Elliptic Integral

An elliptic integral is an Integral of the form

$$\int {A(x)+B(x)\sqrt{S(x)}\over C(x)+D(x)\sqrt{S(x)}}\, dx,$$ (1)

or

$$\int {A(x)\,dx\over B(x)\sqrt{S(x)}},$$ (2)

where $A$, $B$, $C$, and $D$ are Polynomials in $x$ and $S$ is a Polynomial of degree 3 or 4. Another form is

$$\int R(w,x)\,dx,$$ (3)

where $R$ is a Rational Function of $x$ and $w$, $w^2$ is a function of $x$ Cubic or Quadratic in $x$, $R(w,x)$ contains at least one Odd Power of $w$, and $w^2$ has no repeated factors.

Elliptic integrals can be viewed as generalizations of the inverse Trigonometric Functions and provide solutions to a wider class of problems. For instance, while the Arc Length of a Circle is given as a simple function of the parameter, computing the Arc Length of an Ellipse requires an elliptic integral. Similarly, the position of a pendulum is given by a Trigonometric Function as a function of time for small angle oscillations, but the full solution for arbitrarily large displacements requires the use of elliptic integrals. Many other problems in electromagnetism and gravitation are solved by elliptic integrals.

A very useful class of functions known as Elliptic Functions is obtained by inverting elliptic integrals to obtain generalizations of the trigonometric functions. Elliptic Functions (among which the Jacobi Elliptic Functions and Weierstraß Elliptic Function are the two most common forms) provide a powerful tool for analyzing many deep problems in Number Theory, as well as other areas of mathematics.

All elliptic integrals can be written in terms of three ``standard'' types. To see this, write

$$R(w,x)\equiv {P(w,x)\over Q(w,x)} = {wP(w,x)Q(-w,x)\over wQ(w,x)Q(-w,x)}.$$ (4)

But since $w^2=f(x)$,

$$Q(w,x)Q(-w,x)\equiv Q_1(w,x)=Q_1(-w,x),$$ (5)

then

$wP(w,x)Q(-w,x) = A+Bx+Cw+Dx^2+Ewx+Fw^2+Gw^2x+Hw^3x$
$= (A+Bx+Dx^2+Fw^2+Gw^2x)+w(c+Ex+Hw^2x+\ldots) = P_1(x)+wP_2(x),\quad$	(6)

so

$$R(w,x)={P_1(x)+wP_2(x)\over wQ_1(w)} = {R_1(x)\over w}+R_2(x).$$ (7)

But any function $\int R_2(x)\,dx$ can be evaluated in terms of elementary functions, so the only portion that need be considered is

$${\int R_1(x)\over w}\,dx.$$ (8)

Now, any quartic can be expressed as $S_1S_2$ where

$$S_1 \equiv a_1x^2+2b_1x+c_1$$ (9)

$$S_2 \equiv a_2x^2+2b_2x+c_2.$$ (10)

The Coefficients here are real, since pairs of Complex Roots are Complex Conjugates

$${[}x-(R+Ii)][x-(R-Ii)] = x^2+x(-R+Ii-R-Ii)+(R^2-I^2i)$$

$$= x^2-2Rx+(R^2+I^2).$$ (11)

If all four Roots are real, they must be arranged so as not to interleave (Whittaker and Watson 1990, p. 514). Now define a quantity $\lambda$ such that $S_1+\lambda S_2$

$$(a_1-\lambda a_2)x^2-(2b_1-2b_2\lambda)x+(c_1-\lambda c_2)$$ (12)

is a Square Number and

$$2\sqrt{(a_1-\lambda a_2)(c_1-\lambda_2)} = 2(b_1-b_2\lambda)$$ (13)

$$(a_1-\lambda a_2)(c_1-\lambda c_2)-(b_1-\lambda b_2)^2=0.$$ (14)

