On infinite many integral formulas for reciprocals of real, continuous functions

On infinite many integral formulas for reciprocals of real, continuous functions Christian Pruvost Department of Mathematics, University of Hamburg, Bundesstraße 55, Hamburg, 20146, Germany, christian.pruv[email protected]burg.de Abstract In this paper, we derive infinite many formulas for definite Riemann integrals of reciprocals of real, continuous functions on compact, nondegenerated intervals. This extends results that the author presented in a previous work. Keywords: continuous functions, Riemann integral, integral formula Mathematics Subject Classification (2020): 26A36, 26A42 1 Introduction In a previous paper, the author presented a formula for definite Riemann integrals of reciprocals of bounded, Riemann-integrable, real functions that are bounded away from 0 and which are either strictly positive or strictly negative [1]. In this paper, we further suppose that our underlying functions are continuous in the classical sense. Using elementary results for series and differentiation (see [2], [3]), we obtain infinitely many integral formulas from the main integral identity derived in [1]. We make the convention that 1 ∈N should be the least natural number. We denote R+as the set of all positive real numbers. 2 Main part In the main part of our paper, we want to prove our main theorem, namely Theorem 2.1. Let a, b ∈Rwith a < b and let f∈C([a, b],R). Assume x1, x2∈[a, b]. Suppose f(x)= 0 for all x∈[a, b]. Then, for all w∈R\ {0} with 0<inf x∈[a,b]f(x) w≤sup x∈[a,b]f(x) w<1 and all n∈N0, we have Zx2 x1 1 f(x)dx =1 n!wn+1 ∞ X k=0 Zx2 x1 (k+n)! k!fn(x)1−f(x) wk dx. We prepare the proof of this theorem by means of four lemmas: Lemma 2.2. Let M∈R. Then, we have ∞ X k=0 |M|(k+n)! k!|y|k<∞ for all n∈N0,y∈(−1,1). Proof. Let M∈R. We prove the statement via induction over n: Induction base (n= 0): Let y∈(−1,1). It is well-known that ∞> ∞ X k=0 |M||y|k= ∞ X k=0 |M|(k+ 0)! k!|y|k holds. Hence, the induction base follows. Induction step (n→n+ 1): Let n∈N0and suppose ∞ X k=0 |M|(k+n)! k!|y|k<∞(2.1) 1 for every y∈(−1,1). We have to show ∞ X k=0 |M|(k+n+ 1)! k!|y|k<∞ for all y∈(−1,1) then. We consider the function g: (−1,1) →R, g(y) := ∞ X k=0 |M|(k+n)! k!yk. Because of (2.1) and the differentiability of power series functions with nonzero radius of convergence in the open interior of the disk of convergence, g is well-defined and we get ˙g(y) = ∞ X k=1 |M|(k+n)! (k−1)! yk−1= ∞ X k=0 |M|(k+n+ 1)! k!yk for all y∈(−1,1) with ˙gas the first derivative of g. This leads to the statement then (see [2]). Lemma 2.3. Let a, b ∈Rwith a<band let f∈C([a, b],R). Then, the following statements are equivalent: (i) ∀x∈[a, b] : f(x)= 0. (ii) There is a non-degenerated interval I⊆R\ {0}such that all w∈I satisfy 0<inf x∈[a,b]f(x) w≤sup x∈[a,b]f(x) w<1. Proof. We prove (i)⇒(ii) and (ii)⇒(i). (i)⇒(ii): Suppose f(x)= 0 for every x∈[a, b]. As f∈C([a, b],R) holds, we conclude from the Intermediate Value Theorem and the Extreme Value Theorem that there is a σ∈R+with f(x)≥σfor all x∈[a, b]orf(x)≤ −σ for all x∈[a, b]. Since [a, b]⊆Ris a compact, non-degenerated interval and as f∈C([a, b],R) holds, we infer that there is an M∈R+with σ≤ |f(x)|≤M for every x∈[a, b]. Let ρ∈Rwith ρ>M. If σ≤f(x)≤Mholds for all x∈[a, b], we can define our desired interval I⊆R\ {0}as I:= {r∈R|r > ρ} and if −M≤f(x)≤ −σis valid for every x∈[a, b], we can set I:= {r∈R|r < −ρ}. 