Infinite Derivative's Series Expansion of Indefinite and Definite Integral

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3. Lemma of Integration of complex function Lemma 1. Let u, v are functions. If ∀i ∈ {0, N} : ∃u(i), ∃V (i) : exists all i-order derivatives of u function and exists all i-order integrals of v function then ∞ (5) udv = (−1)i u(i)V (i) i=0 Z X Proof. Knowing ( [1]) that

d f (x) dx = f (x) dx Z and dx df df = df = dx dx dx Z Z Z and using (1) and (3), it’s easy to show that following equality of integral vdu is du (6) vdu = d vdx ⇐⇒ V (0)du(0) = u(1)dV (1) R dx Z Z Z Z Z Furthermore, integral with i-order derivative and i-order integra is du(i) (7) V (i)du(i) = d V (i)dx = u(i+1)dV (i+1) dx Z Z Z Z Using ( [1])

(8) udv = uv − vdu Z Z and substituting (6) into it, we get

(9) u(0)dV (0) = u(0)V (0) − u(1)dV (1) Z Z and generalized case, using (7):

(10) u(i)dV (i) = u(i)V (i) − u(i+1)dV (i+1) Z Z Let’s evaluate the integral udv with (9) and (10) using mathematical induction:

udv = u(0)V (0) − uR(1)dV (1) = u(0)V (0) − u(1)V (1) + u(2)dV (2) = ... Z Z Z (11) = u(0)V (0) − u(1)V (1) + u(2)V (2) − u(3)V (3) + u(4)V (4) − u(5)V (5) + ... Thus, ∞ (12) udv = (−1)i u(i)V (i) i=0 Z X INFINITE DERIVATIVE’S SERIES EXPANSION OF INDEFINITE AND DEFINITE INTEGRAL3

Forms. The formula (5) could be written in diﬀerent terms. Standard form: ∞ diu (13) udv = (−1)i ··· v(dx)i dxi i=0 Z X Z Z i Form with partial expantion: | {z } n−1 (14) udv = (−1)iu(i)V (i) + (−1)n u(n)dV (n) i=0 Z X Z

ex Example 1. Let’s ﬁnd integral of x dx : By the Lemma 1 with (5) R ∞ x x i x e v = e , V ( ) = e , C =0, C∞ = C i! dx i ex C (15) = 1 (i) i 1 = i+1 + x u = , u = (−1) i! i x Z x x +1 i=0 X

ex Example 2. Let’s ﬁnd integral of xn dx : By the Lemma 1 with (5) R ex dx = xn Z x (i) x v = e , V = e , Ci =0, C∞ = C 1 (i) i (n−1+i)! 1 = u = xn , u = (−1) (n−1)! xi+n

∞ ex (n − 1+ i)! (16) + C (n − 1)! xi+n i=0 X

4. Formulas of Integration Theorem 1 (Main). Let f(x) is function. If ∀i ∈{0, N} : ∃f (i)(x) exists all i-order derivatives of this function, then ∞ xi+1 (17) f(x)dx = (−1)if (i)(x) + C (i + 1)! i=0 Z X

Proof. We can easily ﬁnd integral by Lemma 1 and (5)

i i x +1 v = x V ( ) = , C =0, C∞ = C f(x)dx = F (1)(x)= (i+1)! i = u = f u(i) = f (i) Z ∞ i+1 i x (18) (−1) f (i)(x) + C (i + 1)! i=0 X 4 VOLOSHYN VICTOR

Forms. The formula (17) could be written in diﬀerent terms. Standard form: ∞ dif(x) xi+1 (19) f(x)dx = f(x)+ (−1)i + C dxi (i + 1)! i=1 Z X

Checking. Let’s check, if ( f(x)dx)′ = f(x): From (17) : R ∞ ′ xi+1 (20) (−1)i f (i) + C = (i + 1)! i=0 ! X ′ ′ ′ = |(fg) = f g + fg | = ∞ ∞ (i + 1)xi xi+1 = (−1)i f (i) + (−1)i f (i+1) (i + 1)! (i + 1)! i=0 i=0 X ∞ X∞ xi xi = f(x)+ (−1)i f (i) − (−1)i f (i) i! i! i=1 i=1 X X (21) = f(x)

Example 3. Let’s ﬁnd integral of xdx : By the Theorem 1 with (17) R f (0) = x x2 x2 (22) xdx = f (1) =1 = xx − + C = + C 2 2 Z f (i) =0,i> 1,i ∈ N

Example 4. Let’s ﬁnd integral of exdx : By the Theorem 1 with (17) R ∞ xi+1 (23) exdx = f (i) = ex,i ∈ N = ex (−1)i + C (i + 1)! Z i=0 X

Using Taylor series ( [1]) ∞ i x − (24) (−1)i+1 = e x(ex − 1) i! i=1 X integral (23) could be rewritten in

− ′ (25) exdx = exe x(ex − 1)+ C = ex + C Z INFINITE DERIVATIVE’S SERIES EXPANSION OF INDEFINITE AND DEFINITE INTEGRAL5

Theorem 2. Let f(x) is function. If ∀i ∈ {0, N} : ∃f (i)(x) exists all i-order derivatives of this function, then ∞ b bi+1f (i) (b) − ai+1f (i)(a) (26) f(x)dx = (−1)i (i + 1)! a i=0 Z X

Proof. By First Fundamental Theorem of Calculus [1] and by Teorem 1 with (17) we get b f(x)dx = Za b (1) (1) = a f(x)dx = F (b) − F (a) ∞ ∞ R bi+1 ai+1 = (−1)i f (i)(b) + C − (−1)i f (i)(a) + C (i + 1)! (i + 1)! i=0 i=0 ! X∞ X bi+1f (i)(b) − ai+1f (i)(a) (27) = (−1)i (i + 1)! i=0 X

References

[1] Milton Abramowitz, Irene A. Stegun, Handbook of Mathematical Functions: with Formulas, Graphs, and Mathematical Tables, 9th printing, Dover Publications (1972). [2] Philip J. Davis,. Philip Rabinowitz, Methods of Numerical Integration: Second Edition, Dover Publications, Inc. (2007). [3] V. I. Krylov, Approximate Calculation of Integrals, Dover Publications, Inc. (2006). [4] Manuel Bronstein, Symbolic Integration I: Transcendental Functions, second edition, Springer (2004). [5] A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Integrals and Series, vol. 2, Gordon and Breach, Amsterdam (1998). [6] Alan Jeﬀrey, Daniel Zwillinger, Table of Integrals, Series, and Products, second edition, Aca- demic Press, 2007 E-mail address: [email protected]