Call the Roots of this equation $\lambda_1$ and $\lambda_2$, then

$$S_1-\lambda_1S_2 = \left[{\sqrt{(a_1-\lambda_1a_2)x}+\sqrt{c_1-\lambda c_2}\,}\right]^2$$

$$= (a_1-\lambda_1a_2)\left({x+\sqrt{c_1-\lambda_1c_2\over a_1-\lambda_1 a_2}\,}\right)$$

$$\equiv (a_1-\lambda_1a_2)(x-\alpha)^2$$ (15)

$$S_1-\lambda_2S_2 = \left[{\sqrt{(a_1-\lambda_1a_2)x}+\sqrt{c_1-\lambda c_2}\,}\right]^2$$

$$= (a_1-\lambda_1a_2)\left({x+\sqrt{c_1-\lambda_2c_2\over a_1-\lambda_2 a_2}\,}\right)$$

$$\equiv (a_1-\lambda_2a_2)(x-\beta)^2.$$ (16)

Taking (15)-(16) and $\lambda_2(1)-\lambda_1(2)$ gives

$$S_2(\lambda_2-\lambda_1) = (a_1-\lambda_1a_2)(x-\alpha)^2$$

$$\phantom{=} -(a_1-\lambda_2a_2)(x-\beta)^2$$ (17)

$$S_1(\lambda_2-\lambda_1) = \lambda_2(a_1-\lambda_1a_2)(x-\alpha)^2$$

$$\phantom{=} -\lambda_1(a_1-\lambda_2a_2)(x-\beta^2).$$ (18)

Solving gives

$$S_1 = {a_1-\lambda_1a_2\over \lambda_2-\lambda_1} (x-\alpha)^2 -{a_1-\lambda_2a_2\over\lambda_2-\lambda_1}(x-\beta)^2$$

$$\equiv A_1(x-\alpha)^2+B_1(x-\beta)^2$$ (19)

$$S_2 = {\lambda_2(a_1-\lambda_1a_2)\over \lambda_2-\lambda_1} (x-\alpha)^2-{\lambda_1(a_1-\lambda_2a_2)\over \lambda_2-\lambda_1} (x-\beta)^2$$

$$\equiv A_2(x-\alpha)^2+B_2(x-\beta)^2,$$ (20)

so we have

$$w^2=S_1S_2 = [A_1(x-\alpha)^2+B_1(x-\beta)^2][A^2(x-\alpha)^2+B^2(x-\beta)^2].$$ (21)

Now let

$$t \equiv {x-\alpha\over x-\beta}$$ (22)

$$dy = [(x-\beta)^{-1}-(x-\alpha)(x-\beta)^{-2}]\,dx$$

$$= {(x-\beta)-(x-\alpha)\over (x-\beta)^2}\,dx$$

$$= {\alpha-\beta\over (x-\beta)^2}\,dx,$$ (23)

so

$$w^2 = (x-\beta)^4\left[{A_1\left({x-\alpha\over x-\beta}\right)^2+B_1}\right]\left[{A_2\left({x-\alpha\over x-\beta}\right)+B_2}\right]$$

$$= (x-\beta)^4(A_1t^2+B_1)(A_2t^2+B_2),$$ (24)

and

$$w = (x-\beta)^2\sqrt{(A_1t^2+B_1)(A_2t^2+B_2)}$$ (25)

$${dx\over w} = \left[{{(x-\beta)^2\over \alpha-\beta} dt}\right]{1\over (x-\beta)^2\sqrt{(A_1t^2+B_1)(A_2t^2+B_2)}}$$

$$= {dt\over (\alpha-\beta)\sqrt{(A_1t^2+B_1)(A_2t^2+B_2)}}.$$ (26)

Now let

$$R_3(t)\equiv {R_1(x)\over\alpha-\beta},$$ (27)

so

$$\int {R_1(x)\,dx\over w} = \int {R_3(t)\,dt\over \sqrt{(A_1t^2+B_1)(A_2t^2+B_2)}}.$$ (28)

Rewriting the Even and Odd parts

$$R_3(t)+R_3(-t) \equiv 2R_4(t^2)$$ (29)

$$R_3(t)-R_3(-t) \equiv 2tR_5(t^2),$$ (30)

gives

$$R_3(t)\equiv {\textstyle{1\over 2}}(R_{\rm even}-R_{\rm odd}) = R_4(t^2)+t R_5(t^2),$$ (31)

so we have

$$\int {R_1(x)\,dx\over w} = \int {R_4(t^2)\,dt\over \sqrt{(A_... ...}}+ \int {R_5(t^2)t\,dt\over \sqrt{(A_1t^2+B_1)(A_2t^2+B_2)}}.$$ (32)

