+
λb
n
λ
n
− λ
φ(x) = x +
∞

n=1
λb
n
λ
n
− λ
φ
n
(x), for x ≥ 1.
If λ = λ
m
, and x, φ
m
 = 0 then there is the one parameter family of solutions,
φ(x) = x + cφ
m
(x) +
∞


n=1
n=m
λb
n
λ
n
− λ
φ
n
(x), for x ≥ 1.
If λ = λ
m
, and x, φ
m
 = 0 then there is no solution.
Solution 48.24
1.
Kx = L
1
L
2
x = λx
2135
L
1
L
2
(L
1

x) = L
1
(L
1
l
2
− I)x
= L
1
(λx − x)
= (λ − 1)(L
1
x)
L
1
L
2
(L
2
x) = (L
2
L
1
+ I)L
2
x
= L
2
λx + L
2

x
= (λ + 1)(L
2
x)
2.
L
1
L
2
− L
2
L
1
=

d
dt
+
t
2

−
d
dt
+
t
2

−


−
d
dt
+
t
2

d
dt
+
t
2

= −
d
dt
+
t
2
d
dt
+
1
2
I −
t
2
d
dt
+

t
2
4
I −

−
d
dt
−
t
2
d
dt
−
1
2
I +
t
2
d
dt
+
t
2
4
I

= I
L
1

L
2
= −
d
dt
+
1
2
I +
t
2
4
I = K +
1
2
I
We note that
e
−t
2
/4
is an eigenfunction corresponding to the eigenvalue λ = 1/2. Since L
1
e
−t
2
/4
= 0 the result
of this problem does not produce any negative eigenvalues. However, L
n

2
e
−t
2
/4
is the product of
e
−t
2
/4
and a
polynomial of degree n in t. Since this function is square integrable it is and eigenfunction. Thus we have the
eigenvalues and eigenfunctions,
λ
n
= n −
1
2
, φ
n
=

t
2
−
d
dt

n−1
e

−t
2
/4
, for n ∈ N.
Solution 48.25
Since λ
1
is in the residual spectrum of T , there exists a nonzero y such that
(T − λ
1
I)x, y = 0
2136
for all x. Now we apply the definition of the adjoint.
x, (T − λ
1
I)
∗
y = 0, ∀x
x, (T
∗
− λ
1
I)y = 0, ∀x
(T
∗
− λ
1
I)y = 0
y is an eigenfunction of T
∗

corresponding to the eigenvalue λ
1
.
Solution 48.26
1.
u

(t) +

1
0
sin(k(s − t))u(s) ds = f(t), u(0) = u

(0) = 0
u

(t) + cos(kt)

1
0
sin(ks)u(s) ds − sin(kt)

1
0
cos(ks)u(s) ds = f(t)
u

(t) + c
1
cos(kt) − c

2
sin(kt) = f(t)
u

(t) = f(t) −c
1
cos(kt) + c
2
sin(kt)
The solution of
u

(t) = g(t), u(0) = u

(0) = 0
using Green functions is
u(t) =

t
0
(t − τ)g(τ) dτ.
Thus the solution of our problem has the form,
u(t) =

t
0
(t − τ)f(τ) dτ −c
1

t

0
(t − τ) cos(kτ) dτ + c
2

t
0
(t − τ) sin(kτ) dτ
u(t) =

t
0
(t − τ)f(τ) dτ −c
1
1 − cos(kt)
k
2
+ c
2
kt − sin(kt)
k
2
2137
We could determine the constants by multiplying in turn by cos(kt) and sin(kt) and integrating from 0 to 1. This
would yields a set of two linear equations for c
1
and c
2
.
2.
u(x) = λ


π
0
∞

n=1
sin nx sin ns
n
u(s) ds
We expand u(x) in a sine series.
∞

n=1
a
n
sin nx = λ

π
0

∞

n=1
sin nx sin ns
n

∞

m=1
a

m
sin ms

ds
∞

n=1
a
n
sin nx = λ
∞

n=1
sin nx
n
∞

m=1

π
0
a
m
sin ns sin ms ds
∞

n=1
a
n
sin nx = λ

∞

n=1
sin nx
n
∞

m=1
π
2
a
m
δ
mn
∞

n=1
a
n
sin nx =
π
2
λ
∞

n=1
a
n
sin nx
n

The eigenvalues and eigenfunctions are
λ
n
=
2n
π
, u
n
= sin nx, n ∈ N.
3.
φ(θ) = λ

