October 01, 2025
In this paper, we evaluate the following families of definite integrals in closed form and we show that they are expressible only in terms of the dilogarithm function and the inverse tangent integral, and elementary functions. \[\int_{0}^{1}\frac{\log\big(x^m+1\big)}{x+1}\thinspace{\rm d}x \quad and\quad \int_{0}^{1}\frac{\log\big(x^m+1\big)}{x^2+1}\thinspace{\rm d}x,\] where \(m\) is a positive odd integer. When \(m\) is a positive even integer, these integrals have been evaluated previously by Sofo and Batır, and the case where \(m\) is an odd integer has been left as open problems. The integrals of the first kind arise in Zagier’s work on the Kronecker limit formula. In addition, we demonstrate that a functional equation satisfied by the Herglotz-Zagier-Novikov function is a very specific case of of a more general formula, and give numerous illustrative examples.
Zagier’s article [1], 2? on the Kronecker limit formula has been a source of inspiration for many researchers working in number theory. One of them, Novikov [2] obtained a Kronecker limit formula containing the integral \[\label{e01} \mathscr{F}(x;u,v) =\int_{0}^{1}\frac{\log\big(1-ut^x\big)}{v^{-1}-t}{\rm d}t,\tag{1}\] where \(u\in\mathbb{C}\backslash{(1,\infty)}\), \(u\neq1\) and \(v\in\mathbb{C}\backslash[1,\infty)\), \(v\neq1\). Choie and Kumar [3] named this function the Herglotz-Zagier-Novikov function and discovered many of its properties. As an example, they proved that \[\label{e02} \mathscr{F}(x;u,v) +\mathscr{F}\left(\frac{1}{x};v,u\right)=-\log(1-u)\log(1-v);\tag{2}\] see [3]. The function given in 1 is a generalization of some functions previously studied by Herglotz [4], Radchenko and Zagier [1], and Muzaffar and Williams [5]. These functions are as follows: \[F(x)=\sum_{n=1}^{\infty}\frac{\psi(nx)-\log(nx)}{n},\] where \(\psi(x)=\frac{\Gamma^\prime(x)}{\Gamma(x)}\) is the digamma function, \[J(x)=\int_{0}^{1}\frac{\log\big(1+t^x\big)}{1+t}{\rm d}t\] and \[T(x)=\int_{0}^{1}\frac{\tan^{-1}\big(t^x\big)}{1+t^2}{\rm d}t\] for \(\Re(x)>0\). Very recently Radchenko and Zagier [1] studied \(F(x)\) and \(J(x)\), and they found the following connection between them: \[J(x)=F(2x)-2F(x)+F\left(\frac{x}{2}\right)+\frac{\pi^2}{12x}.\] As special cases of \(\mathscr{F}(x;u,v)\), by 2 \(J(x)\) and \(T(x)\) satisfy the following simple functional relations:
\[\label{e03} J(x)+J\left(\frac{1}{x}\right)=\log^22\tag{3}\] and \[T(x)+T\left(\frac{1}{x}\right)=\frac{\pi^2}{16}.\] In [1] and [3] the authors evaluated \(J(x)\) for many particular values of \(x\). For example, in [1] the authors showed that \[J\big(4+\sqrt{17}\thinspace\big)=-\frac{\pi^2}{6}+\frac{1}{2}\log^22+\log(2)\log\big(4+\sqrt{17}\big)\] and \[J\left(\frac{2}{5}\right)=\frac{11\pi^2}{240}+\frac{3}{4}\log^22-2\log\bigg(\frac{\sqrt{5}+1}{2}\bigg).