Letting

$$u \equiv t^2$$ (33)

$$du = 2t\,dt$$ (34)

reduces the second integral to

$${1\over 2}\int {R_5(u)\,du\over \sqrt{(A_1u+B_1)(A_2u+B_2)}},$$ (35)

which can be evaluated using elementary functions. The first integral can then be reduced by Integration by Parts to one of the three Legendre elliptic integrals (also called Legendre-Jacobi Elliptic Integrals), known as incomplete elliptic integrals of the first, second, and third kind, denoted $F(\phi,k)$, $E(\phi, k)$, and $\Pi(n; \phi, k)$, respectively (von Kármán and Biot 1940, Whittaker and Watson 1990, p. 515). If $\phi=\pi/2$, then the integrals are called complete elliptic integrals and are denoted $K(k)$, $E(k)$, $\Pi(n; k)$.

Incomplete elliptic integrals are denoted using a Modulus $k$, Parameter $m\equiv k^2$, or Modular Angle $\alpha \equiv \sin^{-1} k$. An elliptic integral is written $I(\phi \vert m)$ when the Parameter is used, $I(\phi,k)$ when the Modulus is used, and $I(\phi \backslash \alpha)$ when the Modular Angle is used. Complete elliptic integrals are defined when $\phi=\pi/2$ and can be expressed using the expansion

$$(1-k^2\sin^2 \theta)^{-1/2} = \sum_{n=0}^\infty {(2n-1)!!\over (2n)!!} k^{2n}\sin^{2n} \theta.$$ (36)

An elliptic integral in standard form

$$\int_a^x {dx\over \sqrt{f(x)}},$$ (37)

where

$$f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$$ (38)

can be computed analytically (Whittaker and Watson 1990, p. 453) in terms of the Weierstraß Elliptic Function with invariants

$$g_2 = a_0a_4-4a_1a_3+3{a_2}^2$$ (39)

$$g_3 = a_0a_2a_4-2a_1a_2a_3-a_4{a_1}^2-{a_3}^2a_0.$$ (40)

If $a\equiv x_0$ is a root of $f(x)=0$, then the solution is

$$x=x_0+{\textstyle{1\over 4}}f'(x_0)[\wp(z;g_2,g_3)-{\textstyle{1\over 24}}f''(x_0)]^{-1}.$$ (41)

For an arbitrary lower bound,

$x=a+$
${\sqrt{f(a)}\wp'(z)+{\textstyle{1\over 2}}f'(a)[\wp(z)-{\textstyle{1\over 24}}f... ...\wp(z)-{\textstyle{1\over 24}}f''(a)]^2-{\textstyle{1\over 48}}f(a)f^{(a)}(a)},$
	(42)

where $\wp(z)\equiv \wp(z;g_2,g_3)$ is a Weierstraß Elliptic Function.

A generalized elliptic integral can be defined by the function

$$T(a,b) \equiv {2\over\pi} \int^{\pi/2}_0 {d\theta\over\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}$$ (43)

$$= {2\over\pi} \int^{\pi/2}_0 \int{d\theta\over\cos\theta\sqrt{a^2+b^2\tan^2\theta}}$$ (44)

(Borwein and Borwein 1987). Now let

$$t \equiv b\tan\theta$$ (45)

$$dt = b\sec^2\theta\,d\theta.$$ (46)

But

$$\sec\theta =\sqrt{1+\tan^2\theta}\,,$$ (47)

so

$$dt = {b\over \cos\theta} \sec\theta\,d\theta = {b\over \cos\theta} \sqrt{1+\tan^2\theta}\,d\theta$$

$$= {b\over \cos\theta} \sqrt{1+\left({t\over b}\right)^2}\,d\theta$$

$$= {d\theta \over \cos \theta }\sqrt{b^2+t^2},$$ (48)

and

$${d\theta\over\cos\theta} ={dt\over\sqrt{b^2+t^2}},$$ (49)

and the equation becomes

$$T(a,b) = {2\over \pi} \int_0^\infty {dt\over \sqrt{(a^2+t^2)(b^2+t^2)}}$$

$$= {1\over \pi} \int_{-\infty}^\infty {dt\over \sqrt{(a^2+t^2)(b^2+t^2)}}.$$ (50)

