The Euler Substitutions Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah last update: Nov 10, 2016

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1 The Euler Substitutions Phil Luht Rimrok Digital Tehnology, Salt Lake City, Utah 8403 last update: Nov 0, 206 The material in this doument is opyrighted by the author.. Introdution Comments about R = a + bx + x A substitution that fails, then three that sueed: A,B and C Substitution A Substitution B Substitution C More forms for the integral of R -/2...6 Appendix A: About Euler's original paper...2 Appendix B. Summary of the three Euler substitutions...26 Referenes Introdution There are many tehniques for the exat analyti evaluation of indefinite integrals. For example, Zwillinger provides a solid list of methods in Setion III of his Handbook of Integration. In any suh list, the very first method is usually "hange of variables" by means of a "substitution". To take a trivial example whih establishes the variable names we use below, onsider how the substitution t = sinx helps in evaluating a trigonometri integral, t(x) = sinx dt = osx dx x sinx osx dx = t(x) t dt = (/2) t 2 t(x) = (/2) sin 2 x. (.) The phrase "Euler substitutions" refers to three substitutions used for evaluating ertain integrals involving powers of x along with powers of the radial a+bx+x 2. These substitutions are briefly reviewed in Setion 2.25 (p 92) of Gradshteyn and Ryzhik [GR7]. Wiki suggests that these substitutions appears in many Russian alulus texts, and we have found them mentioned in a book by Piskunov. Our very elementary purpose in this doument is to flesh out the details of using these Euler substitutions and to state the results in a systemati manner. As an illustration, dx / a+bx+x 2 is evaluated using eah of the substitutions. The results are then transformed into other forms, and all the results are olleted in (7.7). This integral is of partiular

2 interest to the author beause it is used in Goldstein's Classial Mehanis to derive equations for planetary orbits, and there seems to be a benign sign error in that development as noted below. Physiists spend a lot of time worrying about signs of things. Sometimes suh errors are related to onfusion about how branh uts are "taken off" for funtions of a omplex variable, the hoie of Riemann sheets, and other esoteri matters, but more often than not the sign error is aused by a trivial mistake in grade shool algebra. A ertain lass of funtions of t are alled rational funtions and have the form R(t) = poly (t) poly 2 (t) (.2) where the numerator and denominator are just polynomials of t with non-negative powers. In fat negative powers are allowed and an be leared by multiplying top and bottom by some t n. Any funtion having the form R(t) an easily be integrated using the method of partial frations, and this topi is learly outlined in Setion 2.0 of GR7. In the disussion below, the Euler substitutions result in integrands of the form R(t) and then we know that the integration from that point on is just turning a rank. Rational funtions of two variables x and y are defined analogously to the above, R(x,y) = poly (x,y) poly 2 (x,y) (.3) where now the numerator and denominator are polynomials in x and y, suh as 3x 2-2xy y. The Euler substitutions apply to a lass of funtions of a single variable x whih have this form f(x) = R(x, a+bx+x 2 ). (.4) For example, a typial suh funtion might be R(x, a+bx+x 2 ) = Ax a+bx+x2 + Bx 7 ( a+bx+x 2 ) 3 + x 3 2x 3 a+bx+x 2 + x 2. (.5) Sine the numerator terms an be treated separately, one an limit one's interest to the following form, R(x, a+bx+x 2 x m ( a+bx+x 2 ) n ) = poly(x, a+bx+x 2 ). (.6) As noted, the speifi example we shall study is the following, shown here in one of its forms (4.9), 2

3 dx a+bx+x 2 = ln [ 2x + b + 2 a+bx+x 2 ] + onstant > 0. (.7) The above example serves as a good model to look at while reading the omments below onerning the general integrand shown in (.4). Comments: In general, any indefinite integral should really be written x dx f(x) = F(x) + onstant where the onstant is arbitrary. One an always test a andidate F(x) by omputing df/dx to see if the result is f(x). This task is made easy using Maple or another omputer alulus program. When there are parameters suh as a,b, shown above, that integration onstant an be any funtion g(a,b,) whih is of ourse not a funtion of x. We shall use the symbol A = B to mean A = B + onstant in the above sense. Thus we an write dx a+bx+x 2 = ln [ 2x + b + 2 a+bx+x 2 ] = ln [ 2x + b + 2 a+bx+x2 4a-b 2 ]. (.8) On the seond line we have added a denominator whih in effet reates the additive onstant g(a,b,) = - (/ ) ln( 4a-b 2 ). Both forms shown above are "orret" and well-defined when > 0 and 4a-b 2 > 0. Sine ln(-z) = ln(-) + ln(z) = ±iπ + ln(z), one an always write ln(-z) = ln(z). If one thinks of x having units of distane L, then dim(x) = L, dim(a) = L 2, dim(b) = L and dim() = make the integral be dimensionless. Then the seond form above involves the log of a dimensionless ratio, whereas the first form does not, somewhat larifying the dimensionless nature of the integral. As disussed more below, a+bx+x 2 is real for ertain ranges of a,b,,x and one an imagine that the integral being evaluated is over a range of x where a+bx+x 2 is real. One normally thinks of a,b, as real parameters. One an integral has been evaluated for "reasonable" values of the parameters like a,b,, one an extend one or more of these parameters to the omplex plane allowing one to analytially ontinue both sides of an integral evaluation. We shall give an example below in Setion 7. Having stated these general omments, we now look speifially at the objet a+bx+x 2. 3

