On definite integrals and infinite series *

Transcription

1 On definite integrals and infinite series * Ernst Eduard Kummer In the paper published in this journal, volume p. 44, we gave theorems, by means of which infinite series, which contain the complete integral of the Riccati equation, can be expressed in terms of definite integrals, and there we also claimed that we will exhibit many other similar theorems. We will now present these theorems. For, since infinite series and definite integrals are one of the most simple and most frequently used forms, by which transcendental functions can be expressed, the transformation of one form into the other is to be considered of greatest importance. The transformation of definite integrals into series, by expansion of the function to be integrated, is very easy in most cases, but the other problem, on the transformation of infinite series into definite integrals, is a lot harder. For, in the solution of this problem a general method, which has a certain success, is missing, and only in a few cases peculiar artifices, or singular methods, lead to the propounded goal. Our theorems also contain certain special methods, and they can only be applied to certain classes of series, but they all have the same origin; and they are examples of a general method, which we will explain first. For this aim, we recall the definite integral b a U f u, kdu, *Original Title: De integralibus definitis et seriebus infinitis, first published in Crelle Journal für reine und angewandte Mathematik 7, ; reprinted in Ernst Eduard Kummer Collected Papers II, pp , translated by: Alexander Aycock for the project Euler-Kreis Mainz

2 in which U is a function of the variable u, f u, k is a function of the quantities u and k, and k is an integer number, and let us assume that this integral satisfies this equation. b a U f u, kdu = B k b a U f u, du in which B k is a given function of the number k. Further, we assume the infinite series:. A f u, + A f u, k + A f u, + etc. = φu whose sum φu we assume to be expressible in terms of known functions. Having constituted these, it will be b = A and by formula. a U f u, du + A b a b a Uφudu U f u, du + A b a U f u, du + etc. or b a Uφudu = b a U f u, [A B + A B + A B + etc.] 3. A B + A B + A B + = b a b a Uφudu U f u, du By this method, if a certain series to be expressed in terms of definite integrals is propounded, the factors A, A, A etc. are to separated from the single terms, and they are to be assumed in such a way that the sum of the series. can be found easily; hence also the function f u, k is to be chosen appropriately, after this it remains to find the function U satisfying equation.. Therefore, we would be able to express the sum of a certain series in.

3 terms of definite integrals in general, if the appropriate determination of the function U would succeed. But we discussed this problem, which is to be considered to be one of the more difficult ones, and requires singular methods for its solution, in another paper, but here will find certain forms of infinite series from several known integrals satisfying equation., and we will deduce several theorems by means of the method just explained. First, let us take the integral e u u α+k du, which Gauss denoted by Πα + k, which has the following known reduction: 4. e u u α+k du = αα + α + k e u u α du, which compared to equation. gives U = e u u α, Fu, k = u k, B k = αα + α + k, having substituted which value in the equations. and 3., we have. Theorem I. "If the sum of the following series is known one has 5. A + A u + A u + A 3 u 3 + = φu 6. A + αa + αα + A + αα + α + A 3 + = or e u u α φudu e u u α du 7. A + αa + αα + A + + etc. = e u u α φudu. Πα We will explain the use of this theorem in some examples. If one puts φu = cosu tan x, then A =, A =, A = tan x, A 3 =, A 4 = + tan4 x 3 4 etc. 3

4 and hence by equation 7. αα + = tan x + αα + α + α e u u α cosu tan xdu, Πα tan 4 x substituting the known sum cos x α cosαx of this series, we have 8. e u u α cosu tan xdu = Πα cos x α cosαx. In like manner, if one takes φu = sinu tan x, whence A =, A = tan x A =, A 3 = tan3 x 3 etc., by equation 7. α αα + α + tan x tan 3 x + = 3, e u u α sinu tan xdu, Πα and since the sum of this series is known, cos x α sin αx, it follows: 9. e u u α sinu tan xdu = Πα cos x α sinαx. The two integrals, 8. and 9., we found by means of the theorem, were found by mathematicians a long time ago, and by various methods, but nevertheless the proof we gave seems to be better than others, since it does not require imaginary quantities, and holds for any arbitrary positive value, integer and fractional, of the number α. Generally, if the quantities A, A etc. are assumed in such a way that not only φu, but also the sum of the series A + αa + αα + A + can be expressed in terms of known functions, the values of the definite integrals are found. Hence, if one puts φu = cos xu, we have A =, A = x, A = x etc. and hence 3. α x + αα + x3 = Πα e u u α cos xudu 4

