# An Identity Involving the Integral of the First-Kind Chebyshev Polynomials

Authors: Xiao Wang; Jiayuan Hu
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

We used the algebraic manipulations and the properties of Chebyshev polynomials to obtain an interesting identity involving the power sums of the integral of the first-kind Chebyshev polynomials and solved an open problem proposed by Wenpeng Zhang and Tingting Wang.


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## Body

## 1. Introduction

As we all know, the famous Chebyshev polynomials of the first and second kindTnx and Unx are defined by the second-order linear recursive formulae T0x=1, T1x=x, and Tn+1=2xTnx-Tn-1x for all n≥1. U0x=1, U1x=2x, and Un+1=2xUnx-Un-1x for all n≥1.The general term formulae ofTnx and Unx are(1)Tnx=n2∑k=0n/2-1kn-k-1!k!n-2k!2xn-2kand(2)Unx=∑k=0n/2-1kn-k!k!n-2k!2xn-2kMany authors had studied the elementary properties of Chebyshev polynomials and obtained a series of interesting conclusions. For example, C. Cesarano [1], C.-L. Lee and K. B. Wong [2], and Wenpeng Zhang and Tingting Wang [3] proved a series of identities involving Chebyshev polynomials. A. H. Bhrawy et al. (see [4–7]) and N. Bircan and C. Pommerenke [8] obtained many important applications of the Chebyshev polynomials. Xiaoxue Li [9] obtained some identities involving power sums of Tnx and Unx. At the same time, she also proposed the following open problem.Whether there exists an exact expression for the derivative or integral of the Chebyshev polynomials of the first kind in terms of the Chebyshev polynomials of the first kind (and vice versa)?Tingting Wang and Han Zhang [10] partly proved this problem. That is, they proved the identities(3)∑m=1hTm′x2n=-1n4nx2-1n·2n!n!2·∑m=1hm2n+∑i=1n+1∑k=1n2nn-k-1n-kri,hT2kix+si,hT2ki-1x4n+ix2-1n+iUk-12ix+∑i=1n+1∑k=1n2nn-k-1n-kpi,hT2kh+ix+qi,hT2kh+i-1x4n+ix2-1n+iUk-12ixwhere ri,h, si,h, pi,h, and qi,h are computable constants.They also gave the exact expressions for all constantsri,h, si,h, pi,h, and qi,h with 1≤i≤n+1, if n is a small positive integer. If n is large enough, then they only gave an exact computational method for these constants, but the computation is more complex.For the power sums∑m=1h∫0xTmydyn, they have not given any results in [10]. Wenpeng Zhang and Tingting Wang [3] proved that(4)∑n=1h∫0xT2nydy=T2h+1x22h+1-x2,∑n=1h∫0xTn2ydy=14T2h+1x2h+1+2h-1xand (5)∑m=1∞∫01T2m2n+1ydy=-12+π4n+1∑k=0n2n+1n-k2k+1·cot⁡π4k+2Wenpeng Zhang and Tingting Wang [3] also proposed the following two open problems. Whether there exists a exact calculation formula for(6)∑k=1h∫0xTkydynor∑m=0∞∫01T2m+12n+1ydy.In this paper, as a note of [3, 9], we shall use the analytic and elementary method to give an interesting computational formula for the second sums of (6). That is, we shall prove the following conclusion.Theorem 1.
For any positive integern, one has the identity(7)∑m=0∞∫01T2m+12n+1ydy=π4n+1∑k=0n-1k·2n+1n-k2k+1·cot⁡π4+π4·-1k2k+1.
Takingn=1 and 2, noting that cot⁡π/2=0, cot⁡π/6=3, and cot⁡3π/10=tan⁡π/5, from our theorem we may immediately deduce the following two identities:(8)∑m=0∞∫01T2m+13ydy=-π48·3and(9)∑m=0∞∫01T2m+15ydy=π320·tan⁡π5-5π192·3.


