is also satisfied by 0=0, so that this is a double root. The equation in fact is

{03d + Oa − (l3a' + m2b′ + n2c')} (0 + a') (0 + b') (0 + c')

+ {l2a'2(0+b') (0+c') +m2 b'2 (0+c') (0+a') + n2c2 (0+ a') (0+b')}=0,

or, expanding and dividing by 0, this is

d (0 + a') (0 + b') (0 + c')

+ a {02 + 0 (a' + b' + c') + b'c' + c'a' + a'b'}

d (a'x2 + b'y2+c'z2) + (lx + my + dz + dw)2 = 0,

19

and that a,, B., Y,, 8,, a, b, c, denoting as before, the equation of the inverse surface (referred to a different origin) is

2* = 4w3 {a ̧x2 + ẞ ̧y2+y,22 + &,w2 + 2w (ax+b1y+c,z)}.

ON CERTAIN DEFINITE INTEGRALS.

By WILLIAM WALTON, M.A., Fellow of Trinity Hall. THE object of this article is to give new and more simple demonstrations of properties of certain definite integrals which form the principal subject of the Suite du Mémoire sur les Intégrales définies, published by Poisson in the tenth volume of the Journal Polytechnique.

then, differentiating y with regard to a, and integrating by parts, he finally arrives at the differential equation

He then selects, as a particular case where integration can be effected, the value 2n for p: his equation then becomes

He then, in virtue of the general nature of the expansion of the expression for y under the integral sign, shews that C'=0, and then, by equating a to zero, proves that

2

C=[" (sin x)" dæ. Thus he shews that

Of course, if a be greater than unity, we at once infer from this result that

The following method of establishing these results is more simple and points out more clearly the intrinsic origin of the simplicity of the case where p = 2n.

a being less than unity. Let a-cosx-r coso, sinx = = r sin &;

a result which shews that tan(+x) is negative while x varies from 0 to 7, and zero at both these limits. Hence, while x varies from 0 to π, & varies from π to 0. Thus we see that, representing 4+x,

and therefore, the former term of the right-hand member of the equation being zero,

y

= [" de.log(1

2. To evaluate the integral

dx.log(1-2a cosx+a3), a <1.

Poisson evaluates this integral by first differentiating the equation

(sin x)" dx

(1 − 2a cosx+a2)”

0

with regard to n and afterwards putting n = 0. For the sake of verification he effects the evaluation also by the aid of the series

1-a2

1-2a cosx+ a2

1+ a cosx+a" cos 2x + a3 cos 3x+....

The following method is easy and independent, and does not require the aid of series.

By the relations connecting x, p, r, above given, we see that

and

2

y = 2 [" de logr
S

= 2 [* dæ (log sinæ – log sin 4)

<=2 2 log sinx.de-2

0

+2 log sin 4.dp.

But log sin 4 dy = [ log (sin). dy = 0,

20

π

log sin4.dp=2 [ log sin 4.dp=-2 [” log sinæ.dx;

π

If a>1, of course it follows, from this result, that y=2π loga.

3. To give a demonstration of the following relation between two definite integrals, viz.

Poisson's demonstration of this "rapport remarquable" is grounded upon his primary differential equation between y and a, and is certainly abstruse.

The following investigation is comparatively easy.

By the aid of the relations between x, p, r, we see that the relation is equivalent to