5 furthermore, if one puts α =, this series goes over into the expansion of e x, and Πα = Π =, therefore, or having put u = v and x = z : e u u cos xudu = e x, e z e v coszvdv =. We will deduce another theorem from the integral which can be reduced by this formula: u α log u β du = α β Πβ. u α+k log u β du = α β α + k β u α log β du. u By comparison of this formula and equation. for this case we have f u, k = u k, U = u α log u β, Bk = α α+k β, therefore, from the equations. and 3. it follows: Theorem II: "From the known sum of the series it follows A + A u + A u + = φu or α β α β 4. A + A α + + A + = α + u α log u β φudu u α log u β du 5

6 5. A a β + A a + β + A a + β + = u α log β φudu. Πβ u For the sake of an example, let us take φu = x cos ω x u xu cos ω + x u = x cos ω + x u cos ω + x 3 u cos 3ω + whence A = x cos ω, A = x cos ω, A 3 = x 3 cos 3ω etc.; therefore, by equation 5. = 6. x cos ω α β + x cos ω α + β + x3 cos 3ω α + β + u α log β u x cos ω x udu Πβ xu cos ω + x u. The same integral, by means of which theorem II. was found, will give us another theorem by the following formula 7. u α+k log β+k du = u ββ + β + k αβ α + k β+k u α log u β du, which, having compared it to equation., gives: f u, k = u log k, U = u α log β, B u u k = From these, having substituted them in eqs.. and 3., it follows: ββ + β + k αβ α + k β+k. Theorem III.: "By means of the known series: 8. A + A u log u + A u log + = φ u log, u u one also has the sum of this series expressed in terms of definite integrals: 6

7 9. A + β αβ α + β+ A β β + αβ + α + β+ A + = or. A + β αβ α + β+ A β β + αβ + α + β+ A + u α log β φ u log du. Πβ u u u α log u β φ u log u du u α log u β φdu We can use this theorem to find the sum of the series + x + x + x3 + etc.; for, if one takes α =, β = and φ u log u = e xu log u = u ux, we have A =, A = x, A = x etc., therefore, by equation.. + x + x x etc. = u ux du. Here, we will also obtain that theorem, which we propounded already in this journal, volume, p. 44, and used to find several integrals. For this aim let us consider the integral u α u β α du, which is expressed in terms of the function Π this way u α u β α du = and one has this reduction formula: Πα Πβ α Πβ. u α+k u β α du = αα + α + k ββ + β + k u α u β α du. 7

8 If this formula is compared to formula., we have Theorem IV: "From the known sum of the series this sum of the series follows A + A + u + A u + A 3 u 3 + = φu or 4. A + α β A + αα + ββ + A + = u α+ u β α φudu u α+ u β α du = 5. A + α β A + αα + ββ + A + Πβ u α+ u β α φudu. Πα Πβ α Putting u = sin u we can represent equation 5. in this form = 6. A + α β A + αα + ββ + A + Πβ sin v α cos v β α φsin vdv. ΠαΠβ α Now, if one puts φsin v = cosβv, from the known expansion cosβv = β β sin v + ββ + ββ sin 4 v it follows A =, A = β β, A = + ββ+ββ etc., having substituted which in equation 6., we find 7. α β + αα + ββ 8

9 = Πβ sin v α cos v β α φsin vdv, ΠαΠβ α but the sum of this series can be assigned in terms of the function Π see Gauss Disquisit. gen. c. seriem inf. etc. p. 8 α β + αα + ββ = Π Π β α Π β, Π α hence equation 7. goes over into this one: sin v α cos v β α φsin vdv = Π Π β α Πα Πβ α Πβ Π β, Π α this expression is simplified tremendously by the fundamental formulas of the function Π, and obtains the simplest form, if α is changed into α, β into α+β, having done which, it results: 8. sin v α cos v β cosα + βvdv = cos α Πα Πβ. Πα + β In like manner, if in formula 6. one sets one will find the integral φsin v = sinβ v β sin v, 9. sin v α cos v β sinα + βvdv = sin α Πα Πβ. Πα + β But this integral is easily deduced from that one; for, if in that one v is changed into v, α into β and β into α, it results cos α + β sin v α cos v β cosα + βvdv 9