2π
0
1
2π
1 − r
2
1 − 2r cos(θ −t) + r
2
φ(t) dt, |r| < 1
We use Poisson’s formula.
φ(θ) = λu(r, θ),
2138
where u(r, θ) is harmonic in the unit disk and satisfies, u(1, θ) = φ(θ). For a solution we need λ = 1 and that
u(r, θ) is independent of r. In this case u(θ) satisfies
u

(θ) = 0, u(θ) = φ(θ).
The solution is φ(θ) = c

1
+ c
2
θ. There is only one eigenvalue and correspondi ng eigenfunc tion,
λ = 1, φ = c
1
+ c
2
θ.
4.
φ(x) = λ

π
−π
cos
n
(x − ξ)φ(ξ) dξ
We expand the kernel in a Fourier series. We could find the expansion by integrating to find the Fourier coefficients,
but it is easier to expand cos
n
(x) directly.
cos
n
(x) =

1
2
(
e
ıx

+
e
−ıx
)

n
=
1
2
n

n
0

e
ınx
+

n
1

e
ı(n−2)x
+ ··· +

n
n − 1

e
−ı(n−2)x

+

n
n

e
−ınx

2139
If n is odd,
cos
n
(x) =
1
2
n


n
0

(
e
ınx
+
e
−ınx
) +

n

1

(
e
ı(n−2)x
+
e
−ı(n−2)x
) + ···
+

n
(n − 1)/2

(
e
ıx
+
e
−ıx
)

=
1
2
n

n
0


2 cos(nx) +

n
1

2 cos((n − 2)x) + ··· +

n
(n − 1)/2

2 cos(x)

=
1
2
n−1
(n−1)/2

m=0

n
m

cos((n − 2m)x)
=
1
2
n−1
n


k=1
odd k

n
(n − k)/2

cos(kx).
2140
If n is even,
cos
n
(x) =
1
2
n


n
0

(
e
ınx
+
e
−ınx
) +

n
1


(
e
ı(n−2)x
+
e
−ı(n−2)x
) + ···
+

n
n/2 − 1

(
e
i2x
+
e
−i2x
) +

n
n/2


=
1
2
n


n
0

2 cos(nx) +

n
1

2 cos((n − 2)x) + ··· +

n
n/2 − 1

2 cos(2x) +

n
n/2

=
1
2
n

n
n/2

+
1
2
n−1

(n−2)/2

m=0

n
m

cos((n − 2m)x)
=
1
2
n

n
n/2

+
1
2
n−1
n

k=2
even k

n
(n − k)/2

cos(kx).
We will denote,

cos
n
(x − ξ) =
a
0
2
n

k=1
a
k
cos(k(x − ξ)),
where
a
k
=
1 + (−1)
n−k
2
1
2
n−1

n
(n − k)/2

.
We substitute this into the integral equation.
φ(x) = λ


π
−π

a
0
2
n

k=1
a
k
cos(k(x − ξ))

φ(ξ) dξ
φ(x) = λ
a
0
2

π
−π
φ(ξ) dξ + λ
n

k=1
a
k

cos(kx)


π
−π
cos(kξ)φ(ξ) dξ + sin(kx)

π
−π
sin(kξ)φ(ξ) dξ

2141
For even n, su bstituting φ(x) = 1 yields λ =
1
πa
0
. For n and m both even or odd, sub stituting φ(x) = cos(mx)
or φ(x) = sin(mx) yields λ =
1
πa
m
. For even n we have the eigenvalues and eigenvectors,
λ
0
=
1
πa
0
, φ
0
= 1,
λ
m

=
1
πa
2m
, φ
(1)
m
= cos(2mx), φ
(2)
m
= sin(2mx), m = 1, 2, . . . , n/2.
For odd n we have the eigenvalues and eigenvectors,
λ
m
=
1
πa
2m−1
, φ
(1)
m
= cos((2m − 1)x), φ
(2)
m
= sin((2m − 1)x), m = 1, 2, . . . , (n + 1)/2.
Solution 48.27
1. First we shift the range of integration to rewrite the kernel.
φ(x) = λ

2π

0

2π
2
− 6π|x − s| + 3(x − s)
2

φ(s) ds
φ(x) = λ

−x+2π
−x

2π
2
− 6π|y| + 3y
2

φ(x + y) dy
We expand the kernel in a Fourier series.
K(y) = 2π
2
− 6π|y| + 3y
2
=
∞

n=−∞
c
n

e
ıny
c
n
=
1
2π

−x+2π
−x
K(y)
e
−ıny
dy =

6
n
2
, n = 0,
0, n = 0
K(y) =
∞

n=−∞
n=0
6
n
2
e
ıny

=
∞

n=1
12
n
2
cos(ny)
2142
K(x, s) =
∞

n=1
12
n
2
cos(n(x − s)) =
∞

n=1
12
n
2

cos(nx) cos(nx) + sin(nx) sin(ns)