4 2. Comments about R = a + bx + x 2 The letters a,b, are defined onsistently with GR7. It is probably more standard to write ax 2 +bx+, in whih ase one has the familiar rote solution for the roots [-b ± b 2-4a ]/(2a), so in our urrent ontext one must remember that in fat the roots of a+bx+x 2 = 0 are given by α ± [-b ± b 2-4a ]/(2). (2.) The quantity b 2-4a is often alled "the disriminant". GR7 define Δ to be the negative of this disriminant, Δ 4a - b 2. (2.2) Also onsistent with GR7 we define R by R a+bx+x 2 = [ x 2 + (b/)x + (a/) ] = (x-α + )(x-α - ) (2.3) and it is for this reason that we have used R above for the ratio of polynomials. Note that α ± are the roots of R for > 0, for = e iθ, and for < 0. Speial ase: b 2 = 4a (Δ = 0) α ± = -b/(2) R = (x-α + ) 2 R = [x + b/(2)] (2.4) Geometry If b 2-4a > 0 then the roots in (2.) are real. This means that the graphed funtion f(x) = a + bx + x 2 has two intersetions with the x axis. If > 0, the parabola ups up. Therefore to the right of the upper root, x > α +, one has a + bx + x 2 > 0 and so a+bx+x 2 is real and well defined. It is also real for x < α -. If < 0, the parabola ups down. In this ase a+bx+x 2 is real and well defined for α - < x < α +. By well defined we simply mean that the square root implies a positive real number and we don't worry about branhes of the square root funtion. When talking about integrals involving x where a + bx + x 2 is positive, so then a+bx+x 2, it seems best to start with an integral over a range of a+bx+x 2 is real and positive. There are four ases of interest: 4

5 Case : > 0, b 2-4a > 0 (real roots), ups up, x > α + or x < α - to have real a+bx+x 2 Case 2: > 0, b 2-4a < 0 (imag roots), ups up, all real values of x give real a+bx+x 2 Case 3: < 0, b 2-4a > 0 (real roots), ups down, must have α - < x < α + to have real a+bx+x 2 Case 4: < 0, b 2-4a < 0 (imag roots), ups down, no real value of x gives real a+bx+x 2 (2.5) (2.6) The red bars show the range of x for whih a+bx+x 2 is real. We shall initially work in Case 2 below. This means > 0 and b 2-4a < A substitution that fails, then three that sueed: A,B and C One's first inlination in evaluating integrals inluding t(x) = a+bx+x 2 a+bx+x 2 might be to make the substitution (3.) so that R(x, a+bx+x 2 ) R(x(t),t). (3.2) But, t 2 = a+bx+x 2 x 2 + bx + (a-t 2 ) = 0 x(t) = -b ± b2-4(a-t 2 ) 2. (3.3) The result then is, R(x, a+bx+x 2 ) R( -b ± b2-4(a-t 2 ) 2, t). (3.4) 5

6 The goal is to remove square roots from the integrand, but this substitution just replaes one square root with another square root and is thus not very useful, so we rejet this substitution. We are then led to three substitutions whih are redited to Leonhard (brave lion) Euler ( ) and are now known as "the Euler substitutions". There is some disagreement about how these are numbered,2 and 3 so we instead all them substitutions A,B and C favoring the ordering of Piskunov : Piskunov GR7 Boyadzhiev A t + x = a+bx+x 2 2 B xt + a = a+bx+x C t (x-α) = a+bx+x (3.5) Euler below is squinting at e iπ + = 0 written on his blakboard and is wondering what it all means. (3.6) Gradshteyn and Ryzhik Another of Zwillinger's "methods" for doing an indefinite integral is "looking it up" in a table of integrals. The astounding Table of Integrals, Series, and Produts assoiated with Gradshteyn and Ryzhik ontains, as a small fration of its ontent, about 200 pages of indefinite integrals of elementary funtions whih have aumulated over three enturies. Currently in the editorial hands of Dan Zwillinger and Vitor Moll, the book is in its 8th edition, though we ontinue to use the 7th edition. During the period of eah edition, new integrals and errata for old integrals are olleted to be inorporated into the next edition. The first edition was published by Russian mathematiian Ryzhik in 94, and he was joined by Gradshteyn in 95 for the 3rd edition, see wiki. 6

