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(Chebyshev showed that therewere no other cases in which the integral (5.194) could be expressed in elementaryfunctions.)5. Elliptic integrals.a) Any polynomial of degree three with real coefficients has a real root x0 , andcan be reduced to a polynomial of the form t 2 (at 4 + bt 3 + ct 2 + dt + e), wherea = 0, by the substitution x − x0 = t 2 .b) If R(u,√ of degree 3 or 4, the func√ v) is a rational function and P a polynomialtion R(x, P (x)) can be reduced to the form R1 (t, at 4 + bt 3 + · · · + e), wherea = 0.5.7 Primitives327c) A fourth-degree polynomial ax 4 + bx 3 + · · · + e can be represented as aproduct a(x 2 + p1 x + q1 )(x 2 + p2 x + q2 ) and can always be brought into the form(M1 +N1 t 2 )(M2 +N2 t 2 )by a substitution x = αt+βγ t+1 .(γ t+1)2√d) A function R(x, ax 4 + bx 3 + · · · + e) can be reduced to the form R1 t, A 1 + m1 t 2 1 + m2 t 2by a substitution x =αt+βγ t+1 .√(x,y)e) A function R(x, y) can be represented as a sum R1 (x, y) + R2√y , whereR1 and R2 are rational functions.f) Any rational function can be represented as the sum of even and odd rationalfunctions.g) If the rational function R(x) is even, it has the form r(x 2 ); if odd, it has theform xr(x 2 ), where r(x) is a rational function.√h) Any function R(x, y) can be reduced to the formR1 (x, y) +R2 (x 2 , y) R3 (x 2 , y)+x.√√yy√i) Up to a sum of elementary terms, any integral of the form R(x, P (x)) dx,where P (x) is a polynomial of degree four, can be reduced to an integralr(t 2 ) dt#,A(1 + m1 t 2 )(1 + m2 t 2 )where r(t) is a rational function and A = ±1.√√√j) If |m1 | > |m2 | > 0, one of the substitutions m1 t = x, m1 t = 1 − x 2 ,√√r(t 2 ) dtm1 t = √ x 2 , and m1 t = √ 1 2 will reduce the integral √A(1+m1 t 2 )(1+m2 t 2 )1−x1−x2 ) dxr̃(xto the form √, where 0 < k < 1 and r̃ is a rational function.22 2(1−x )(1−k x )k) Derive a formula for lowering the exponents 2n and m for the integralsx 2n dxdx##,.2222m(1 − x )(1 − k x )(x − a) · (1 − x 2 )(1 − k 2 x 2 )l) Any elliptic integral #R x, P (x) dx,where P is a fourth-degree polynomial, can be reduced to one of the canonical forms(5.185), (5.186), (5.187), upto a sum of terms consisting of elementary functions.m) Express the integral √ dx 3 in terms of canonical elliptic integrals.1+xn) Express the primitives of the functionselliptic integrals.√ 1cos 2xand√1cos α−cos xin terms of3285Differential Calculus6.
Using the notation introduced below, find primitives of the following nonelementary special functions, up to a linear function Ax + B; xea) Ei(x) =dx (the exponential integral); xsin xb) Si(x) =dx (the sine integral); xcos xdx (the cosine integral);c) Ci(x) = xsinh xd) Shi(x) =dx (the hyperbolic sine integral);xcosh xe) Chi(x) =dx (the hyperbolic cosine integral);x ⎫⎬f) S(x) = sin x 2 dx ⎪(the Fresnel integrals);⎭g) C(x) = cos x 2 dx ⎪2h) Φ(x) = e−x dx (the Euler–Poisson integral);dxi) li(x) =(the logarithmic integral).ln x7. Verify that the following equalities hold, up to a constant:a)b)c)d)e)Ei(x) = li(x);Chi(x) = 12 [Ei(x) + Ei(−x)];Shi(x) = 12 [Ei(x) − Ei(−x)];Ei(ix) = Ci(x) + i Si(x);eiπ/4 Φ(xe−iπ/4 ) = C(x) + iS(x).8.
A differential equation of the formdy f (x)=dxg(y)is called an equation with variables separable, since it can be rewritten in the formg(y) dy = f (x) dx,in which the variables x and y are separated. Once this is done, the equation can besolved asg(y) dy = f (x) dx + c,by computing the corresponding integrals.Solve the following equations:2a) 2x 3 yy +√ y = 2;b) xyy = 1 + x 2 ;5.7 Primitives329c) y = cos(y + x), setting u(x) = y(x) + x;d) x 2 y − cos 2y = 1, and exhibit the solution satisfying the condition y(x) → 0as x → +∞.e) x1 y (x) = Si(x);f)y (x)cos x= C(x).9. A parachutist has jumped from an altitude of 1.5 km and opened the parachute atan altitude of 0.5 km. For how long a time did he fall before opening the parachute?Assume the limiting velocity of fall for a human being in air of normal density is50 m/s. Solve this problem assuming that the air resistance is proportional to:a) the velocity;b) the square of the velocity.Neglect the variation of pressure with altitude.10.
