Зображення сторінки
[ocr errors][subsumed][ocr errors][ocr errors][ocr errors][subsumed]

Cauchy, Mémoires des Savans Etrangers, Vol. 1. p. 638. (10) Find the value of 6* dx e –aor* cos 2rx.

Calling the definite integral u, and differentiating with respect to r, we have


= - 26" dx re *a*r* sin 2rx.

[ocr errors]

On integrating the second side with respect to a by parts, the equation becomes

du 21
dr=- u,

since the integrated part vanishes at both limits, and the unintegrated part when taken between the limits is equal to u. This equation on integration gives

[ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][merged small][ocr errors][ocr errors]

Laplace, Mémoires de l'Institut, 1810, p. 290.

From this may be deduced the following integrals :
() $*da cos aRa* cos 2r2 = cos ( - ).

[merged small][ocr errors]

COS at


(11) To find the value of u = S Rdv 19 ta segunda
Differentiating twice with respect to a, we have
"c, a cos a x

= -fo* dx cos ax + u.

1 + gula By formula (g) of Ex. (8) f*dw cos a x = 0; therefore

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

The integral of this is

U = C€ + C, e-". To determine the constants, we observe that u cannot increase continually with a, and therefore the term involving € must vanish, or C = 0. This being the case we have, when a = 0,

» = 0; -8, 14 Therefore (a) sada cos 4* - On differentiating this with respect to a, there results

po, w sin ax

Integrating with respect to a and determining the constant so as to inake the integral vanish with a, we find

(c) rödä sin am - 5 (1 - 6-9).

Jo x 1 + x 2

Laplace, Mémoires de l'Académie, 1782.

Sin a

It is to be observed that the formula (a) is discontinuous, as the integral is equal to 6" when a is positive, and to

Te when a is negative. Libri* has accordingly expressed the value of the integral in the following manner : poo cos ad /

) 1 + x2 2 1 + 0-a 1 + 0°)
By a similar method we find

[ocr errors]

(a) Sodo costante « * (cosa + sin ).

Laplace, Mémoires de l'Institut, 1810, p. 295.

[ocr errors][merged small][merged small][ocr errors]

Cah. xvi. p. 225 (Poisson), and for tl

, cos a x | dr

(1 + x?)* see Jour. de Mathématiques, Vol. v. p. 110 (Catalan). (12) To find the value of /*

pa dx cos rx

lo 1 – 2a cos x + a!" If we expand the denominator we have the series

[ocr errors][ocr errors]

For the general expressions for po cos ra J. a 20. COSI *)» and Jodx cos rx (1 – 2a cos x + a®)".

the reader may consult Legendre, Exercices, Vol. 1. p. 373.

Crelle's Journal, Vol. x. p. 309.

manner W

By a similar expansion it is easily seen that (6) Som log (1 - 2a cos x + a®) = 0, or 25 log a, according as a is less or greater than 1. Poisson, Journal de l'Ecole Polytechnique, Cah. xvii. p. 617.

Also in like manner we find (c) So*dx cos r ~ log (1 – 2a cos x+a’) = -a' or - Ia-'. according as a < or > 1.

Integrating (6) by parts, we find (d) 1*_dx x sin a

do 1 - 2a cos x + ax according as a < or > 1.

[ocr errors]
[merged small][ocr errors][merged small][merged small][merged small][ocr errors]

Expand the second factor as before, and integrate each term separately by (11, a); then on summing the result, we have

1 1 1 + a eJ. 1 + x? 1 - 2a cos rx +

a 2 1 - a' 1 - ae-'

(a) So do neanche

[merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors]
[ocr errors]

Hence multiplying by 4 m, and integrating from 0 to a by (11, b), we find

roo dx x sin rex
19 J. 1 + x 1 - 2a cos r x + a22€ - a
In (6) make a = 1, then

[ocr errors]

(a) 5,14 og (sin "T) - ().

Changing the sign of a and then making a = 1, we have

[ocr errors][ocr errors][ocr errors][subsumed][ocr errors][merged small][ocr errors][ocr errors][merged small][ocr errors][ocr errors][merged small][merged small][ocr errors]

Changing the sign of a and then making a = 1, we have

[blocks in formation]

The formulæ (a), (b), (c) are due to Legendre: see his Exercices, Vol. 11. p. 123.

The formulæ (d), (e), (f), (g), (h) were first given by Georges Bidone, in the Mémoires de Turin, Vol. xx. (14) To find the value of

805* dx log (sin x) = fofa log (cos x). By Cotes's Theorem we have z?"- 1

= (x+ – 2 % cos – T + 1)...(x? - 2x cos — + 1).

[ocr errors]
[ocr errors]


[ocr errors]
« НазадПродовжити »