7 4. Substitution A Instead of using the substitution (3.), onsider t(x) = a+bx+x 2 - x or t + x = a+bx+x 2. (4.) When we finish this setion we an replae - everywhere and thereby generate an alternate result. One now has, a + bx + x 2 = t t x + x 2. (4.2) The key idea is that the two x 2 terms anel out, giving a + bx = t t x (b - 2 t )x = (t 2 -a) x(t) = t 2 -a b - 2 t, (4.3) an expression with no messy square roots, unlike (3.3). Then, a+bx+x 2 t 2 -a tb - 2 t2 t 2 - a = t + x = t + = + b - 2 t b - 2 t b - 2 t = - t2 + bt - a. (4.4) b - 2 t Then our general replaement beomes R(x, a+bx+x 2 ) R( One may then ompute t 2 -a b - 2 t, - t2 + bt - a b - 2 t ). (4.5) dx dt = d dt ( t 2 -a b - 2 t ) = (b - 2 t )(2t) - (t2 -a) (-2 ) (b - 2 t ) 2 = 2bt - 4 t2 + 2 t 2-2a (b - 2 t ) 2 = 2 bt - t2 - a (b - 2 t ) 2 so that dx = 2 - t2 + bt - a (b - 2 t ) 2 dt. (4.6) 7

8 Our integral evaluation then beomes x dx R(x, a+bx+x 2 ) = t(x) - t 2 + bt - a t 2 -a dt 2 (b - 2 t ) 2 * R( b - 2 t, - t2 + bt - a ) b - 2 t where t(x) = a+bx+x 2 - x. (4.7) Notie that there are no messy square roots anywhere in the dt integrand. The integrand is now a rational funtion in the variable t : integrand = poly (t)/poly 2 (t). As noted earlier, one may replae - (and ) to obtain the following alternative form, x dx R(x, a+bx+x 2 ) = t(x) dt 2 t 2 + bt + a (b + 2 t ) 2 * R( t 2 -a b + 2 t, t 2 + bt + a b + 2 t where t(x) = a+bx+x 2 + x. (4.8) ) Example: Let R(x, a+bx+x 2 ) = / a+bx+x 2. Then using (4.8), x dx a+bx+x 2 = t(x) dt 2 t 2 + bt + a (b + 2 t ) 2 * b + 2 t t 2 + bt + a = 2 t(x) dt b + 2 t = t(x) dt t + [b/2 ] = ln (t + [b/2 ] } t(x) = ln ( a+bx+x2 + x + [b/2 ] = ln ( R + x + [b/2 ] ) = ln [2 R + 2x + b ] (4.9) in agreement with (.7) stated earlier without proof. To get the last line, we multiplied by 2 top and bottom inside the log, then dropped the onstant term - (/ ) ln (2 ), hene the = sign. From (4.7) we would have gotten instead x dx a+bx+x 2 = - ln [- 2 R + 2x + b ]. (4.0) These seemingly different results are both valid sine they differ by a onstant independent of x : 8

9 ln [2 R + 2x + b ] - { - ln [- 2 R + 2x + b ] } = ln [ (2 R + 2x + b)(- 2 R + 2x + b) ] = ln [ (2x+b) 2-4R ] = ln [ (2x+b) 2-4(a+bx+x 2 ] = ln [ 4 2 x 2 + 4bx + b 2-4a + 4bx x 2 ] = ln [ b 2-4a ]. (4.) We then write, x dx a+bx+x 2 = ln [2 R + 2x + b ] = - ln [- 2 R + 2x + b ]. (4.2) Just for the reord, Maple omes up with the first of these forms (again apart from a onstant), whih is (4.3) ln [ b+2x + 2 R 2 ] = ln [2 R + 2x + b ] - ln (2 ). In the speial ase that 4a = b 2 we know from (2.4) that R = [x + b/(2)] so (4.9) beomes x dx R = = = = ln ( 2 R + 2x + b ) ln ( 2 [ (x + b/(2)] + 2x + b ) ln (2x +b + 2x + b ) = ln (4x +2b ) ln (2x +b ). > 0, 4a = b2 (4.4) 9

