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π/2

                            λ
                                                              k
               30.        sin t cos m  ty(ξ) dt = f(x),  ξ = x sin t.
                      0
                     1 . Let λ > –1, m > –1, and k > 0. The transformation
                      ◦
                                            2                      λ–1   k
                                                       2
                                       z = x k ,  ζ = z sin t,  w(ζ)= ζ 2 y ζ 2
                     leads to an equation of the form 1.1.43:
                                     z
                                          m–1                          λ+m   k
                                     (z – ζ)  2 w(ζ) dζ = F(z),  F(z)=2z  2 f z 2 .
                                   0
                      ◦
                     2 . Solution with –1< m <1:
                                                                  π/2
                                  2k     π(1 – m)     k–λ–1 d  λ+1     λ+1   m
                            y(x)=    sin         x  k      x k      sin  t tan tf(ξ) dt ,
                                   π        2          dx       0
                                 k
                     where ξ = x sin t.

                 3.5-7. A Singular Equation


                         2π    t – x
               31.       cot       y(t) dt = f(x),  0 ≤ x ≤ 2π.
                      0        2
                     Here the integral is understood in the sense of the Cauchy principal value and the right-hand
                                                       2π
                     side is assumed to satisfy the condition  f(t) dt =0.
                                                     0
                        Solution:
                                                 1     2π     t – x
                                         y(x)= –        cot      f(t) dt + C,
                                                4π 2  0      2
                     where C is an arbitrary constant.
                                                     2π
                        It follows from the solution that  y(t) dt =2πC.
                                                   0
                        The equation and its solution form a Hilbert transform pair (in the asymmetric form).
                     •
                       Reference: F. D. Gakhov (1977).

               3.6. Equations Whose Kernels Contain Combinations of
                      Elementary Functions

                 3.6-1. Kernels Containing Hyperbolic and Logarithmic Functions


                         b

               1.       ln cosh(λx) – cosh(λt) y(t) dt = f(x).


                      a
                     This is a special case of equation 1.8.9 with g(x) = cosh(λx).
                         b

               2.       ln sinh(λx) – sinh(λt) y(t) dt = f(x).


                      a
                     This is a special case of equation 1.8.9 with g(x) = sinh(λx).

                 © 1998 by CRC Press LLC









               © 1998 by CRC Press LLC
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