10 + sin α + β from which by formula 8. is follows sin v α cos v β sinα + βvdv = cos β Πα Πβ Πα + β sin v α cos v β sinα + βvdv cos β = cos α cos α + β Πα Πβ sin α + β, Πα + β which formula is identical to equation 9., since cos β cos α cosα + β sinα + β = sin α. Another integral is found by means of the fourth theorem by putting α =, φsin v = cosγv, hence by the known expansion γ γ cosγv = sin v + having substituted which, it results γ γ + γ γ 3, etc. γ γ γ γ β + + γ γ ββ + Πβ = Π Π β cos v β cosγvdv, and since the sum of this series can be expressed in terms of the function Π this way: we have Πβ Πβ Π β + γ Π β γ cos v β cosγvdv = Π Πβ Π β Π β + γ Π β γ,

11 finally, if β is changed into β+, and Π β Π β Πβ, this integral will take on the form: β is transformed into 3. cos v β cosγv = β+ Π Πβ β+γ By the same method one can find the two integrals sin v β cosγvdv and Π. β γ sin v β sinγvdv but they are deduced more easily from the one we just found, which for this purpose can be represented this way and from this cos v β cosγvdv = β Π Πβ Π β+γ β γ cos v β sinγvdv =, whose validity is obvious. For, if that one is multiplied by cos γ, by addition: sin γ, this one by cos v β cos γv γ dv = β Π cos γ Πβ Π β+γ β γ, finally, if v is changed into v, we have sin v β cosγvdv = sin v β sinγvdv = cos γ Πβ β β+γ Π Π sin γ Πβ β Π Π β+γ, β γ. β γ

12 From the found value of the integral 3. or in other ways only easily deduces the reduction formula: cos u β+k cosγudu = B k cos u β cosγudu, in which B k denotes this expression: B k = ββ + β + β + k β + γ + β + γ + 3 β + γ + k β γ + β γ + 3 β γ + k, this formula yields the theorem: and Theorem V: "If one puts A + A cos u + A cos 4 u + A 3 cos 6 u + = φcos u we find 34. R = A + + ββ + A β + γ + β γ + ββ + β + β + 3A β + γ + β + γ + β γ + β γ + + or 35. R = cos u β cosγuφcos udu cos u β cosγudu 36. R = β Π β+γ β γ Π Πβ cos u β cosγuφcos udu.

13 We have a simple example of this theorem for β =, γ = α, φcos u = cosz cos u, whence A =, A = z, A = + 4 z etc., having substituted which values in equation 36.: 37. z + α α + z 4 + α + α α α = α cosαu cosz cos udu. sin α Another theorem is deduced from the formula cos v β α cosβ + α + k vdv = B k where k αα + α + k B k = ββ + β + k, cos v β α cosβ + α vdv, which is easily demonstrated applying equation 3. or by other methods. By comparison of this formula to equation. we see that in this case one has to set f u, k = cosβ + α + k u, U = cos u β α, which values substituted in the equations. and 3. give: Theorem VI. "If the sum of the following series is known 38. A cosα + β u + A cosα + β + u + A cosα + β + 3u + = φu one also has the sum of this series 39. A α β A + αα + ββ + A = cos u β α φudu cos u β α du 3

14 or 4. A α β A + αα + ββ + A = β Πβ Π α cos u β α φudu. Πβ α If we use this theorem to find the sum of the series it has to be α = and A =, A = the sum of this series + x β + x ββ + +, x, A = + x etc., one has to find φu = cos β u x cos β + u 3 + x cos β which by known methods is derived from of course cos x = x + x 3 ; φu = e x sin u cos β u x cos u + e x sin u cos β u + x cos u having substituted which in equation 4. we have + x β + x ββ + + = β Πβ Π β cos u β φudu. The function φu consists of two parts, which only differ by the sign of the quantity x and it is easily clear, if we integrate each of both parts separately and write v instead of v in one of them, that these two integrals differ only in regard to their limits, which are and = for the one, and + for the other. Therefore, having combined these integrals, we have 4

15 = β Πβ Π β 4. + x β + x ββ cos u β e x sin u cos Let us return to the integral found above of equation. e v coszvdv = β u x cos u du. e z which, having put z = α log q, is transformed into e v cos αv log dq = q qα, hence, if one writes α + k instead of α, one deduces this formula 4. = q k +αk e v cos e v cos α + kv αv log q log q This formula will give us a new most remarkable theorem. For, by comparison with equation. we have f u, k = cos α + kv log q, U = e v, B k = q k +αk ; therefore, from the equations. and 3. it follows: dv dv. Theorem VII. "If the sum of the following series is known 43. A cos αv log q + A cos α + v log q + A cos α + v log q + = φv 5