10 5. Substitution B Here we mimi the previous setion as losely as possible, using mathing equation numbers. Instead of using the substitution (3.), onsider t(x) = ( a+bx+x 2 - a ) / x or xt + a = a+bx+x 2. (5.) When we finish this setion we an replae a - a everywhere and thereby generate an alternate result. One now has, a + bx + x 2 = x 2 t 2 + 2xt a + a. (5.2) The key idea is that the two a terms anel out, giving bx + x 2 = x 2 t a x t b + x = xt a t x(-t 2 ) = 2 a t - b x(t) = 2 a t - b -t 2, (5.3) an expression with no messy square roots, unlike (3.3). Then, a+bx+x 2 = xt + a = 2 a t - b -t 2 t + a = (2 a t - b)t -t 2 + a (-t 2 ) -t 2 = a t 2 - bt + a -t 2. (5.4) Then our general replaement beomes R(x, a+bx+x 2 ) R( 2 a t - b -t 2, a t 2 - bt + a -t 2 ). (5.5) One may then ompute dx dt = d dt ( 2 a t - b -t 2 ) = (-t2 )2 a - (2 a t - b)(-2t) (-t 2 ) 2 = 2 a - 2t2 a + 4t 2 a -2bt (-t 2 ) 2 so that = 2 a t2 - bt + a (-t 2 ) 2 0

11 dx = 2 a t 2 - bt + a (-t 2 2 ) dt. (5.6) Our integral evaluation then beomes x dx R(x, a+bx+x 2 ) = t(x) dt 2 a t 2 - bt + a (-t 2 ) 2 * R( 2 a t - b -t 2, a t 2 - bt + a -t 2 ) where t(x) = ( a+bx+x 2 - a ) / x. (5.7) Notie that there are no messy square roots anywhere in the dt integrand. The integrand is now a rational funtion in the variable t : integrand = poly (t)/poly 2 (t). As noted earlier, one may replae a - a (and a a) to obtain the following alternative form. x dx R(x, a+bx+x 2 ) = t(x) - a t 2 - bt - a -2 a t - b dt 2 (-t 2 2 ) * R( -t 2, - a t2 - bt - a -t 2 ) where t(x) = ( a+bx+x 2 + a ) / x. (5.8) Example: Let R(x, a+bx+x 2 ) = / a+bx+x 2. Then using (5.8), x dx a+bx+x 2 = t(x) dt 2 - a t 2 - bt - a (-t 2 ) 2 * -t 2 - a t 2 - bt - a = 2 t(x) dt -t 2 = -2 t(x) dt t 2 -. Maple kindly omputes this integral, so we ontinue, (5.8a) x dx a+bx+x 2 = + 2 tanh- ( t ) t = t(x) = 2 tanh-[ ( a+bx+x2 + a ) / x]

12 = 2 tanh- ( R + a x ). (5.9) From (5.7) we would have instead found x dx a+bx+x 2 = 2 tanh- ( R - a x ). (5.0) One might reasonably wonder how both these results an be orret sine there is a sign differene. The answer again is that the two forms differ by a onstant independent of x. To show this, one an use with tanh - u = 2 u = R - a x +u ln ( -u ) u < // Spiegel 8.57 (5.) ± u = x x ± R - a x = x ±( R - a ) x so that tanh - ( R - a x ) = 2 ln [ x + R - a x - R + a ] tanh - ( R + a x ) = 2 ln [ x + R + a x - R - a ]. The differene between these two artangents is then tanh - ( R + a x ) - tanh - ( R - a x ) = 2 ln [ x + R + a x - R - a ] - 2 ln [ x + R - a x - R + a ] = 2 ln [ x + R + a x - R - a * x - R + a x + R - a ] = 2 ln [ (x + a )2 - R (x - a ) 2 - R ] = 2 ln [ x2 + a x + a - (a + bx + x 2 ) x 2 - a x + a - (a + bx + x 2 ) ] = 2 ln [ a x - bx - a x - bx ] Therefore = 2 ln [ a - b - a - b ]. 2 tanh- ( R + a x ) - 2 tanh- ( R - a x ) = 2 2 ln [ a - b - a - b ] (5.2) 2

13 whih is a onstant independent of x. Therefore we write x dx a+bx+x 2 = 2 tanh- ( R + a x ) = 2 tanh- ( R - a x ) (5.3) and we have now aumulated two more forms for this integral. Both forms an be verified by diret differentiation as Maple shows, (5.4) In the Maple language, a olon suppresses output from a ommand, while symbol % refers to the last omputed quantity. In the diff(j,x) line we suppress output and simplify to get / R, but for J2 we show the typially messy expression Maple generates, followed by the simplified result. 3