Now we make the further substitution $u\equiv {\textstyle{1\over 2}}(t-ab/t)$. The differential becomes

$$du={\textstyle{1\over 2}}(1+ab/t^2)\,dt,$$ (51)

but $2u=t-ab/t$, so

$$2u/t=1-ab/t^2$$ (52)

$$ab/t^2=1-2u/t$$ (53)

and

$$1+ab/t^2=2-2u/t = 2(1-u/t).$$ (54)

However, the left side is always positive, so

$$1+ab/t^2=2-2u/t = 2\vert 1-u/t\vert$$ (55)

and the differential is

$$dt={du\over \left\vert 1-{u\over t}\right\vert}.$$ (56)

We need to take some care with the limits of integration. Write (50) as

$$\int_{-\infty}^\infty f(t)\,dt=\int_{-\infty}^{0^-} f(t)\,dt+\int_{0^+}^\infty f(t)\,dt.$$ (57)

Now change the limits to those appropriate for the $u$ integration

$$\int_{-\infty}^\infty g(u)\,du+\int_{-\infty}^\infty g(u)\,du = 2\int_{-\infty}^\infty g(u)\,du,$$ (58)

so we have picked up a factor of 2 which must be included. Using this fact and plugging (56) in (50) therefore gives

$$T(a,b)={2\over \pi} \int_{-\infty}^\infty {du\over \left\vert 1-{u\over t}\right\vert\sqrt{a^2b^2+(a^2+b^2)t^2+t^4}}.$$ (59)

Now note that

$$u^2 = {t^4-2abt^2+a^2b^2\over 4t^2}$$ (60)

$$4u^2t^2 = t^4-2abt^2+2abt^2$$ (61)

$$a^2b^2+t^4 = 4u^2t^2+2abt^2.$$ (62)

Plug (62) into (59) to obtain

$$T(a,b) = {2\over \pi} \int_{-\infty}^\infty {du\over \left\vert 1-{u\over t}\right\vert\sqrt{4u^2t^2+2abt^2+(a^2+b^2)t^2}}$$

$$= {2\over \pi} \int_{-\infty}^\infty {du\over \vert t-u\vert \sqrt{4u^2+(a+b)^2}}.$$ (63)

But

$\quad 2ut=t^2-ab$	(64)
$\quad t^2-2ut-ab=0$	(65)
$\quad t={\textstyle{1\over 2}}(2u\pm\sqrt{4u^2+4ab}) = u\pm\sqrt{u^2+ab},$	(66)

so

$$t-u=\pm\sqrt{u^2+ab},$$ (67)

and (63) becomes

$$T(a,b) = {2\over \pi} \int_{-\infty}^\infty {du\over \sqrt{[4u^2+(a+b)^2](u^2+ab)}}$$

$$= {1\over \pi} \int_{-\infty}^\infty {du\over \sqrt{\left[{u^2+\left({a+b\over 2}\right)^2}\right](u^2+ab)}}.$$ (68)

We have therefore demonstrated that

$$T(a,b) = T({\textstyle{1\over 2}}(a+b),\sqrt{ab}).$$ (69)

We can thus iterate

$$a_{i+1} = {\textstyle{1\over 2}}(a_i+b_i)$$ (70)

$$b_{i+1} = \sqrt{a_ib_i},$$ (71)

as many times as we wish, without changing the value of the integral. But this iteration is the same as and therefore converges to the Arithmetic-Geometric Mean, so the iteration terminates at $a_i=b_i=M(a_0,b_0)$, and we have

$$T(a_0,b_0) = T(M(a_0,b_0),M(a_0,b_0))$$

$$= {1\over \pi} \int_{-\infty}^\infty {dt\over M^2(a_0,b_0)+t^2}$$

$$= {1\over \pi M(a_0,b_0)} \left[{\tan^{-1}\left({t\over M(a_0,b_0)}\right)}\right]^\infty_{-\infty}$$

$$= {1\over \pi M(a_0,b_0)} \left[{{\pi\over 2}-\left({\pi\over 2}\right)}\right]$$

$$= {1\over M(a_0,b_0)}.$$ (72)

