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# 23 if s 1 then z 1 2 π sin θ s dθ z 1 x s dx 1 x 2

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23. If s > - 1 then Z 1 2 π 0 (sin θ ) s = Z 1 0 x s dx 1 - x 2 = 1 2 Z 1 0 x 1 2 ( s - 1) dx 1 - x = 1 2 Z 1 0 (1 - x ) 1 2 ( s - 1) dx x . 24. Establish the formulae Z 1 0 f ( x ) dx 1 - x 2 = Z 1 2 π 0 f (sin θ ) dθ, Z b a f ( x ) dx p ( x - a )( b - x ) = 2 Z 1 2 π 0 f ( a cos 2 θ + b sin 2 θ ) dθ, Z a - a f r a - x a + x dx = 4 a Z 1 2 π 0 f (tan θ ) cos θ sin θ dθ. 25. Prove that Z 1 0 dx (1 + x )(2 + x ) p x (1 - x ) = π 1 2 - 1 6 . [Put x = sin 2 θ and use Ex. lxiii . 8.] ( Math. Trip. 1912.) 182. Some care has occasionally to be exercised in applying the rule for transformation by substitution. The following example affords a good illustra- tion of this. Let J = Z 7 1 ( x 2 - 6 x + 13) dx. We find by direct integration that J = 48. Now let us apply the substitution y = x 2 - 6 x + 13 , which gives x = 3 ± y - 4. Since y = 8 when x = 1 and y = 20 when x = 7, we appear to be led to the result J = Z 20 8 y dx dy dy = ± 1 2 Z 20 8 y dy y - 4 .

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[VIII : 183] THE CONVERGENCE OF INFINITE SERIES, ETC. 416 The indefinite integral is 1 3 ( y - 4) 3 / 2 + 4( y - 4) 1 / 2 , and so we obtain the value ± 80 3 , which is certainly wrong whichever sign we choose. The explanation is to be found in a closer consideration of the relation be- tween x and y . The function x 2 - 6 x +13 has a minimum for x = 3, when y = 4. As x increases from 1 to 3, y decreases from 8 to 4, and dx/dy is negative, so that dx dy = - 1 2 y - 4 . As x increases from 3 to 7, y increases from 4 to 20, and the other sign must be chosen. Thus J = Z 7 1 y dx = Z 4 8 - y 2 y - 4 dy + Z 20 4 y 2 y - 4 dy, a formula which will be found to lead to the correct result. Similarly, if we transform the integral Z π 0 dx = π by the substitution x = arc sin y , we must observe that dx/dy = 1 / p 1 - y 2 or dx/dy = - 1 / p 1 - y 2 according as 0 5 x < 1 2 π or 1 2 π < x 5 π . Example. Verify the results of transforming the integrals Z 1 0 (4 x 2 - x + 1 16 ) dx, Z π 0 cos 2 x dx by the substitutions 4 x 2 - x + 1 16 = y , x = arc sin y respectively. 183. Series of positive and negative terms. Our definitions of the sum of an infinite series, and the value of an infinite integral, whether of the first or the second kind, apply to series of terms or integrals of functions whose values may be either positive or negative. But the special tests for convergence or divergence which we have established in this chapter, and the examples by which we have illustrated them, have had reference almost entirely to the case in which all these values are positive. Of course the
[VIII : 184] THE CONVERGENCE OF INFINITE SERIES, ETC. 417 case in which they are all negative is not essentially different, as it can be reduced to the former by changing u n into - u n or φ ( x ) into - φ ( x ). In the case of a series it has always been explicitly or tacitly assumed that any conditions imposed upon u n may be violated for a finite number of terms: all that is necessary is that such a condition ( e.g. that all the terms are positive) should be satisfied from some definite term onwards . Similarly in the case of an infinite integral the conditions have been supposed to be satisfied for all values of x greater than some definite value , or for all values of x within some definite interval [ a, a + δ

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