16 one also has the sum of this series 44. A + A q +α + A q 4+4α + A 3 q 9+6α + = or e v φvdv e v cos αv log q dv 45. A + A q +α + A q 4+4α + A 3 q 9+6α + = q α e v φvdv. By means of this theorem one can find the sums of series containing powers of the quantity q, whose exponents proceed in an arithmetic progression of second order. On finds some very simple series in the theory of elliptic functions, to which we will mainly apply our method here. For this aim let us take φv = cos βv log q + x cos β v log q the sum of which series is φv = cos βv log q x cos + cos β v log q + x cos v β + v log q + x log q whence A =, A = x, A = x etc., α = β, having substituted the values in equation 45., we have = q β and, if one puts x = zq β, xq β + x q 4 4β + x 3 q 9 6β + [ ] e v cos βv log q x cos β + v log q dv x cos v log q + x 6

17 = q β zq + z q 4 + z 3 q 9 + z 4 q 6 [ + ] e v cos βv log q zq β cos β + v log q dv. zq β cos v log q + z q 4β The quantity β, which is not contained in the one side of the equation, can be chosen arbitrarily, but nevertheless one has to note, that it is taken in such a way that x = zq β is smaller than for φv not to become a divergent series. The integral would have the simplest form for β = ; but then it would not be possible to attribute the value to the quantity z. For this reason let us take β =, and furthermore, let us set z = + and z = ; hence we will obtain these series q + q 4 + q 9 + q 6 [ + ] e v cos v log = q q cos 3v log q dv, 4 q q cos v log q + q 49. q + q 4 q 9 + q 6 [ ] e v cos v log = q + q cos 3v log q dv. 4 q + q cos v log q + q Jacobi Fund. nova theor. f. ell. p. 84 found the same two series + q + q 4 + q 9 + q 6 + = + K, q + q 4 q 9 + q 6 = + k K, where the quantities q, K and k depend on the variable k in such a way that k = k, K = dϕ, K = k sin ϕ q = e K K. dϕ, k sin ϕ 7

18 In like manner in theorem VII. let us take φv = cos β v log q + x cos β 3v log q + x 3 cos β 5v log q + whence it follows α = β and A =, A =, A =, A 3 = x, A 4 =, A 5 = x 3 etc. The sum of the series φv is easily found φv = cos β v hence by equation 45. we have = q β log q x cos x cos 4v log q β + v log q + x 5. q β + xq 9 6β + x 3 q 5 β + x 3 q 49 4β + [ ] e v cos β v log q x cos β + v log q dv x cos 4v log q + x putting x = zq 4β and β = and multiplying by q, we have 5. q + zq 9 + z q 5 + z 3 q 49 + [ ] = q e v zq 4 cos 4v log q dv, zq 4 cos 4v log q + z q 8 if q is changed into 4 q, this formula goes over into this one 5. 4 q + z 4 q 9 + z 4 q 5 + z q = 4 q [ ] e v zq cos 4v log q dv, zq cos v log q + z q finally, putting z = and z =, we have these two series 8

19 53. 4 q + 4 q q 5 + q = 4 q [ ] e v q cos 4v log q dv, q cos v log q + q q 4 q = 4 q q 5 q [ ] e v + q cos 4v log q dv, + q cos v log q + q of which the first was found by Jacobi in his book mentioned above 4 q + 4 kk q 9 + q q 49 + =. By the same method one can also find the sums of the more general series Kx θ = q cos x + q 4 cos 4x q 9 cos 6x +, Kx H = 4 q sin x 4 q 9 sin 3x + 4 q 5 sin 5x, which occur frequently in the theory of elliptic functions, whose expressions in terms of definite integrals, since they become less simple, we omit here. From the known integral it immediately follows: Theorem VIII. "If one puts we have u α u β α du = log log u β A + A u + A u + A 3 u 3 + = φu 9

20 57. A log α + A log β α + + A log β + α + + = β + We obtain a remarkable example of this theorem by putting u α u β φudu. log u whence: φu = u + u u u n = + un+ + u, 58. log α log β α + + log β + α + + log β + α + n = β + n this sum of logarithms is collected into one logarithm of this product αα + α + nβ + β + 3 β + n ββ + β + nα + α + 3 α + n and, if we, following Gauss, put Πk, z = this product can be represented this way therefore, n β α n + 3 k k z z + z + z + k n Π +, β Π n +, α Π Π n, α, n, β u α u β + u n+ du, + u log u 59. u α u β + u n+ + u log u n du = log n + β α n Π +, β Π n +, α Π Π n, α, n, β if the number n becomes infinitely large, u n+ in the integral vanishes and Πn, z goes over into Πz, whence: 6. u α u β + u log u du = log β Π Π α Π Π α. β

21 From this integral, which we believe to be new, also this special case follows 6. u α u α + u log u = log tan α Many other theorems can be deduced from other values and reductions of definite integrals applying the general method mentioned above; but to collect them all, would take a lot of time. And those, we gave here, shall suffice as examples of the general method. But it is convenient to add a theorem of another kind here, which is connected to the others and is demonstrated by applying them. Theorem IX. "If a certain function has an expansion of this form 6. φcos u = A + A cos u + A cos 4 u + A 3 cos 6 u + and the same function is expanded into a series of this form: 63. φcos u = B + B cos u cos u + B cos u cos u + B 3 cos 3u cos u 3 + the coefficients B, B, B etc. are determined by definite integrals in such a way that in general. 64. B h = 4 cos u k cosh + uφcos udu. The proof of this theorem is based on the formula 65. cos u k = 3 5 k 4 6 k + k k + cos u kk + + cos u k + k + cos u cos u +, which is easily deduced from a more general formula, I propounded in this journal Volume 5 p. 6 form. 3. For, if in the equation φcos u = A + A cos u + A cos 4 u + one substitutes the expressions, which equation 65. yields, for the powers of the cosines, we find

22 65. φcos u = A + A A A cos u cos u cos u cos u cos u cos u which expansion compared to this one: cos u cos u cos u cos u cos u cos u cos 3u cos u 3 + cos 3u cos u 3 + cos 3u cos u 3 + gives φcos cos u u = B + B cos u + B cos u cos u + B cos 3u 3 cos u B = A + A A A 3 +, 67. B = A A A 3 +, 68. B = A A A 3 +, and it is easily seen that in general 69. b h = A h the general term of which term is: A h + 4h + 5 A 3 +, p h + p + h + p + h + p 4 6 p p + p + p which is easily transformed into this simpler form h + p + h + p + h + p p A p, 3 p

23 whence the series B h takes on this form: 7. B h = A + h A + h + 4h A 3 + The sum of this series is deduced by means of theorem V. from this one ψcos u = A + A cos u + A 3 cos 4 u + for, if in the equations of this theorem V. one puts βh +, γ = h +, changes A into A, A into A, A into A 3 etc. and φcos u into ψcos u, it results: A + h + 3 A + therefore, h + 4h A 3 + = h+ cos u h+ cosh + uψcos udu B h = h+ but since ψcos u = φcos u A, we have cos u cos u h+ cosh + uψcos udu; B h = h+ A h+ cos u h cosh + uφcos udu cos u h cosh + udu, and since for each positive value of the number h cos u h cosh + udu = But it seems that the case h = is to be excluded from this determination of the coefficients; for, the series B differs from the remaining ones, since it contains the terms A ; but by means of theorem IV. or V. one easily finds that this series is expressed by the integral 3

24 B = φcos udu whence it is clear that the first coefficient follows the same law as the other ones. There is a remarkable relation among the coefficients of these series 7. φcos cos u u = B = + B cos u + B cos u cos u + B cos 3u 3 cos u 3 + and 73. φcos u = C + C cos u + C cos 4u + C 3 cos 6u +, for, if one substitutes this series for φcos u in equation 7., we find 74. B h = 4 cos u h cosh + uc + C cos u + C cos 4u + du, where the single terms of this form are to be integrated 4 cos u h cosh + u coskudu, which integral is divided into these two cos u h cosh + k + udu + cos u h cosh k + udu, which according to formula 3. is expressed in terms of the function Π this way Πh Πh + kπ k + Πh Πh kπk 4

25 the first part always vanishes for each arbitrary integer number k, since then Π k =, the other part vanishes, if h k is a positive number, or if k > h, but if k h, it goes over into therefore, h h h k + ; k 4 cos u h cosh + u coskudu = hence equation 74. goes over into this one 75. B k = C + h C + h h h k +, 3 k h h C C h ; furthermore, since the first term of the series 73. is expressed by the integral C = φcos udu it follows B = C, therefore, the coefficients of the expansion 7. can be found most easily from the coefficients of the series 73., Lignietz, in the month of May, 836 5