Now we substitute the Fourier series expression for the kernel into the eigenvalue problem.
φ(x) = 12λ

2π

0

∞

n=1
1
n
2

cos(nx) cos(ns) + sin(nx) sin(ns)


φ(s) ds
From this we obtain the eigenvalues and eigenfunctions,
λ
n
=
n
2
12π
, φ
(1)
n
=
1
√
π
cos(nx), φ
(2)
n

=
1
√
π
sin(nx), n ∈ N.
2. The set of eigenfunctions do not form a complete set. Only those functions with a vanishing integral on [0, 2π]
can be represented. We consider the equation

2π
0
K(x, s)φ(s) ds = 0

2π
0

∞

n=1
12
n
2

cos(nx) cos(ns) + sin(nx) sin(ns)


φ(s) ds = 0
This has the solutions φ = const. The set of eigenfunctions
φ
0
=

1
√
2π
, φ
(1)
n
=
1
√
π
cos(nx), φ
(2)
n
=
1
√
π
sin(nx), n ∈ N,
is a complete set. We can also write the eigenfunctions as
φ
n
=
1
√
2π
e
ınx
, n ∈ Z.
2143
3. We consider the problem

u − λT u = f.
For λ = λ, (λ not an eigenvalue), we can obtain a unique solution for u.
u(x) = f(x) +

2π
0
Γ(x, s, λ)f(s) ds
Since K(x, s) is self-adjoint and L
2
(0, 2π), we have
Γ(x, s, λ) = λ
∞

n=−∞
n=0
φ
n
(x)φ
n
(s)
λ
n
− λ
= λ
∞

n=−∞
n=0
1
2π

e
ınx
e
−ıns
n
2
12π
− λ
= 6λ
∞

n=−∞
n=0
e
ın(x−s)
n
2
− 12πλ
Γ(x, s, λ) = 12λ
∞

n=1
cos(n(x − s))
n
2
− 12πλ
Solution 48.28
First assume that λ is an eigenvalue of T , T φ = λφ.
p(T )φ =
n


k=0
a
n
T
n
φ
=
n

k=0
a
n
λ
n
φ
= p(λ)φ
2144
p(λ) is an eigenvalue of p(T ).
Now assume that µ is an eigenvalues of p(T ), p(T)φ = µφ. We assume that T has a complete, orthonormal set of
eigenfunctions, {φ
n
} corresponding to the set of eigenvalues {λ
n
}. We expand φ in these eigenfunctions.
p(T )φ = µφ
p(T )

c
n

φ
n
= µ

c
n
φ
n

c
n
p(λ
n
)φ
n
=

c
n
µφ
n
p(λ
n
) = µ, ∀n such that c
n
= 0
Thus all eigenvalues of p(T ) are of the form p(λ) with λ an eigenvalue of T.
Solution 48.29
The Fourier cosine transform is defined,
ˆ

f(ω) =
1
π

∞
0
f(x) cos(ωx) dx,
f(x) = 2

∞
0
ˆ
f(ω) cos(ωx) dω.
We can write the integral equation in terms of the Fourier cosine transform.
φ(x) = f(x) + λ

∞
0
cos(2xs)φ(s) ds
φ(x) = f(x) + λπ
ˆ
φ(2x) (48.8)
2145
We multiply the integral equation by
1
π
cos(2xs) and integrate.
1
π


∞
0
cos(2xs)φ(x) dx =
1
π

∞
0
cos(2xs)f(x) dx + λ

∞
0
cos(2xs)
ˆ
φ(2x) dx
ˆ
φ(2s) =
ˆ
f(2s) +
λ
2

∞
0
cos(xs)
ˆ
φ(x) dx
ˆ
φ(2s) =
ˆ

f(2s) +
λ
4
φ(s)
φ(x) = −
4
λ
ˆ
f(2x) +
4
λ
ˆ
φ(2x) (48.9)
We eliminate
ˆ
φ between (48.8) and (48.9).