14 6. Substitution C Rename the roots of R = 0 to be α = α - and β = α +. Reall that a+bx+x 2 = (x-α)(x-β). (2.3) Now, instead of using the substitution (3.), onsider t(x) = a+bx+x 2 / (x-α) = (x-α)(x-β) / (x-α) = x - β x - α. (6.) When we finish this setion we an do α β everywhere and thereby generate an alternate result. Solving for x one finds (there is no equation (6.2) beause we are mathing the previous setions), t 2 = (x-β)/(x-α) t 2 (x-α) = (x-β) x(t 2 -) = (αt 2 - β) x(t) = αt2 - β t 2 -, (6.3) an expression with no messy square roots, unlike (3.3). Then, a+bx+x 2 = (x-α)t = ( αt2 - β t 2 - -α)t = t(αt2 - β) t αt(t2 -) t 2 - = αt 3 - βt - αt 3 + αt t 2 - = (α-β)t t 2 -. (6.4) Then our general replaement beomes R(x, a+bx+x 2 ) R( αt2 - β t 2 -, (α-β)t t 2 - ). (6.5) One may then ompute dx dt = d dt (αt2 - β t 2 - ) = (t2 -)2αt - (αt 2 -β)2t (t 2 -) 2 = 2αt3-2αt - 2αt 3 + 2βt - αt + βt (t 2 2 -) = 2 (t 2 -) 2 = 2 (β-α)t (t 2 -) 2 so that dx = 2 (β-α)t (t 2 2 -) dt. (6.6) 4

15 Our integral evaluation then beomes x dx R(x, a+bx+x 2 ) = t(x) (β-α)t dt 2 (t 2 -) 2 * - β R(αt2 t 2 -, (α-β)t t 2 - ) where t(x) = a+bx+x 2 / (x-α) = R /(x-α). (6.7) Notie that there are no messy square roots anywhere in the dt integrand. The integrand is now a rational funtion in the variable t : integrand = poly (t)/poly 2 (t). As noted earlier, one may swap α β to obtain the following alternative form, x dx R(x, a+bx+x 2 ) = t(x) (α-β)t dt 2 (t 2 -) 2 * - α R(βt2 t 2 -, (β-α)t t 2 - ) (6.8) where t(x) = a+bx+x 2 / (x-β) = R /(x-β). Example: Let R(x, a+bx+x 2 ) = / a+bx+x 2. Then using (6.7), x dx a+bx+x 2 = t(x) dt 2 (β-α)t (t 2 -) 2 * t 2 - (α-β)t = - 2 t(x) dt t 2 - = + 2 tanh- ( t ) t(x) // using (5.8a) = 2 tanh- ( x - β x - α ). (6.9) Thus we arrive at yet another form for our dx/ R integral. Maple verifies it as follows, whih is just / a+bx+x 2. The result is learly symmetri under α β, so one has x dx a+bx+x 2 = 2 tanh- ( x - β x - α ) = 2 tanh- ( x - α x - β ). (6.0) We leave it to the reader to find the onstant by whih these two forms differ from eah other and from those forms presented earlier. 5

16 7. More forms for the integral of R -/2 Define y 2x + b 4a-b 2. (7.) We wish to use the following identity with the above y, sinh - y = ln(y + y 2 + ) y < // Spiegel 8.55 (7.2) so we need to evaluate y 2 + = ( 2x + b 4a-b 2 ) 2 + = (2x+b)2 4a-b 2 + = (2x+b)2 + 4a-b 2 4a-b 2 = 42 x 2 + 4bx + b 2 + 4a-b 2 4a-b 2 = 4(x2 +bx +a) 4a - b 2 = 4R 4a - b 2. (7.3) Then y + y 2 + = 2x + b 4a-b R 4a-b 2 = 2x + b + 2 R 4a-b 2. Then from (7.2), sinh - ( 2x + b 4a-b 2 ) = ln ( 2x + b + 2 R 4a-b 2 ) = ln (2x + b + 2 R ) (7.4) where as usual we have thrown out a onstant g(a,b,). Comparing this result to (4.9), x dx a+bx+x 2 = ln [2 R + 2x + b ], (4.9) we may onlude that x dx a+bx+x 2 = sinh- ( 2x + b 4a-b 2 ) (7.5) giving a ommonly appearing form for the integral valid for > 0 and 4a - b 2 > 0. 6

17 Analyti Continuation. Consider this z-plane showing omplex vetors z and z-α for z > α > 0, We are interested in the funtion f(z) = (z-α) r where 0 < r <. The funtion has a branh point at z = α and we draw the ut off to the left as shown in blak. We delare that f(z) is real and positive for z > α, so we are viewing the "priniple sheet" of our funtion. Angles θ and φ are phases of the vetors z and z-α, z = z e iθ and (z-α) = z-α e iφ. (7.7) At point A, one has θ = 0, φ = 0, z > 0 and z-α > 0 so, z = z e i0 = z and (z-α) = z-α e i0 = (z-α). (7.8) At point B one has θ = +π, φ = +π, z < 0 and z-α < 0 so (z is just above the ut), (7.6) so z = z e iπ = (-z)e iπ and (z-α) = z-α e iπ = (α-z)e iπ z r = (-z) r e iπr and (z-α) r = (α-z) r e iπr. (7.9) For r = /2 the last line says z = -z e iπ/2 = + i -z and z-α = α-z e iπ/2 = + i α-z. (7.0) Therefore at point B the phases of z and z-α are the same, indiated by e iπ/2 = +i. If we move point B below the ut and redraw the piture so point B has θ = -π and φ = -π, both phases are -i instead of both +i. In either ase the two phases are the same, and this fat follows from doing proper analyti ontinuation over a smooth path in the z-plane from z = A to z = B. Appliation. Assume a > 0 and let z = 4a and α = b 2. Then taking point B above the ut, or 4a = +i -4a and 4a-b 2 = +i b 2-4a = +i - and 4a-b 2 = +i b 2-4a. (7.) We an then analytially ontinue our integral (7.5) using these rules to get 7

