An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 124-125. Reduction of elliptic integrals to normal forms.

213

A - q- 2 1 +, A 2- 2q2 + 6,
B 2(pq -- A(p + q) + ), ' = 2(pq - Q(p + q) + ~ ),
C =p2 - 22:p + C' p2- 29p + 6;
p and q can always be determined so that B and B' may vanish.
For from the equations:
pq- (p + q) + = o, )0 q - Q(p + q) + o 0,
unless A -- = 0, we have:
-ffi' PQ=R zand hence:
r/ ~ ^q)2 ^ ~o) ~ ((~g _4 ( --- 2 -4 (a - 2) (tt - X2)
(p-j)2 (! - ) 4( (I +-2)2-( (- -2)
Since the values 4 + q and pq are real, the values of p and q will
separately be real unless (p - q)2 be negative.
When two of the four roots are real and two complex, '2 -,t and
2' - o are different in sign, hence the numerator is positive.
When the four roots are real, developing the numerator we find:
( y6 ( - L))2(Y _4 ) 4(y8 ( )2) (P - ( +j (36)
=(, + (a + ( + ))2 _ )2( - P)2.
This expression is symmetrical in a and 3, y and 6, it vanishes for
a =, thus has (a -y) as a factor, it has therefore also the factors
( - 6), ( - y), ([ 3 -- ). Being of the fourth degree, it consists
of their product multiplied only by a numerical factor, which is found
to be 1 as the term a2 P2 occurs in both with that factor, thus:
(ra/ +Y - (-+ ) (~+8))2 y8)2( )2=(2-y)(. 8)(/3-y)( 8').
This product is positive since a > 13 > y > d.
When all the four roots are complex, let them be:
a =; -- ia', / = - - ia', y = Q-iy', 6= - iy7'
then:
a - ( - Q) + i(a -y'), - -= ( F -) + i( 'q- ),
3 -- Y = (a -- Q) i(' + y) -- - 13 (a - ) - ( - )
Thus the product is ((A - Q)2 + (,' - ')2) ((A -- Q)2 + (a' + y')2),
i. e. positive as before.
When I - =9:
R = e(x2 -22x + c) (x2 - 2 lx + 6),
if we put x == - + A:
= e((2 + - a2) (2 + 6- _ 2).
By the assumed substitution the reduced elliptic integral becomes: