The name is absent



70 Lectures on Scientific Subjects

so that

Further, we may differentiate once under the integral sign
so that (ʃ2-ti)2 is replaced by the derivative of this ex-
pression. If we write ʃ =tu we see, however, that if и =s∕t > 1,
ds         du

Hence we obtain the final solution for m = 1 :

(4)      A1W= ⅛


(w=→l).
2>


Thus our general conclusion in the case of a harmonic
distribution of the first order is as follows: In the case of a
given harmonic density function of the first order = 1)

Fi(r, θ) ≈=Fι(r) cos 0÷Gι(r) sin θ,

where ZZ1W = Ffr) +iGfr) remains finite near r = 0,3 there
exists a corresponding continuous distribution function,

/i(ʃ, v) =√ιW cos φ+i gι(j) sin φ,

likewise bounded near j=0 and harmonic of the first order, if
and only if the integral in (4) represents a function with a
continuous derivative. In this event the unique solution hfs)
is exhibited in (4), where of course,

AiW =∕ιW+⅛ιW∙

9∙ THE SOLUTION IN HIGHER HARMONIC CASES m > 1

The method of solution used above applies essentially to
any order m>l. Here we start by writing the identity

•We are assuming that F{r, θ} is continuous for r>0 only.



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