18 x dx (/ R) = (/ ) sinh - [ (2x +b)/ 4a-b 2 ] // (7.5) = (/[i - ]) sinh - [ (2x +b)/(i b 2-4a )] = (-i)(/ - ) sinh - [ -i(2x +b)/ b 2-4a ] = - (-i)(/ - ) sinh - [ i(2x +b)/ b 2-4a ] // Spiegel 8.64 = - i (-i)(/ - ) sin - [ (2x +b)/ b 2-4a ] // Spiegel 8.93 = - - sin- [ 2x +b b 2-4a ] = + - sin- [ -2x -b b 2-4a ], (7.2) giving forms valid for < 0 and b 2-4a > 0. Next we use this relation sin - (z) = - os - (z) + π/2 // Spiegel 5.74 = - os - (z) (7.3) to obtain two more forms, x dx (/ R) = + - os- [ 2x +b b 2-4a ] = - - os- [ -2x -b b 2-4a ]. (7.4) Trust but verify, (7.5) In these last integrals, we started with a > 0, but the results an be ontinued to part of the range a < 0 where we have b 2-4a > 0 b 2 +4a > 0 4a > -b 2 a > -(b 2 / ). (7.6) 8

19 Goldstein Classial Mehanis In the disussion of orbits with an inverse-square fore law, Goldstein (950) on page 77 writes the seond evaluation in (7.4) omitting the leading minus sign. This error is repeated on page 93 of the later 200 third edition of the book (Goldstein, Poole and Safko, all deeased), from whih we quote, where a,b, = α,β,γ, This results in another sign error in (3.54), but as it turns out, this error makes no differene in the key final result (3.55) due to the fat that os(θ-θ') = os(θ'-θ). That final result is this. The inverse-square fore law is F = -k/r 2, a partile has mass m, energy E, and angular momentum l. This last result shows that the orbits are oni setions expressed in polar oordinates r,θ where the radial is the orbit eentriity ε. For a sun-planet system, m,r,θ refer to an equivalent one-body problem where m is the redued mass, and r,θ are relative to the enter of mass. A summary of forms appearing in this doument : (7.7) x dx R = R = a + bx + x 2 ln (2x + b + 2 R ) // (4.9) > 0 2 = - ln (2x + b - 2 R ) // (4.0) > 0 3 = ln (2x +b ) // (4.4) > 0, b2-4a = 0 4 = ln [ 2x + b + 2 R 4a-b 2 ] // (.8) > 0, b 2-4a < 0 5 = 2 tanh- ( R + a x ) // (5.9) > 0, a > 0 6 = 2 tanh- ( R - a x ) // (5.0) > 0, a > 0 9

20 7 = 2 tanh- ( x - β x - α ) // (6.9) > 0, α,β roots of R 8 = ln [ x-α + x-β x-α - x-β ] // item 7 above with identity (5.) 9 = sinh - ( 2x + b 4a-b 2 ) // (7.5) > 0, b 2-4a < 0 0 = - = + 2 = - - sin- ( 2x + b b 2-4a ) // (7.2) < 0, b2-4a > 0 - os- ( 2x + b b 2-4a ) // (7.4) < 0, b2-4a > 0 - os- ( -2x - b b 2-4a ) // (7.4) < 0, b2-4a > 0 The evaluations 4, 9, 3, 0 appear in GR7 page 94, (7.8) See also Spiegel whih, however, uses R = ax 2 +bx+. Both Spiegel and GR7 present many integrals of the form x m ( R ) n for m and odd n being various positive and negative integers. Footnote onerning TI above. Adrian Fedorovih Timofeev ( ) led a ompliated life in Russia and wrote a few non-mathematial books about it (e.g., My Prison Diary). 20

