Page:EB1911 - Volume 11.djvu/330

$\phi ^{(\mu )}(a_{1})=\sum _{\lambda _{1}=0}^{m_{1}}{\frac {\phi ^{(\mu +\lambda _{1})}(0)}{\lambda _{1}!}}({\frac {x}{n}})^{\lambda _{1}}+\alpha _{\mu }$

with

αμ < g/ρμ 2^m1,

provided (μ + m₁ + 1)μ < (2/3)m1 + 1; in fact for μ ⋜ 2n2n−2 it is sufficient to take m₁ = n²ⁿ; by another application of the same inequality, replacing x, z respectively by a₁ and x/n, we have

φ(μ) (a2) = Σ m2λ2=0	φ(μ + λ2) (a1)	(	x	) λ2 + β′μ ,
	λ2!		n

where

|β′μ| < g / ρμ 2^m₂

provided (μ + m₂ + 1)μ < (3/2)m₂ + 1; we take m₂ = n2n − 2, supposing μ < 2n2n−4. So long as λ₂ ⋜ m₂ ⋜ n2n−2 and μ < 2n2n−4 we have μ + λ₂ < 2n2n−2, and we can use the previous inequality to substitute here for φ(μ + λ₂) (a₁). When this is done we find

φ(μ) (a2) = Σ m2λ2=0 Σ m1λ1=0	φ(μ + λ1 + λ2) (0)	(	x	) λ1 + λ2 + βμ ,
	λ1! λ2!		n

where |βμ| < 2g/ρμ 2m₂, the numbers m₁, m₂ being respectively n²ⁿ and n2n−2.

Applying then the original inequality to φ(μ) (a₃) = φ(μ) (a₂ + x/n), and then using the series just obtained, we find a series for φ(μ) (a₃). This process being continued, we finally obtain

φ(x) = Σ m1λ1=0 Σ m2λ2=0 ... Σ mnλn=0	φh (0)	(	x	) h + ε ,
	K		n

where h = λ₁ + λ₂ + ... + λ_n, K = λ₁! λ₂! ... λ_n!, m₁ = n²ⁿ, m₂ = n2n−2, ..., m_n= n², |ε| < 2g/2^m_n.

By this formula φ(x) is represented, with any required degree of accuracy, by a polynomial, within the region in question; and thence can be expressed as before by a series of polynomials converging uniformly (and absolutely) within this region.

§ 13. Application of Cauchy’s Theorem to the Determination of Definite Integrals.—Some reference must be made to a method whereby real definite integrals may frequently be evaluated by use of the theorem of the vanishing of the integral of a function of a complex variable round a contour within which the function is single valued and non singular.

We are to evaluate an integral ∫ ba ƒ(x)dx; we form a closed contour of which the portion of the real axis from x = a to x = b forms a part, and consider the integral ∫ƒ(z)dz round this contour, supposing that the value of this integral can be determined along the curve forming the completion of the contour. The contour being supposed such that, within it, ƒ(z) is a single valued and finite function of the complex variable z save at a finite number of isolated interior points, the contour integral is equal to the sum of the values of ∫ƒ(z)dz taken round these points. Two instances will suffice to explain the method. (1) The integral ∫ ∞0 [(tan x)/x] dx is convergent if it be understood to mean the limit when ε, ζ, σ, . . . all vanish of the sum of the integrals

$\int _{0}^{{\frac {1}{2}}\pi -\epsilon }$	tan x	dx,  ∫ 3/2π−ζ1/2π+ε	tan x	dx,  ∫ 5/2π−σ3/2π+ζ	tan x	dx, ...
	x		x		x

Now draw a contour consisting in part of the whole of the positive and negative real axis from x = −nπ to x = +nπ, where n is a positive integer, broken by semicircles of small radius whose centres are the points x = ±1/2π, x = ±3/4π, ... , the contour containing also the lines x = nπ and x = −nπ for values of y between 0 and nπ tan α, where α is a small fixed angle, the contour being completed by the portion of a semicircle of radius nπ sec α which lies in the upper half of the plane and is terminated at the points x = ±nπ, y = nπ tan α. Round this contour the integral ∫ [(tan z / z)] dz has the value zero. The contributions to this contour integral arising from the semicircles of centres −1/2(2s − 1)π, + 1/2(2s − 1)π, supposed of the same radius, are at once seen to have a sum which ultimately vanishes when the radius of the semicircles diminishes to zero. The part of the contour lying on the real axis gives what is meant by 2 ∫ nπ0 [(tan x / x)] dx. The contribution to the contour integral from the two straight portions at x = ±nπ is

∫ nπ tan α0 idy (	tan iy	−	tan iy	)
	nπ + iy		−nπ + iy

where i tan iy, = −[exp(y) − exp(−y)]/[exp(y) + exp(−y)], is a real quantity which is numerically less than unity, so that the contribution in question is numerically less than

∫ nπ tan α0 dy	2nπ	, that is than 2α.
	n2π2 + y2

Finally, for the remaining part of the contour, for which, with R = nπ sec α, we have z = R(cos θ + i sin θ) = RE(iθ), we have

dz	= idθ, i tan z =	exp(−R sin θ) E(iR cos θ) − exp(R sin θ) E(−iR cos θ)	;
z		exp(−R sin θ) E(iR cos θ) + exp(R sin θ) E(−iR cos θ)

when n and therefore R is very large, the limit of this contribution to the contour integral is thus

− ∫ π−αα dθ = − (π − 2α).

