We accept these conditions and call such an arc a smooth arc.
In further discussion, the positions of start and end points may change depending on a real parameter α. (In applied problems the parameter is commonly the time.) Then γ
a
=γ
a
(α), γ
b
=γ
b
(α), a=a(α), b=b(α). The curve is smooth for each value of α.
We assume that the density
g(
τ) has (
k−1)-th Holder continuous derivative with respect to
τ. In the non-trivial case when
t∈(
a,
b), the hypersingular integral is defined as [
13]:
$$\begin{aligned} I_{k}(t)=\int_{a}^{b}\frac{g(\tau)}{(\tau-t)^{k}}d\tau =&\lim_{\varepsilon\rightarrow0} \Biggl[\int_{a}^{t_{1}}\frac{g(\tau)}{(\tau-t)^{k}}d\tau+\int_{t_{2}}^{b} \frac{g(\tau)}{(\tau-t)^{k}}d\tau \\ &{}-\sum_{m=1}^{k-1}\frac{(m-1)!}{(k-1)!}g^{(k-1-m)}(t) \frac{1-(-1)^{m}}{\varepsilon^{m}}e^{-im\varphi} \Biggr], \end{aligned}$$
(2)
where
t
1 and
t
2 are points on the arc located at the distance
ε from the point
t before and after this point, respectively,
φ is the angle of the tangent at the point
t with the
x-axis (Fig.
1). In the case
k=1, the sum on the r.h.s. of (
2) is not present and the integral (
1) is Cauchy principal value integral; in the case
k=2, it is Hadamard finite-part integral. Equation (
2) is obtained in the way, which provides the common Cauchy principal value integral in the case
k=1. When
k=2, it follows the line of introducing the Hadamard
finite-part integral. Specifically, (i) a small
ε-vicinity
L
ε
is excluded from the integration contour
L, so that integration is performed over the part
L−
L
ε
, where the kernel is non-singular; (ii) successive integration by parts is used for the proper integral over
L−
L
ε
until arriving at the integral with logarithmic kernel; (iii) only
finite parts of the resulting out-of-integral terms are left (they do not depend on
ε). This actually means subtraction of the terms going to infinity, when
ε→0, from the integral over
L−
L
ε
, what is expressed by the sum in the brackets of (
2). From the said it is obvious that (
2) may be also written as:
$$\begin{aligned} I_{k}(t) =\int_{a}^{b}\frac{g(\tau)}{(\tau-t)^{k}}d\tau =&\frac{1}{(k-1)!}g^{(k-1)}(t)i\pi \\ &{}+\frac{1}{(k-1)!} \bigl[g^{(k-1)}(b)\ln(b-t) -g^{(k-1)}(a)\ln(a-t) \bigr] \\ &{}-\sum_{m=1}^{k-1}\frac{(m-1)!}{(k-1)!} \biggl[\frac{g^{(k-1-m)}(b)}{(b-t)^{j}} -\frac{g^{(k-1-m)}(a)}{(a-t)^{j}}\biggr] \\ &{}-\frac{1}{(k-1)!}\int_{a}^{b}g^{(k)}(\tau)\ln(\tau-t)d\tau. \end{aligned}$$
(3)
Equation (
3) shows that the integral
I
k
(
t) is a Holder continuous function of
t on (
a,
b) (for a singular integral (
k=1), the sum on the r.h.s. of (
3) is omitted).
Consider a density
g(
α,
τ) depending on the same parameter
α as the limits of integration. The integral becomes a function of
α:
$$ I_{k}(\alpha,t)=\int _{a(\alpha)}^{b(\alpha)}\frac{g(\alpha,\tau)}{(\tau-t)^{k}}d\tau. $$
(4)
Assume that the density and its derivatives up to the order
k−1 with respect to
τ have Holder continuous derivative with respect to
α for any
τ∈(
a,
b). In this case, from (
3) it follows that the integral has Holder continuous partial derivative with respect to
α, which may be evaluated by direct differentiation of the r.h.s. of (
3) with respect to
α. In the case, when the limits do not depend on the parameter, the inspection of the result of differentiation leads to the conclusion that partial differentiation with respect to the parameter may be performed under the integral sign.