21 Appendix A: About Euler's original paper Euler's study of integrals of the form dx x n / R appears in this paper, L. Euler, "Speulationes super formula integrali (x n dx)/ (aa-2bx+xx), ubi simul egregiae observationes ira frationes ontinuas ourrunt", Ata Aademiae Sientarum Imperialis Petropolitinae 782, 786, pp or in English L. Euler, "Speulations onerning the integral formula (x n dx)/ (aa-2bx+xx), where at the same time exeptional observations about ontinued frations our", Transations of the Imperial Aademy of Sienes in St. Petersburg 782, 786, pp In 775 Euler wrote about 60 (!) papers inluding the one above. The paper was formally presented in 782 (a year before his death) but was not was published until 786. The original paper an be viewed in the Euler Arhive, by looking up Subjet / Mathematis / Integration / index number 606. His paper does not atually use any of our substitutions A,B,C. Instead, he plants the seed of the idea and someone at some later time extrated the three substitutions from his work. Below we shall display parts of the original paper, but first it is helpful to desribe what he is doing. Here is a long-winded interpretation of Setion of his paper :. I. Consider the quantity a 2-2bx + x 2. Make the substitution, so that x = b+z z = x - b and dx = dz (A.) (a 2-2bx + x 2 ) = a 2-2b (b+z)/ + (b+z) 2 / 2 = (/ 2 )[a 2 2-2b (b+z) + (b+z) 2 ] = (/ 2 )[a 2 2-2b 2-2bz + b 2 + 2bz + z 2 ] = (/ 2 )[a b 2 + z 2 ] = (a 2 - b 2 + z 2 )/. // no linear z term (A.2) Then, // Euler uses = f 2 but we keep it as dx a 2-2bx+x 2 = dz a 2 - b 2 + z 2 = dz a 2 - b 2 + z 2. (A.3) 2

22 Euler's substitution (A.) has not onverted the integrand to a rational funtion R(t) of the form (.2) as we did earlier with all the offiial "Euler substitutions". However, Euler was familiar with the following integral (having more or less invented the natural logarithm and e), dz A 2 +z 2 = ln [z + A 2 +z 2 ] + onstant so he ontinues the above, dz a 2 - b 2 + z 2 = ln [ z + a2 - b 2 + z 2 C ] (A.4) where C is a onstant. He then replaes z = x - b and a 2 - b 2 + z 2 = a 2-2bx+x 2 to get x dx a 2-2bx+x 2 = ln [ x - b + a2-2bx+x 2 C and then 0 x dx a 2-2bx+x 2 = ln [ x - b + a2-2bx+x 2 -b+a He next onsiders the ase < 0 (whih he writes as = - g 2 ) and finds, ] > 0 (A.5) ]. > 0 (A.6) 0 x dx a 2-2bx+x 2 = g sin- ( x-b b 2 - a 2 ) + g sin- ( b b 2 - a 2 ). < 0 (A.7) Comparison with GR7: With a 2 a and b -b/2 one has a 2-2bx + x 2 a + bx + x 2 = R. Then (A.5) and (A.7) beome dx a+bx+x 2 = 2x + b + 2 ln [ a+bx+x2 2C ] > 0 (A.5)' dx a+bx+x 2 = g sin- ( 2x +b b 2-4a ) = - g2 < 0. (A.7') Eq. (A.5)' agrees with (7.8) line (apart from an additive onstant), and (A.7)' agrees with (7.8) line 4 if we take g = - -. Euler does not really define the meaning of his g when he writes = -g 2, nor was the notion of analyti ontinuation muh developed in 775. Here then is the original of Setion, taken from the Euler Arhive noted above, annotated with our equation numbers and a few small repairs, 22

23 (A.8) As was the ustom of the time, the paper is written in Latin ("We begin with the simplest ase, where n = 0, and we seek an integral formula for..."). The letter v is u, the number is I, and s in most non-final positions is written f, known as a "long s", a usage that was dropped after 800. Quadrati powers are written aa instead of a 2, though higher powers are later written with exponents. A root expression is written (expression). The ln symbol is a large itali lower-ase letter l, and Arsin(α) is written A fin. α The Arhive has this paper translated into German with lean typesetting (no English yet). One an only marvel at how printers of the day hand-typeset Euler's many equations for printing. The photoopy lips above and below are of modest quality, and we had to manually do some derotation of the text (Visio). As his paper title shows, Euler was in pursuit of the integral dx x n / a 2-2bx+x 2 and in the first Setion above he has handled the ase n = 0. He goes on to make more substitutions to obtain the ases n =,2,3... Here is his entire Setion 4 with a new substitution s = a 2-2bx+x 2 - a to get the n = integral expressed in terms of the n = 0 integral, 23

24 (A.9) Note that Π x dx/ a 2-2bx+x 2 and then Δ x=α dx/ a 2-2bx+x 2 where α = (b± b 2 -a 2 )/ is either root of a 2-2bx+x 2 = 0. He does this to obtain the simple result x dx x/ a 2-2bx+x 2 = b Δ - a. Later Euler summarizes his results for n = 0,,2,3 and 4 : where I.3.5 means *3*5 = 5. He then obtains a reursion relation (A.0) 24