Complete elliptic integrals arise in finding the arc length of an Ellipse and the period of a pendulum. They also arise in a natural way from the theory of Theta Functions. Complete elliptic integrals can be computed using a procedure involving the Arithmetic-Geometric Mean. Note that

$$T(a,b) \equiv {2\over\pi} \int^{\pi/2}_0 {d\theta\over\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}$$

$$= {2\over\pi}\int^{\pi/2}_0 {d\theta \over a\sqrt{\cos ^2\theta +\left({b\over a}\right)^2 \sin^2 \theta}}$$

$$= {2\over a\pi}\int^{\pi/2}_0 {d\theta \over \sqrt{1-\left({1-{b^2\over a^2}}\right)^2\sin^2 \theta}}.$$ (73)

So we have

$$T(a,b)= {2\over a\pi}K\left({1-{b^2\over a^2}}\right)={1\over M(a,b)},$$ (74)

where $K(k)$ is the complete Elliptic Integral of the First Kind. We are free to let $a\equiv a_0\equiv 1$ and $b\equiv b_0\equiv k'$, so

$${2\over \pi}K(\sqrt{1-k'^2}\,) = {2\over \pi}K(k)={1\over M(1,k')},$$ (75)

since $k\equiv\sqrt{1-k'^2}$, so

$$K(k)={\pi\over 2M(1,k')}.$$ (76)

But the Arithmetic-Geometric Mean is defined by

$$a_i = {\textstyle{1\over 2}}(a_{i-1}+b_{i-1})$$ (77)

$$b_i = \sqrt{a_{i-1}b_{i-1}}$$ (78)

$$c_i = \left\{\begin{array}{ll} {\textstyle{1\over 2}}(a_{i-1}-b_{i-1}) & \mbox{$i>0$}\\ \sqrt{{a_0}^2-{b_0}^2} & \mbox{$i=0$,}\end{array}\right.$$ (79)

where

$$c_{n-1}={\textstyle{1\over 2}}{a_n-b_n} = {{c_n}^2\over 4a_{n+1}} \leq {{c_n}^2\over 4 M(a_0,b_0)},$$ (80)

so we have

$$K(k)= {\pi\over 2a_N},$$ (81)

where $a_N$ is the value to which $a_n$ converges. Similarly, taking instead $a_0'=1$ and $b_0'=k$ gives

$$K'(k)={\pi\over 2a_N'}.$$ (82)

Borwein and Borwein (1987) also show that defining

$$U(a,b)\equiv {\pi\over 2} \int_0^{\pi/2} \sqrt{a^2\cos^2+b^2\sin^2\theta}\,d\theta = aE'\left({b\over a}\right)$$ (83)

leads to

$$2U(a_{n+1},b_{n+1})-U(a_n,b_n)=a_nb_nT(a_n,b_n),$$ (84)

so

$${K(k)-E(k)\over K(k)}={\textstyle{1\over 2}}({c_0}^2+2{c_1}^2 +2^2{c_2}^2+\ldots+2^n{c_n}^2)$$ (85)

for $a_0\equiv 1$ and $b_0\equiv k'$, and

$${K'(k)-E'(k)\over K'(k)}={\textstyle{1\over 2}}({c_0'}^2+2{c_1'}^2 +2^2{c_2'}^2+\ldots+2^n{c_n'}^2).$$ (86)

The elliptic integrals satisfy a large number of identities. The complementary functions and moduli are defined by

$$K'(k)\equiv K(\sqrt{1-k^2}\,)=K(k').$$ (87)

Use the identity of generalized elliptic integrals

$$T(a,b)=T({\textstyle{1\over 2}}(a+b), \sqrt{ab}\,)$$ (88)

to write

$${1\over a}K\left({\sqrt{1-{b^2\over a^2}}\,}\right) = {2\over a+b}K\left({\sqrt{1-{4ab\over (a+b)^2}}\,}\right)$$

$$= {2\over a+b} K\left({\sqrt{a^2+b^2-2ab\over(a+b)^2}\,}\right)$$

$$= {2\over a+b}K\left({a-b\over a+b}\right)$$ (89)

$$K\left({\sqrt{1-{b^2\over a^2}}\,}\right)= {2\over 1+{b\over a}}K\left({1-{b\over a}\over 1+{b\over a}}\right).$$ (90)

Define

$$k'\equiv {b\over a},$$ (91)

and use

$$k\equiv \sqrt{1-k'^2},$$ (92)

so

$$K(k)={2\over 1+k'} K\left({1-k'\over 1+k'}\right).$$ (93)

