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Differentiating (g) with respect to a and then making

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Expanding this last, and equating the coefficients of like

powers of p,

(m) dx log (cos a) cos 2rx = (−)'−1

π 1
4 r

(23) Swanberg has proved the following theorems more general than those of Poisson.

If

2 M = f (a + au3, b + ß u“...) +f(a + ava, b + ßv“.....), 2 (−)3N = ƒ (a + au3, b + ẞu“...) − f(a + av1, b + ßv“.....), u and having the same signification as before; then

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Then, changing a into 2x, and therefore the limit into we obtain from the first of these expressions

c da (cos λx)' (cos ux)"...cos (lλ + mμ +...) x

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(24)

theorems.

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The same writer (p. 233) has proved the following
If

P = f(a + au, b + ßu“...) + f (a + av1, b + ßv“...), (-)1 Q = f(a + au^, b + ẞu"...) − f(a + av1, b + ßv“...), where u and have the same signification as before,

• Nova Acta Reg. Soc. Upsaliensis, Vol. x. p. 271.

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These expressions are easily proved by expansion, with the assistance of the formula in Ex. (11): they are evidently subject to the same cases of exception as the formula of Poisson. f (a, b...) a'. b"......

Let

=

μ

and a = a = b = B1; then changing A into 2λ, u into 2u, &c., we have

* dx (cos λx)' (cos μx)"...cos (lλ + mμ +.....) x

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h2 + x2

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(b)

* dx x (cos λx)' (cos u x)"...sin (lλ + mμ +.....) x

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If in (a) m, &c. be made equal to zero, the expression

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mb

Let f (a, b) = a. e, considering two terms only, and α = α = B = 1, b = 0. Then as before, changing A into 2X, we find by formula (1),

[o dx (cos \x)'. em cos μ x cos (1λx + m sin μx)

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h2 + x2

τμή

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expression put = 0, then

dre

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cos (msin a r

h2 + a"

π

=

2h

The student may exercise himself in deducing other integrals from the general formulæ by assuming other forms for the functions, and other values for the constants a, b... a, B....

(25) Jacobi has proved the following remarkable transformation of a definite integral:

* dxf (cosa) (sin x)2=1.3.5.7... (2r-1) * da ƒ (cos x) cosræ;

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dz

f(x), and all the differential coeffi

To demonstrate

cients up to the (1)th inclusive remain continuous from
1, or from x = 0 to x = π.
this formula we must premise the following.

8 = 1 to 8 =

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This is easily proved by the formula (A) p. 16, for writing

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2r 1

2

for n, and making a = 1, b = 0,

1, it becomes

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= (-)'-' 3. 5 ... (2r − 1) {(1 − x2)3 xr-1

-3

(r−1)(r−2)(r−3)(r−4) (1−x2)a xr-3_ &c. }·

1.2.3.4

Or, putting cosa for and sin x for (1 - x2)3,

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Now if u be a function which, with its differentials up to the (r− 1)th, vanishes at the limits, we have, on integrating r times by parts between the limits,

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27-1
2

27-1

f'd≈ (1 - x) ƒ" (x) = (-)' _'' d≈ f (x)

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d

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daf (cosa). (sin x)2=1.3.5... (2r-1) f* dxf(cos.x). cos rx.

(26) I shall conclude this Chapter on Definite Integrals with some examples of their application to the solution of partial differential equations. This mode of expressing the integrals of such equations was introduced by Laplace*, and has been much employed by later writers, particularly Poissont, Fouriert, Cauchy, and Brisson.

It is particularly applicable to linear equations of orders higher than the first with constant coefficients, and it is useful because the solutions are put into a shape which facilitates the determination of the arbitrary functions. The principle of the method is to transform an explicit function not expressible in finite terms into an ordinary function involved under a

Mémoires de l'Académie, 1779.

Mémoires de l'Institut. 1818, and Journal Polytech. Cah. x11.
Journal Polytech. Cah. XII. and XIV.

Théorie de la Chaleur.

definite integral; but the mode of transformation must be determined in each particular case by the nature of the function to be transformed, as will be seen in the following examples.

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(see Chap. vi. Sect. 1. Ex. 4. of the Integ. Calc.), and our object is to transform the operative function into one involving

only the first power of

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d

Therefore, putting a

dx

t for b, and multiplying the

two sides of the integral by the two sides of this equation,

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therefore

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π1x = [_+ dwe="2 ƒ (x + 2 wat3).

This transformation is due to Laplace, Jour. Polytech

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2v =

atD

(εa1D – € ̃atD) & (x, y, ≈) + (ea1D + e−a¡D) ↓ (x, y, ≈),

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