1 −
πλ
2
4

φ(x) = f(x) + λπ
ˆ
f(2x)
φ(x) =
f(x) + λ

∞
0

f(s) cos(2xs) ds
1 − πλ
2
/4
Solution 48.30

D
vLu dx dy =

D
v(u
xx
+ u
yy
+ au
x
+ bu
y
+ cu) dx dy
=

D
(v∇
2
u + avu
x
+ bvu
y
+ cuv) dx dy
=


D
(u∇
2
v + avu
x
+ bvu
y
+ cuv) dx dy +

C
(v∇u − u∇v) · n ds
=

D
(u∇
2
v −auv
x
− buv
y
− uva
x
− uvb
y
+ cuv) dx dy +

C

auv

∂x
∂n
+ buv
∂y
∂n

ds +

C

v
∂u
∂n
− u
∂v
∂n

ds
2146
Thus we see that

D
(vLu − uL
∗
v) dx dy =

C
H(u, v) ds,
where
L

∗
v = v
xx
+ v
yy
− av
x
− bv
y
+ (c − a
x
− b
y
)v
and
H(u, v) =

v
∂u
∂n
− u
∂v
∂n
+ auv
∂x
∂n
+ buv
∂y
∂n


.
Let G be the harmonic Green function, which satisfies,
∆G = δ in D, G = 0 on C.
Let u satisfy Lu = 0.

D
(GLu − uL
∗
G) dx dy =

C
H(u, G) ds
−

D
uL
∗
G dx dy =

C
H(u, G) ds
−

D
u∆G dx dy −

D
u(L
∗
− ∆)G dx dy =


C
H(u, G) ds
−

D
uδ(x − ξ)δ(y −η) dx dy −

D
u(L
∗
− ∆)G dx dy =

C
H(u, G) ds
−u(ξ, η) −

D
u(L
∗
− ∆)G dx dy =

C
H(u, G) ds
2147
We expand the operators to obtain the first form.
u +

D
u(−aG

x
− bG
y
+ (c − a
x
− b
y
)G) dx dy = −

C

G
∂u
∂n
− u
∂G
∂n
+ auG
∂x
∂n
+ buG
∂y
∂n

ds
u +

D
((c − a
x

− b
y
)G − aG
x
− bG
y
)u dx dy =

C
u
∂G
∂n
ds
u +

D
((c − a
x
− b
y
)G − aG
x
− bG
y
)u dx dy = U
Here U is the harmonic function that satisfies U = f on C.
We use integration by parts to obtain the second form.
u +

D

(cuG − a
x
uG − b
y
uG − auG
x
− buG
y
) dx dy = U
u +

D
(cuG − a
x
uG − b
y
uG + (au)
x
G + (bu)
y
G) dx dy −

C

auG
∂y
∂n
+ buG
∂x
∂n


ds = U
u +

D
(cuG − a
x
uG − b
y
uG + a
x
uG + au
x
G + b
y
uG + bu
y
G) dx dy = U
u +

D
(au
x
+ bu
y
+ cu)G dx dy = U
Solution 48.31
1. First we differentiate to obtain a differential equation.
φ(x) = λ


1
0
min(x, s)φ(s) ds = λ


x
0
e
s
φ(s) ds +

1
x
e
x
φ(s) ds

φ

(x) = λ

xφ(x) +

1
x
φ(s) ds − xφ(x)

= λ

1

x
φ(s) ds
φ

(x) = −λφ(x)
2148
We note that that φ(x) satisfies the constraints,
φ(0) = λ

1
0
0 · φ(s) ds = 0,
φ

(1) = λ

1
1
φ(s) ds = 0.
Thus we have the problem,
φ

+ λφ = 0, φ(0) = φ

(1) = 0.
The general solution of the differential equation is
φ(x) =








a + bx for λ = 0
a cos

√
λx

+ b sin

√
λx

for λ > 0
a cosh

√
−λx

+ b sinh

√
−λx

for λ < 0
We see that for λ = 0 and λ < 0 only the trivial solution satisfies the homogeneous boundary conditions. For
positive λ the left boundary condition demands that a = 0. The right boun dary condition is then
b

√
λ cos

√
λ

= 0
The eigenvalues and eigenfunctions are
λ
n
=

(2n − 1)π
2

2
, φ
n
(x) = sin

(2n − 1)π
2
x

, n ∈ N
2149
2. First we differentiate the integral equation.
φ(x) = λ



x
0
e
s
φ(s) ds +

1
x
e
x
φ(s) ds

φ

(x) = λ

e
x
φ(x) +
e
x

1
x
φ(s) ds −
e
x
φ(x)