Making n very large the result obtained for the whole contour is

2 ∫ ∞0	tan x	dx − (π − 2α) − 2αε = 0,
	x

where ε is numerically less than unity. Now supposing α to diminish to zero we finally obtain

∫ ∞0	tan x	dx =	π	.
	x		2

(2) For another case, to illustrate a different point, we may take the integral

∫	za−1	dz,
	1 + z

wherein a is real quantity such that 0 < a < 1, and the contour consists of a small circle, z = rE(iθ), terminated at the points x = r cos α, y = ± r sin α, where α is small, of the two lines y = ± r sin α for r cos α ⋜ x ⋜ R cos β, where R sin β = r sin α, and finally of a large circle z = RE(iφ), terminated at the points x = R cos β, y = ±R sin β. We suppose α and β both zero, and that the phase of z is zero for r cos a ⋜ x ⋜ R cos β, y = r sin α = R sin β. Then on r cos α ⋜ x ⋜ R cos β, y = −r sin α, the phase of z will be 2π, and zα − 1 will be equal to xα − 1 exp [2πi(a − 1)], where x is real and positive. The two straight portions of the contour will thus together give a contribution

[1 − exp(2πiα)] ∫ R cos βr cos α	xa−1	dx.
	1 + x

It can easily be shown that if the limit of zƒ(z) for z = 0 is zero, the integral ∫ƒ(z)dz taken round an arc, of given angle, of a small circle enclosing the origin is ultimately zero when the radius of the circle diminishes to zero, and if the limit of zƒ(z) for z = ∞ is zero, the same integral taken round an arc, of given angle, of a large circle whose centre is the origin is ultimately zero when the radius of the circle increases indefinitely; in our case with ƒ(z) = zα−1/(1 + z), we have zƒ(z) = z^a/(1 + z), which, for 0 < a < 1, diminishes to zero both for z = 0 and for z = ∞. Thus, finally the limit of the contour integral when r = 0, R = ∞ is

$[1-{\text{exp}}(2\pi ia)]\int _{0}^{\infty }{\frac {x^{a-1}}{1+x}}dx.$

Within the contour ƒ(z) is single valued, and has a pole at z = 1; at this point the phase of z is π and z^a−1 is exp [iπ(a − 1)] or − exp(iπa); this is then the residue of ƒ(z) at z = −1; we thus have

$[1-{\text{exp}}(2\pi ia)]\int _{0}^{\infty }{\frac {x^{a-1}}{1+x}}dx=-2\pi i{\text{exp}}(i\pi a)$,

that is

$\int _{0}^{\infty }{\frac {x^{a-1}}{1+x}}dx=\pi {\text{ cosec }}(a\pi ).$

§ 14. Doubly Periodic Functions.—An excellent illustration of the preceding principles is furnished by the theory of single valued functions having in the finite part of the plane no singularities but poles, which have two periods.

Before passing to this it may be convenient to make here a few remarks as to the periodicity of (single valued) monogenic functions. To say that ƒ(z) is periodic is to say that there exists a constant ω such that for every point z of the interior of the region of existence of ƒ(z) we have ƒ(z + ω) = ƒ(z). This involves, considering all existing periods ω = ρ + iσ, that there exists a lower limit of ρ² + σ² other than zero; for otherwise all the differential coefficients of ƒ(z) would be zero, and ƒ(z) a constant; we can then suppose that not both ρ and σ are numerically less than ε, where ε > σ. Hence, if g be any real quantity, since the range (−g, ... g) contains only a finite number of intervals of length ε, and there cannot be two periods ω = ρ + iσ such that με ⋜ ρ < (μ + 1)ε, νε ⋜ σ < (ν + 1)ε, where μ, ν are integers, it follows that there is only a finite number of periods for which both ρ and σ are in the interval (−g ... g). Considering then all the periods of the function which are real multiples of one period ω, and in particular those periods λω wherein 0 < λ ⋜ 1, there is a lower limit for λ, greater than zero, and therefore, since there is only a finite number of such periods for which the real and imaginary parts both lie between −g and g, a least value of λ, say λ₀. If Ω = λ₀ω and λ = Mλ₀ + λ′, where M is an integer and 0 ⋜ λ′ < λ₀, any period λω is of the form MΩ + λ′ω; since, however, Ω, MΩ and λω are periods, so also is λ′ω, and hence, by the construction of λ₀, we have λ′ = 0; thus all periods which are real multiples of ω are expressible in the form MΩ where M is an integer, and Ω a period.

If beside ω the functions have a period ω′ which is not a real multiple of ω, consider all existing periods of the form μω + νω′ wherein μ, ν are real, and of these those for which 0 ⋜ μ ⋜ 1, 0 < ν ⋜ 1;