Now letting $l\equiv (1-k')/(1+k')$ gives

$$l(1+k')=1-k' \Rightarrow k'(l+1)=1-l$$ (94)

$$k'={1-l\over 1+l}$$ (95)

$$k = \sqrt{1-k'^2}=\sqrt{1-\left({1-l\over 1+l}\right)^2}$$

$$= \sqrt{(1+l)^2-(1-l)^2\over (1+l)^2} = {2\sqrt{l}\over 1+l},$$ (96)

and

$${\textstyle{1\over 2}}(1+k') = {1\over 2}\left({1+{1-l\over 1+l}}\right)= {1\over 2}\left[{(1+l)+(1-l)\over 1+l}\right]$$

$$= {1\over 1+l}.$$ (97)

Writing $k$ instead of $l$,

$$K(k)={1\over k+1} K\left({2\sqrt{k}\over 1+k}\right).$$ (98)

Similarly, from Borwein and Borwein (1987),

$$E(k)={1+k\over 2}E\left({2\sqrt{k}\over 1+k}\right)+{k'^2\over 2}K(k)$$ (99)

$$E(k)=(1+k')E\left({1-k'\over 1+k'}\right)-k'K(k).$$ (100)

Expressions in terms of the complementary function can be derived from interchanging the moduli and their complements in (93), (98), (99), and (100).

$$K'(k) = K(k') = {2\over 1+k} K\left({1-k\over 1+k}\right)$$

$$= {2\over 1+k}K'\left({\sqrt{1-\left({1-k\over 1+k}\right)^2}\,}\right)$$

$$= {2\over 1+k}K'\left({2\sqrt{k}\over 1+k}\right)$$ (101)

$$K'(k)={1\over 1+k'} K\left({2\sqrt{k'}\over 1+k'}\right)= {1\over 1+k'} K'\left({1-k'\over 1+k'}\right),$$ (102)

and

$$E'(k)=(1+k)E'\left({2\sqrt{k}\over 1+k}\right)-kK'(k)$$ (103)

$$E'(k)=\left({1+k'\over 2}\right)E'\left({1-k'\over 1+k'}\right)+{k^2\over 2}K'(k).$$ (104)

Taking the ratios

$${K'(k)\over K(k)}=2 {K'\left({2\sqrt{k}\,\over 1+k}\right)\o... ...t({1-k'\over 1+k'}\right)\over K\left({1-k'\over 1+k'}\right)}$$ (105)

gives the Modular Equation of degree 2. It is also true that

$$K(x)={4\over (1+\sqrt{x'})^2} K\left({\left[{1-{\root 4\of {1-x^4}}\over 1+{\root 4\of {1-x^4}}\,}\right]^2}\right).$$ (106)

See also Abelian Integral, Amplitude, Argument (Elliptic Integral), Characteristic (Elliptic Integral), Delta Amplitude, Elliptic Function, Elliptic Integral of the First Kind, Elliptic Integral of the Second Kind, Elliptic Integral of the Third Kind, Elliptic Integral Singular Value, Heuman Lambda Function, Jacobi Zeta Function, Modular Angle, Modulus (Elliptic Integral), Nome, Parameter

References

Elliptic Integrals

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Arfken, G. ``Elliptic Integrals.'' §5.8 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 321-327, 1985.

Borwein, J. M. and Borwein, P. B. Pi & the AGM: A Study in Analytic Number Theory and Computational Complexity. New York: Wiley, 1987.

Hancock, H. Elliptic Integrals. New York: Wiley, 1917.

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Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. ``Elliptic Integrals and Jacobi Elliptic Functions.'' §6.11 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp. 254-263, 1992.

Prudnikov, A. P.; Brychkov, Yu. A.; and Marichev, O. I. Integrals and Series, Vol. 1: Elementary Functions. New York: Gordon & Breach, 1986.

Timofeev, A. F. Integration of Functions. Moscow and Leningrad: GTTI, 1948.

von Kármán, T. and Biot, M. A. Mathematical Methods in Engineering: An Introduction to the Mathematical Treatment of Engineering Problems. New York: McGraw-Hill, p. 121, 1940.

Whittaker, E. T. and Watson, G. N. A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, 1990.