## 2. Several Simple Lemmas

In this section, we shall give several simple lemmas, which are necessary in the proof of our theorem. Hereinafter, we shall use a few basic results, including the properties ofsin⁡x, for which we refer the reader to the introductory books by Chengdong Pan and Chengbiao Pan [11]. First we have the following.Lemma 2.
For any integerk≥0, one has the identity(10)Fk≡∑m=0∞1+-1m2m+14k+1+1-1--1m2m+14k+1-1=π2·14k+1·cot⁡π2k+124k+1.Proof.
For any real numberx, from the infinite product of sin⁡πx, we have(11)sin⁡πx=πx∏n=1∞1-x2n2Taking the logarithm for (11) and then differentiating it for x we have(12)πcot⁡πx=1x-∑n=1∞2xn2-x2=1x+∑n=1∞1n+x-1n-x.Let N be a positive integer. Then from the properties of infinite series we have(13)Fk=limN→+∞⁡∑0≤m≤2N1+-1m2m+14k+1+1-1--1m2m+14k+1-1.On the other hand, we have(14)∑0≤m≤2N1+-1m2m+14k+1+1-1--1m2m+14k+1-1=12k+1+∑m≤N24m+14k+1+1-∑m≤N24m-14k+1-1=12k+1+24k+1∑m≤N14m+1+1/4k+1-∑m≤N14m-1-1/4k+1=12k+1+124k+1∑m≤N1m+2k+1/24k+1-∑m≤N1m-2k+1/24k+1=12k+1+124k+1∑m≤N1m+2k+1/24k+1-1m-2k+1/24k+1.Taking N→+∞ in (14) and then combining (12) and (13) we may immediately deduce the identity(15)Fk=∑m=0∞1+-1m2m+14k+1+1-1--1m2m+14k+1-1=12k+1+124k+1∑m=1∞1m+2k+1/24k+1-1m-2k+1/24k+1=π2·14k+1·cot⁡π2k+124k+1.This proves Lemma 2.Lemma 3.
For any positive integerk, one has the identity(16)∑m=0∞1--1m2m+14k-1+1-1+-1m2m+14k-1-1=-π2·14k-1cot⁡π2·2k-14k-1.Proof.
For any positive integerN, we have the identity(17)∑0≤m≤2N1--1m2m+14k-1+1-1+-1m2m+14k-1-1=-12k-1+24k-1∑m≤N14m-1+1/4k-1-24k-1∑m≤N14m+1-1/4k-1=-12k-1+124k-1∑0≤m≤N14m-2k-1/24k-1-14m+2k-1/24k-1.Taking N→+∞ in (17), from the method of proving Lemma 2, we have(18)∑m=0∞1--1m2m+14k-1+1-1+-1m2m+14k-1-1=-12k-1+124k-1∑m=1∞14m-2k-1/24k-1-14m+2k-1/24k-1=-π2·14k-1cot⁡π2·2k-14k-1.This proves Lemma 3.Lemma 4.
For any positive integern, one has the identity(19)∫0xT2n+1ydy=12T2n+2x2n+2+-1n2n+12nn+1-T2nx2n.Proof.
See Lemma 1 in [3].Lemma 5.
For any nonnegative integern, one has the expressions of xn in the following form:(20)x2n+1=2n+1!4n∑k=0n1n-k!n+k+1!T2k+1xProof.
In fact this is Lemma 4 of [12].


## 3. Proof of the Theorem

Now we shall complete the proof of our theorem. Takingx as Tmx in Lemma 5 and noting that T2k+1Tmx=T2k+1mx (see Lemma 3 of [13]) we have(21)Tm2n+1x=2n+1!4n∑k=0n1n-k!n+k+1!T2k+1mxApplying (1) with x=±1 we have Tn±1=±1n, and then from (21), Lemma 2, Lemma 3, and Lemma 4 we can deduce the identity(22)∑m=0∞∫01T2m+12n+1ydy=2n+1!4n∑k=0n1n-k!n+k+1!∑m=0∞∫01T2k+12m+1ydy=2n+1!2·4n∑k=0n1n-k!n+k+1!∑m=0∞T4km+2m+2k+214km+2m+2k+2-T4km+2m+2k14km+2m+2k+2n+1!2·4n∑k=0n1n-k!n+k+1!∑m=0∞-1m+k4km+2m+2k+2+-1m+k4km+2m+2k=24n+1∑k=0n2n+1!n-k!n+k+1!∑m=0∞1+-1k+m4km+2m+2k+2-1--1m+k4km+2m+2k=24n+1∑k=0n/22n+1!n-2k!n+2k+1!∑m=0∞1+-1m4k+12m+1+1-1--1m4k+12m+1-1+24n+1∑k=0n+1/22n+1!n-2k+1!n+2k!∑m=0∞1+-1m4k-12m+1+1-1--1m4k-12m+1-1=π4n+1∑k=0n/22n+1!n-2k!n+2k+1!4k+1cot⁡π2·2k+14k+1-π4n+1∑k=1n+1/22n+1!n-2k+1!n+2k!4k-1cot⁡π2·2k-14k-1=π4n+1∑k=0n/22n+1n-2k4k+1cot⁡π2·2k+14k+1-π4n+1∑k=1n+1/22n+1n+2k4k-1cot⁡π2·2k-14k-1=π4n+1∑k=0n-1k2n+1n-k2k+1cot⁡π4+π4·-1k2k+1.This completes the proof of our theorem.



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*Source: 7186940-2018-05-31.xml*