= λ

e
x

1
x
φ(s) ds
φ

(x) = λ

e
x

1
x
φ(s) ds −
e
x
φ(x)

φ(x) satisfies the differential equation
φ

− φ

+ λ
e
x
φ = 0.
We note the boundary conditions,

φ(0) − φ

(0) = 0, φ

(1) = 0.
In self-adjoint form, the problem is

e
−x
φ



+ λφ = 0, φ(0) − φ

(0) = 0, φ

(1) = 0.
The Rayleigh quotient is
λ =
[−
e
−x
φφ

]
1
0
+


1
0
e
−x
(φ

)
2
dx

1
0
φ
2
dx
=
φ(0)φ

(0) +

1
0
e
−x
(φ

)
2
dx


1
0
φ
2
dx
=
(φ(0))
2
+

1
0
e
−x
(φ

)
2
dx

1
0
φ
2
dx
2150
Thus we see that there are only positive eigenvalues. The differential equation has the general solution
φ(x) =
e
x/2


aJ
1

2
√
λ
e
x/2

+ bY
1

2
√
λ
e
x/2

We define the functions,
u(x; λ) =
e
x/2
J
1

2
√
λ
e

x/2

, v(x; λ) =
e
x/2
Y
1

2
√
λ
e
x/2

.
We write the solution to automatically satisfy the right b oundary condition, φ

(1) = 0,
φ(x) = v

(1; λ)u(x; λ) − u

(1; λ)v(x; λ).
We determine the eigenvalues from the left boundary condition, φ(0) − φ

(0) = 0. The first few are
λ
1
≈ 0.678298
λ

2
≈ 7.27931
λ
3
≈ 24.9302
λ
4
≈ 54.2593
λ
5
≈ 95.3057
The eigenfunctions are,
φ
n
(x) = v

(1; λ
n
)u(x; λ
n
) − u

(1; λ
n
)v(x; λ
n
).
Solution 48.32
1. First note that
sin(kx) sin(lx) = sign(kl) sin(ax) sin(bx)

where
a = max(|k|, |l|), b = min(|k|, |l|).
Consider the analytic function,
e
ı(a−b)x
−
e
ı(a+b)
2
= sin(ax) sin(bx) − ı cos(ax) sin(bx).
2151
−

∞
−∞
sin(kx) sin(lx)
x
2
− z
2
dx = sign(kl) −

∞
−∞
sin(ax) sin(bx)
x
2
− z
2
dx

= sign(kl)
1
2z
−

∞
−∞

sin(ax) sin(bx)
x − z
−
sin(ax) sin(bx)
x + z

dx
= −π sign(kl)
1
2z
(−cos(az) s in(bz) + cos(−az) sin(−bz))
−

∞
−∞
sin(kx) sin(lx)
x
2
− z
2
dx = sign(kl)
π

z
cos(az) s in(bz)
2. Consider the analytic function,
e
ı|p|x
−
e
ı|q|x
x
=
cos(|p|x) − cos(|q|x) + ı(sin(|p|x) − sin(|q|x))
x
.
−

∞
−∞
cos(px) − cos(qx)
x
2
dx = −

∞
−∞
cos(|p|x) − cos(|q|x)
x
2
dx
= −π lim
x→0

sin(|p|x) − sin(|q|x)
x
−

∞
−∞
cos(px) − cos(qx)
x
2
dx = π(|q| − |p|)
3. We use the analytic function,
ı(x − ıa)(x − ıb)
e
ıx
(x
2
+ a
2
)(x
2
+ b
2
)
=
−(x
2
− ab) sin x + (a + b)x cos x + ı((x
2
− ab) cos x + (a + b)x sin x)
(x

2
+ a
2
)(x
2
+ b
2
)
−

∞
−∞
−(x
2
− ab) sin x + (a + b)x cos x
x(x
2
+ a
2
)(x
2
+ b
2
)
= −π lim
x→0
(x
2
− ab) cos x + (a + b)x sin x
(x

2
+ a
2
)(x
2
+ b
2
)
= −π
−ab
a
2
b
2
2152
−

∞
−∞
−(x
2
− ab) sin x + (a + b)x cos x
(x
2
+ a
2
)(x
2
+ b
2

)
=
π
ab
Solution 48.33
We consider the function
G(z) =

(1 − z
2
)
1/2
+ ız

log(1 + z).
For (1 − z
2
)
1/2
= (1 − z)
1/2
(1 + z)
1/2
we choose the angles,
−π < arg(1 − z) < π, 0 < arg(1 + z) < 2π,
so that there is a branch cut on the interval (−1, 1). With this choice of branch, G(z) vanishes at infinity. For the
logarithm we choose the principal branch,
−π < arg(1 + z) < π.
For t ∈ (−1, 1),
G

