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so that the value of the triple integral given in (86) is made to depend on that of a double integral. Similarly if I had been a multiple integral of the nth order, its value may be made to depend on that of a double integral.

287.] An important theorem may be derived from (88). Let a, b, c, and u be replaced by 0a, 0b, Oc, Ou respectively; and to shorten the formulæ, let us take only two variables: then, from (86) and (88), the limits being given by μ<x+y<λ,

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XP−1y?−1 e ̄(ax+bv)° F (x + y) dy dx = the real part of

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do,

Let each side of this equation be multiplied by @P+q-1e-wo and let the 6-integral of each side be taken for the limits and 0; then

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F

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du

r(p) r (q) f^r (t) dt [__ (a+u√ −1)" (b + u√ = 1)o !% f*

П

μ

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cos utoe we do.

Hence, since by (250) Art. 122, and by (6), Art. 82,

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therefore for the limits assigned by the inequality, u<x+y<λ,

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wr (p)r (q) πr (p + q) √μ

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du

..(91)

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• {w2 + (ut)2 } (a+u√ − 1)o(b + u√ − 1)a

Now the last integral in the right-hand member admits of evaluation by the following process due to Lejeune Dirichlet.* From (161) and (162), Art. 106, it is evident that

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Also by a further similar process it may be shewn that

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And substituting in (91), and observing that the whole expression

is real, we have

* Crelle's Journal, Vol. IV, p. 94.

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where the limits of integration in the left-hand member are given

by the inequality μ<x+y<λ.

288] This theorem also admits of extension to the case where the limits are given by the inequality,

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The process of reducing this to the form (97) is exactly similar to that of Art. 280; and it is unnecessary to repeat it. The following example will illustrate it.

Let it be required to determine the value of

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where the limits of integration are given by the inequality,

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x2 y2 22

μ2 < + + < 12.

y2

b2

a2 b2 c2

(100)

(101)

= n,

= ; so that for limits assigned

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and thus the triple integral is reduced to a single integral. 289.] Thus far, when the infinitesimal element has contained an arbitrary function, the subject of that function has been the expression by which the inequality assigning the limits of integration has been determined. Let us consider the following

integral in which this is not the case; and here, as before, I will take an integral involving only three variables; viz.

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= ff fr (x + y + z) dz dy da,

where the limits are assigned by the inequality,

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so that the integral includes all values of the variables which lie within the surface of an ellipsoid. Let

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than or greater than 1, so will the real part of

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dt = 1 or 0, according as k is less

2

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= 1, or 0, according as k is less than or greater than 1. Hence if we introduce this factor of discontinuity into the definite integral, the ranges of the x-, y-, and z-integrations may be extended tox and; so that with the limits assigned by the given inequality,

ƒ ƒ ƒ x (x + y + z ) d z dy dx

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8

= the real part of [*[*[* ¥ (8) dz dy dx
[_[_[_ x (8) dz dy dx 2 [* sint et dt.

2

π

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1881∞

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π

Ο

sint at
dt [* ƒ ƒ ® (8) ett√=1dz dy dx. (104)

t

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Now by Fourier's Theorem, (144), Art. 204,

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so that in (104), replacing F (8) by this value, we have

F (x + y + z) dz dy dx = the real

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replacing k and s by their values given in (103).

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and similar values are true for the y- and the z-integrals; so that substituting these, and replacing a2+b2+ c2 by r2, we have

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whereby the triple integral is expressed in terms of a double

integral.

If the integral contains n variables, the limits being given by the inequality,

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0 <

a2

++

22
c2

F(x + y + z + ...)... dz dy dx

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sin t

< 1, then

1

abc... [7 (4) du" in cos ("+") dt; (107) ୮ F dw

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0

and thus the evaluation of the multiple integral depends on that

of the double integral.

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