Lecture Notes in Differential Equations - Bruce E. Shapiro

299
Evaluation of the integral depends on the value of ∆. If ∆ = 0 the first term
is logarithmic; if ∆ ∈ Z then the k = ∆ term in the sum is logarithmic;
and if ∆ /∈ Z, there are no logarithmic terms. We assume in the following
that t > 0, so that we can set |t| = t; the t < 0 case is left as an exercise.
Case 1. ∆ = 0. In this case equation (30.151) becomes
{ ∫ 1 ∞
v(t) = Ku(t)
t dt + ∑
∫ }
d k t k−1 dt
k=1
{
}
∞∑ t k
= Ku(t) ln |t| + d k
k
k=1
(30.152)
(30.153)
Substitution of equation (30.139) gives
v(t) = Kt α
∞ ∑
k=0
= Kt α ln |t|
c k t k ⎡
⎣ln |t| +
⎤
∞∑ t j
d j
⎦ (30.154)
j
j=1
∞∑
c k t k + Kt α
k=0
∞ ∑
∞∑
k=0 j=1
c k d j
t j+k
j
(30.155)
Since the product of two differentiable functions is differentiable, then the
product of two analytic functions is analytic, hence the product of two
power series is also a power series. The last term above is the product
of two power series, which we can re-write as a single power series with
coefficients e j as follows,
v(t) = Kt α ln |t|
∞∑
c k t k + Kt α
k=0
∞ ∑
k=0
e k t k (30.156)
∑
∞
= K ln |t| u(t) + t α e k t k (30.157)
for some sequence of numbers {e 0 , e 1 , ...}. This proves equation 30.135.
k=0
Case 2. ∆ = a positive integer. In this case we rewrite (30.151) as
∞∑
∫
v(t) = Ku(t) d k
k=0
t k−(1+∆) dt (30.158)