+
(t) =

√
1 − t
2
+ ıt

log(1 + t),
G
−
(t) =

−
√
1 − t
2
+ ıt

log(1 + t),
G
+
(t) − G
−
(t) = 2
√
1 − t
2
log(1 + t),
1

2

G
+
(t) + G
−
(t)

= ıt log(1 + t).
For t ∈ (−∞, −1),
G
+
(t) = ı

√
1 − t
2
+ t

(log(−t − 1) + ıπ) ,
G
−
(t) = ı

−
√
1 − t
2
+ t


(log(−t − 1) − ıπ) ,
2153
G
+
(t) − G
−
(t) = −2π

√
t
2
− 1 + t

.
For x ∈ (−1, 1) we have
G(x) =
1
2

G
+
(x) + G
−
(x)

= ıx log(1 + x)
=
1
ı2π


−
∞
−1
−2π(
√
t
2
− 1 + t)
t − x
dt +
1
ı2π

1
−1
2
√
1 − t
2
log(1 + t)
t − x
dt
From this we have

1
−1
√
1 − t
2
log(1 + t)

t − x
dt
= −πx log(1 + x) + π

∞
1
t −
√
t
2
− 1
t + x
dt
= π

x log(1 + x) − 1 +
π
2
√
1 − x
2
−
√
1 − x
2
arcsin(x) + x log(2) + x log(1 + x)


1
−1

√
1 − t
2
log(1 + t)
t − x
dt = π

x log x − 1 +
√
1 − x
2

π
2
− arcsin(x)

Solution 48.34
Let F (z) denote the value of the integral.
F (z) =
1
ıπ
−

C
f(t) dt
t − z
From the Plemelj formula we have,
F
+
(t

0
) + F
−
(t
0
) =
1
ıπ
−

C
f(t)
t − t
0
dt,
f(t
0
) = F
+
(t
0
) − F
−
(t
0
).
2154
With W (z) defined as above, we have
W
+

(t
0
) + W
−
(t
0
) = F
+
(t
0
) − F
−
(t
0
) = f(t
0
),
and also
W
+
(t
0
) + W
−
(t
0
) =
1
ıπ
−


C
W
+
(t) − W
−
(t)
t − t
0
dt
=
1
ıπ
−

C
F
+
(t) + F
−
(t)
t − t
0
dt
=
1
ıπ
−

C

g(t)
t − t
0
dt.
Thus the solution of the integral equation is
f(t
0
) =
1
ıπ
−

C
g(t)
t − t
0
dt.
2155
Solution 48.35
(i)
G(τ) = (τ − β)
−1

τ − β
τ − α

γ
G
+
(ζ) = (ζ −β)

−1

ζ − β
ζ − α

γ
G
−
(ζ) =
e
−ı2πγ
G
+
(ζ)
G
+
(ζ) − G
−
(ζ) = (1 −
e
−ı2πγ
)(ζ − β)
−1

ζ − β
ζ − α

γ
G
+

(ζ) + G
−
(ζ) = (1 +
e
−ı2πγ
)(ζ − β)
−1

ζ − β
ζ − α

γ
G
+
(ζ) + G
−
(ζ) =
1
ıπ
−

C
(1 −
e
−ı2πγ
) dτ
(τ − β)
1−γ
(τ − α)
γ

(τ − ζ)
1
ıπ
−

C
dτ
(τ − β)
1−γ
(τ − α)
γ
(τ − ζ)
= −ı cot(πγ)
(ζ − β)
γ−1
(ζ − α)
γ
(ii) Consider the branch of

z −β
z −α

γ
2156
that tends to unity as z → ∞. We find a series expansion of this function about infinity.

z −β
z −α

γ

=

1 −
β
z

γ

1 −
α
z

−γ
=

∞

j=0
(−1)
j

γ
j

β
z

j

∞


k=0
(−1)
k

−γ
k


α
z

k

=
∞

j=0

j

k=0
(−1)
j

γ
j − k

−γ
k


β
j−k
α
k

z
−j
Define the polynomial
Q(z) =
n

j=0

j

k=0
(−1)
j

γ
j − k

−γ
k

β
j−k
α
k


z
n−j
.
Then the function
G(z) =

z −β
z −α

γ
z
n
− Q(z)
2157