25 (A.) whih appears in GR7 in a more general form (for R = a + bx + x 2 ), If in this GR7 integral we take n 0, m n+, a a 2 and b -2b so R Q = a 2-2bx+x 2, we get or xn+ dx Q = (n+) xn+ dx Q x n (n+) Q (2n+)(-2b) 2(n+) = (2n+)b xn dx Q xn dx Q na 2 xn- dx Q (n)a2 (n+) xn- dx Q + x n Q (A.2) whih gives Euler's reursion result (A.) to the letter. Euler next attempts to write a losed-form expression for the general ase n = n with some partial suess. The paper then ends with a long setion on writing quantities as ontinued frations, for example (A.3) There are many pages showing suh ontinued frations and Euler seems to be fasinated with them. As the title says, this is the seond topi of his paper, "Speulations onerning the integral formula (x n dx)/ (aa-2bx+xx), where at the same time exeptional observations about ontinued frations our " Euler ( ) was Swiss but moved to St. Petersburg in 727, to Berlin in 74, and finally bak to St. Petersburg in 766 where he wrote the above paper. His presene in Russia is probably the reason that the "Euler substitutions" are ommonly assoiated with Russian soures like Piskunov. He had a prolifi and eventful life in eventful times and is often ranked the greatest mathematiian of all time. 25

26 Appendix B. Summary of the three Euler substitutions Substitution A t ± x = a+bx+x 2 (B.) x dx R(x, a+bx+x 2 ) = t(x) - t 2 + bt - a t 2 -a dt 2 (b - 2 t ) 2 * R( b - 2 t, - t2 + bt - a ) b - 2 t where t(x) = a+bx+x 2 - x. (4.7) x dx R(x, a+bx+x 2 ) = t(x) dt 2 t 2 + bt + a (b + 2 t ) 2 * R( t 2 -a b + 2 t, t 2 + bt + a b + 2 t where t(x) = a+bx+x 2 + x. (4.8) ) Substitution B xt ± a = a+bx+x 2 (B.2) x dx R(x, a+bx+x 2 ) = t(x) a t 2 - bt + a dt 2 (-t 2 ) 2 * R( 2 a t - b -t 2, a t 2 - bt + a -t 2 ) where t(x) = ( a+bx+x 2 - a ) / x. (5.7) x dx R(x, a+bx+x 2 ) = t(x) - a t 2 - bt - a -2 a t - b dt 2 (-t 2 2 ) * R( -t 2, - a t2 - bt - a -t 2 ) where t(x) = ( a+bx+x 2 + a ) / x. (5.8) Substitution C : t = ( x - β x - α )± where a+bx+x 2 = (x-α)(x-β) (B.3) x dx R(x, a+bx+x 2 ) = t(x) (β-α)t dt 2 (t 2 -) 2 * - β R(αt2 t 2 -, (α-β)t t 2 - ) where t(x) = a+bx+x 2 / (x-α) = R /(x-α). (6.7) x dx R(x, a+bx+x 2 ) = t(x) (α-β)t dt 2 (t 2 -) 2 * - α R(βt2 t 2 -, (β-α)t t 2 - ) (6.8) where t(x) = a+bx+x 2 / (x-β) = R /(x-β). 26

27 Referenes Links were last heked on 0 Nov K.N. Boyadzhiev, "Euler Substitutions" (Ohio Northern University, 2006), 5p. Has examples. H. Goldstein, Classial Mehanis (Addison-Wesley, Boston, 950). H. Goldstein, C.P. Poole Jr. and J.L. Safko, Classial Mehanis, 3rd Ed. ( Addison-Wesley, New York, 200) now rebranded by Pearson, London. Authors deeased 2005, 205 and 206. [GR7] I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Produts, 7th Ed. (Aademi Press, New York, 2007). The 8th edition ame out Ot 204. Editor Dan Zwillinger has errata for editions 6,7 and 8 at N.S. Piskunov, Differential and Integral Calulus (Mir, Mosow, 969). This 895 page text was translated from the Russian by G. Yankovsky (searh the web). Setion 2 pp mentions the Euler substitutions with a few examples. M.R. Spiegel, Shaum's Outlines: Mathematial Handbook of Formulas and Tables (MGraw-Hill, New York, 968). Our page referenes are to this edition. The urrent version of this book is given below. M.R. Spiegel, S. Lipshutz, M. and J. Liu, Shaum's Outlines: Mathematial Handbook of Formulas and Tables, 4th Ed. (MGraw-Hill, New York, 202). John Liu was added for the 999 2nd Ed, and Seymour Lipshutz joined for the rd Ed. Not to be onfused with a watered-down "Easy Outline" version. This low-ost paperbak is an exellent fast referene for well-known mathematial fats. D. Zwillinger, Handbook of Integration (Jones and Bartlett, Boston, 992). 27