This is a hint for problem 12, p. 280. First, use the definitions of Hν(1)(z) and Hν(2)(z) for fractional order ν, with 0 ≤ arg(z) < 2π, to show that
Hν(1)(z) = i csc(π ν)( e−iπ ν Jν(z) − J−ν(z))
and
Hν(2)(z) = −i csc(π ν)( eiπ ν Jν(z) − J−ν(z))
Next, use these formulas to show that
Hν(1)(−z) = − e−iπ ν Hν(2)(z)
and
Hν(2)(−z) = − e iπ ν Hν(1)(z)
both hold, provided 0 ≤ arg(z) < π. In particular, both formulas will hold if z = x is real and positive. Finally, all the functions above are analytic in ν. By letting ν → 0, we have that
H0(1)(−z) = − H0(2)(z)
and
H0(2)(−z) = − H0(1)(z).
Judiciously using these in the problem will give the solution.