2 (ii)⇒(i): Suppose that (ii) holds. Let I⊆R\ {0}be a non-degenerated interval such that all w∈Isatisfy 0<inf x∈[a,b]f(x) w≤sup x∈[a,b]f(x) w<1.(2.2) Then, (i) follows clearly. Lemma 2.4. Let a, b ∈Rwith a < b and suppose f∈C([a, b],R). Let w∈R\ {0}with 0<inf x∈[a,b]f(x) w≤sup x∈[a,b]f(x) w<1. Then, there exists an ϵ∈R+such that all z∈(w−ϵ, w +ϵ)satisfy z= 0 and 0<inf x∈[a,b]f(x) z≤sup x∈[a,b]f(x) z<1. Proof. As w∈R\{0}holds, there is a δ∈R+such that [w−δ, w+δ]⊆R\{0} is valid. Set J:= [w−δ, w +δ]. Define h∈C([a, b]×J, R) via h: [a, b]×J→R, h(x, y) := f(x) y. Now, let λ∈(0,1) with λ < inf x∈[a,b]f(x) w≤sup x∈[a,b]f(x) w<1−λ. As [a, b]×J⊆R2is a compact rectangle and since h∈C([a, b]×J, R) holds, we conclude that there is an ϵ∈R+with ϵ<δsuch that all (x1, y1),(x2, y2)∈[a, b]×Jwith ∥(x1−x2, y1−y2)∥2< ϵ satisfy |h(x1, y1)−h(x2, y2)|<λ 2. Then, all z∈(w−ϵ, w +ϵ) satisfy z= 0 with 0<λ 2≤inf x∈[a,b]f(x) z≤sup x∈[a,b]f(x) z≤1−λ 2<1. Lemma 2.5. Let a, b ∈Rwith a<band let n∈N0. Let f∈C([a, b],R) with f(x)= 0 for every x∈[a, b]. Then, the set I:= (w∈R\ {0}: 0 <inf x∈[a,b]f(x) w≤sup x∈[a,b]f(x) w<1) 3 is non-empty and open in Rand the function gn:I→R, gn(z) := ∞ X k=0 Zb a (k+n)! k!fn(x)1−f(x) zk dx is well-defined and continuously differentiable with first derivative ˙gn:I→R,˙gn(z) := ∞ X k=0 Zb a (k+n+ 1)! z2k!fn+1(x)1−f(x) zk dx. Proof. We infer from Lemma 2.3 and Lemma 2.4 that Iis non-empty and open in R. We show now that gnis well-defined. Let z∈I. Then, there is a β∈(0,1) with 0<1−f(x) z< β for all x∈[a, b]. Moreover, there is an M∈R+with |f(x)| ≤ Mfor every x∈[a, b] because of f∈C([a, b],R). We infer from Lemma 2.2 that ∞ X k=0 Mn(b−a)(k+n)! k!βk<∞(2.3) holds. We see Zb a (k+n)! k!fn(x)1−f(x) zk dx ≤Zb a (k+n)! k!|f(x)|n 1−f(x) z k dx ≤Zb a (k+n)! k!Mnβkdx =Mn(b−a)(k+n)! k!βk for every k∈N0. This and (2.3) deliver that gnis well-defined. We only need to prove the differentiation claim in our lemma now. Let y∈I. It follows from the proof of Lemma 2.4 that there is an ϵ∈R+ and an α∈(0,1) such that all ˜y∈Rwith |y−˜y|<2ϵsatisfy ˜y∈Iand 0<1−f(x) ˜y< α for all x∈[a, b]. Set y1:= y−ϵand y2:= y+ϵ. Define the functions Tl: [y1, y2]→R, Tl(t) := l X k=0 Zb a (k+n)! k!fn(x)1−f(x) tk dx for every l∈N. Using f∈C([a, b],R) and elementary arguments regarding the interchange of integration and differentiation (see [3]), one can see that Tlis continuously differentiable with first derivative ˙ Tl:I→R,˙ Tl(t) := l X k=0 Zb a (k+n+ 1)! t2k!fn+1(x)1−f(x) tk dx 4 for all l∈N. In order to terminate this proof, we show now that there is a bounded function T: [y1, y2]→Rwith lim l→∞ ∥T−˙ Tl∥∞= 0, as the differentiation claim follows from a well-known result about differentiation and uniform convergence of functions then (see [2]). We prove this by means of the convergence criterion of Weierstrass (see [2]). Let ˜ϵ∈R+. We have to show then that there is an N∈Nsuch that all q, m ∈Nwith N≤m < q satisfy ∥˙ Tq−˙ Tm∥∞<˜ϵ. Let L:= min(|y1|,|y2|). Lemma 2.2 delivers the existence of an N∈Nsuch that all q, m ∈Nwith N≤m < q satisfy q X k=m+1 (k+n+ 1)! k!Mn+1(b−a)αk<˜ϵL2 2.(2.4) Then, all t∈[y1, y2] satisfy |˙ Tq(t)−˙ Tm(t)|= q X k=m+1 Zb a (k+n+ 1)! t2k!fn+1(x)1−f(x) tk dx ≤ q X k=m+1 Zb a (k+n+ 1)! |t|2k!Mn+1  1−f(x) t k dx ≤ q X k=m+1 Zb a (k+n+ 1)! L2k!Mn+1αkdx =1 L2 q X k=m+1 (k+n+ 1)! k!Mn+1(b−a)αk <1 L2·˜ϵL2 2=˜ϵ 2 for all q, m ∈Nwith N≤m < q, where we have used (2.4). We get ∥˙ Tq−˙ Tm∥∞≤˜ϵ 2<˜ϵ for all q, m ∈Nwith N≤m < q. Hence, the lemma follows. Proof of Theorem 1. Take Ifrom Lemma 2.5. We assume x1=aand x2=bwithout loss of generality. We show the integral identities by means of induction over n∈N0. The induction base (n= 0) follows from [1]. Thus, suppose n∈N0with Zb a 1 f(x)dx =1 n!wn+1 ∞ X k=0 Zb a (k+n)! k!fn(x)1−f(x) wk dx (2.5) 5 for all w∈I. We must deduce Zb a 1 f(x)dx =1 (n+ 1)!wn+2 ∞ X k=0 Zb a (k+n+ 1)! k!fn+1(x)1−f(x) wk dx for every w∈Inow. (2.5) yields n!wn+1 Zb a 1 f(x)dx = ∞ X k=0 Zb a (k+n)! k!fn(x)1−f(x) wk dx. (2.6) for all w∈I. (2.6) and Lemma 2.5 deliver that the functions H1:I→R, H1(w) := n!wn+1 Zb a 1 f(x)dx and H2:I→R, H2(w) := ∞ X k=0 Zb a (k+n)! k!fn(x)1−f(x) wk dx are well-defined and differentiable with H1=H2 and (n+ 1)!wnZb a 1 f(x)dx = ∞ X k=0 Zb a (k+n+ 1)! w2k!fn+1(x)1−f(x) wk dx (2.7) for all w∈I. (2.7) yields Zb a 1 f(x)dx =1 (n+ 1)!wn+2 ∞ X k=0 Zb a (k+n+ 1)! k!fn+1(x)1−f(x) wk dx for every w∈Iand the induction step is done. Remark 2.6. Theorem 2.1 provides an integral formula for definite Riemann integrals of reciprocals of real, continuous functions for every n∈N0 and can be contemplated as an ”identity generator”. 3 Conclusion Our work has provided infinite many formulas for definite Riemann integrals of reciprocals of real, continuous functions. The derivation of these identities is an extension of a previous work of the author and can be used in lectures for undergraduate students. Investigations with respect to numerical applications of the integral formulas presented in this paper might be of interest in the future. 6 References [1] Christian Pruvost. An Integral Formula for Reciprocals of Riemannintegrable Functions and Applications. Oct. 2025. doi:10.5281/zenodo. 17488744. (Visited on 12/21/2025). [2] Otto Forster. Analysis 1: Differentialund Integralrechnung einer Ver¨anderlichen. 9., ¨uberarbeitete Auflage. SpringerLink B¨ucher. Wiesbaden: Vieweg+Teubner, 2008. isbn: 978-3-8348-9464-9. doi:10.1007/978-3-8348-9464-9. [3] Otto Forster. Analysis. 2: Differentialrechnung im IRn, gew¨ohnliche Differentialgleichungen. 7., verb. Aufl. Vieweg-Studium Grundkurs Mathematik. Wiesbaden: Vieweg, 2006. isbn: 978-3-8348-0250-7. 7