It is known that the velocity of outflow of water from a small aperture√at thebottom of a vessel can be computed quite precisely from the formula 0.6 2gH ,where g is the acceleration of gravity and H the height of the surface of the waterabove the aperture.A cylindrical vat is set upright and has an opening in its bottom. Half of the waterfrom the full vat flows out in 5 minutes. How long will it take for all the water toflow out?11. What shape should a vessel be, given that it is to be a solid of revolution, inorder for the surface of the water flowing out of the bottom to fall at a constant rateas water flows out its bottom? (For the initial data, see Exercise 10.)12. In a workshop with a capacity of 104 m3 fans deliver 103 m3 of fresh air perminute, containing 0.04 % CO2 , and the same amount of air is vented to the outside.At 9:00 AM the workers arrive and after half an hour, the content of CO2 in the airrises to 0.12 %.
Evaluate the carbon dioxide content of the air by 2:00 PM.Chapter 6Integration6.1 Definition of the Integral and Description of the Setof Integrable Functions6.1.1 The Problem and Introductory ConsiderationsSuppose a point is moving along the real line, with s(t) being its coordinate at time tand v(t) = s (t) its velocity at the same instant t. Assume that we know the positions(t0 ) of the point at time t0 and that we receive information on its velocity. Havingthis information, we wish to compute s(t) for any given value of time t > t0 .If we assume that the velocity v(t) varies continuously, the displacement ofthe point over a small time interval can be computed approximately as the product v(τ )Δt of the velocity at an arbitrary instant τ belonging to that time intervaland the magnitude Δt of the time interval itself.
Taking this observation into account, we partition the interval [t0 , t] by marking some times ti (i = 0, . . . , n) so thatt0 < t1 < · · · < tn = t and so that the intervals [ti−1 , ti ] are small. Let Δti = ti − ti−1and τi ∈ [ti−1 , ti ]. Then we have the approximate equalitys(t) − s(t0 ) ≈nv(τi )Δti .i=1According to our picture of the situation, this approximate equality will becomemore precise if we partition the closed interval [t0 , t] into smaller and smaller intervals. Thus we must conclude that in the limit as the length λ of the largest of theseintervals tends to zero we shall obtain an exact equalitylimλ→0nv(τi )Δti = s(t) − s(t0 ).(6.1)i=1This equality is none other than the Newton–Leibniz formula (fundamental theorem of calculus), which is fundamental in all of analysis.
It enables us on the one© Springer-Verlag Berlin Heidelberg 2015V.A. Zorich, Mathematical Analysis I, Universitext,DOI 10.1007/978-3-662-48792-1_63313326IntegrationFig. 6.1hand to find a primitive s(t) numericallyfrom its derivative v(t), and on the otherhand to find the limit of sums ni=1 v(τi )Δti on the left-hand side from a primitives(t) found by any means whatever.Such sums, called Riemann sums, are encountered in a wide variety of situations.Let us attempt, for example, following Archimedes, to find the area under theparabola y = x 2 above the closed interval [0, 1] (see Fig.
6.1). Without going intodetail here as to the meaning of the area of a figure, which we shall take up later,like Archimedes, we shall work by the method of exhausting the figure with simplefigures – rectangles, whose areas we know how to compute. After partitioning theclosed interval [0, 1] by points 0 = x0 < x1 < · · · < xn = 1 into tiny closed intervals[xi−1 , xi ], we can obviously compute the required area σ as the sum of the areas ofthe rectangles shown in the figure:σ≈n2xi−1Δxi ;i=1here Δxi = xi − xi−1 .
Setting f (x) = x 2 and ξi = xi−1 , we rewrite the formula asσ≈nf (ξi )Δxi .i=1In this notation we have, in the limit,limλ→0nf (ξi )Δxi = σ(6.2)i=1where, as above, λ is the length of the longest interval [xi−1 , xi ] in the partition.Formula (6.2) differs from (6.1) only in the notation. Forgetting for a moment thegeometric meaning of f (ξi )Δxi and regarding x as time and f (x) as velocity, wefind a primitive F (x) for the function f (x) and then, by formula (6.1) we find thatσ = F (1) − F (0).In our case f (x) = x 2 , so that F (x) = 13 x 3 + c, and σ = F (1) − F (0) = 13 .This is Archimedes’ result, which he obtained by a direct computation of thelimit in (6.2).6.1 Definition of the Integral333A limit of integral sums is called an integral.
Thus the Newton–Leibniz formula (6.1) connects the integral with the primitive.We now turn to precise formulations and the verification of what was obtainedabove on the heuristic level from general considerations.6.1.2 Definition of the Riemann Integrala. PartitionsDefinition 1 A partition P of a closed interval [a, b], a < b, is a finite system ofpoints x0 , . . . , xn of the interval such that a = x0 < x1 < · · · < xn = b.The intervals [xi−1 , xi ], (i = 1, . . .
, n) are called the intervals of the partition P .The largest of the lengths of the intervals of the partition P , denoted λ(P ), iscalled the mesh of the partition.Definition 2 We speak of a partition with distinguished points (P , ξ ) on the closedinterval [a, b] if we have a partition P of [a, b] and a point ξi ∈ [xi−1 , xi ] has beenchosen in each of the intervals of the partition [xi−1 , xi ] (i = 1, . . . , n).We denote the set of points (ξ1 , . . .
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