\] In [6] Sofo and Batır established the following formula \[\begin{align} \label{e06} \int_{0}^{1}\frac{\log\big(1+u^{2b}\big)}{1+u}{\rm d}u&=\frac{1}{2}\sum_{k=0}^{2b-1}\log^2\bigg(2\sin\left(\frac{(2k+1)\pi}{4b}\right)\bigg)+\frac{1-2b^2}{8b}\zeta(2)\notag\\ &+\log^22-\frac{1}{2}\sum_{k=0}^{b-1}\log^2\bigg(2\sin\left(\frac{(2k+1)\pi}{2b}\right)\bigg), \quad b \in \mathbb{N} \end{align}\tag{4}\] and in [7] Batır proved the following integral formula \[\int_{0}^{1}\frac{\log\big(x^3+1\big)}{x+1}{\rm d}x=\frac{1}{2}\operatorname{Li}_2\left(-\frac{1}{3}\right)+\frac{1}{2}\log^23-\frac{1}{6}\zeta(2)\] and in [7] he proposed the following problem, as an interesting area of source to study: Whether there exists an explicit computational formula for the logarithmic integral \[\int_{0}^{1}\frac{\log\big(x^m+1\big)}{x+1}{\rm d}x,\] where \(m\) is an odd integer with \(m\geq 5\). We note that a formula equivalent to 4 has been achieved in [3]. In 2023 Choie and Kumar [3] provided the following general formula for \(J(n)\), which is valid for all \(n\in \mathbb{N}\): \[J(n)=\frac{n}{2}\log^2 2-\frac{\pi^2(n^2-1)}{12n}+\sum_{j=1}^{n}\operatorname{Li}_2\left(\frac{1}{2}\left(1+e^{\frac{\pi i}{n}(2j+1}\right)\right).\] This formula looks very nice, but it does not give explicit results even for small values of positive odd integers \(n\). For example, for \(n=3\) we have \[J(3)=\frac{3}{2}\log^2 2-\frac{2\pi^2}{9}+2\Re\bigg\{\operatorname{Li}_2\left(\frac{3+i\sqrt{5}}{4}\right)\bigg\}\] and it is very hard to evalute the values of the dilogarithm function at complex arguments. Our first aim in this work is to evaluate \(J(m)\) for all odd integers \(m\) in closed form in terms of real arguments of the dilogarithm function.
In [7] Batır has evaluated the following family of integrals for even positive integers \(m\): \[\int_{0}^{1}\frac{\log\big(x^m+1\big)}{x^2+1}{\rm d}x.\] He also showed that \[\label{e08461} \int_{0}^{1}\frac{\log\big(x^3+1\big)}{x^2+1}{\rm d}x=\frac{\pi}{8}\log 2-\frac{5}{3}\operatorname{G}+\frac{\pi}{3}\log\big(2+\sqrt{3}\big)\tag{5}\] and left the evaluation of these integrals for odd integers \(m\) with \(m\geq 5\) as an open problem. Our second aim is to provide a solution to this problem and to evaluate these integrals in closed form for all odd integers \(m\geq 3\).
Our third and last aim is to provide a generalization of the Herglotz-Zagier-Novikov function given by 1 and to offer some applications. In this work, an empty sum is conventionally interpreted as zero wherever it appears.
We need the following three lemmas.
Lemma 1. Let \(m\) be an odd integer with \(m\geq3\). Then we have \[\begin{align} \log\bigg[\bigg(\frac{1-u}{1+u}\bigg)^m+1\bigg]=&\frac{m+1}{2}\log 2-m\log(1+u)+\sum_{k=0}^{\frac{m-3}{2}}\log(1+\varphi_k)\\ +&\sum_{k=0}^{\frac{m-3}{2}}\log\bigg[u^2+\frac{1-\varphi_k}{1+\varphi_k}\bigg], \end{align}\] where \(\varphi_k=\cos\left(\frac{(2k+1)\pi}{m}\right)\).
Proof. By [7] we have \[\begin{align} z^m+1&=(z+1)\prod_{k=0}^{\frac{m-3}{2}}\bigg(z-e^{\frac{ i(2k+1)\pi}{m}}\bigg)\bigg(z-e^{-\frac{ i(2k+1)\pi}{m}}\bigg)\\ &=(z+1)\prod_{k=0}^{\frac{m-3}{2}}\big[z^2-2z\varphi_k+1\big]. \end{align}\] Setting \(z=\frac{1-u}{1+u}\), and then taking the logarithm of both sides, we, after an easy computation, get the desired result. ◻
Lemma 2. Let \(q\) be any nonnegative real number. Then we have \[\begin{align} \int_{0}^{1}\frac{\log\big(u^2+q\big)}{1+u}{\rm d}u=\log 2\log(1+q)-\arctan^2\left(\frac{1}{\sqrt{q}}\right)+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{q+1}\right). \end{align}\]
Proof. By the substitution \(u=t\sqrt{q}\), we have \[\int_{0}^{1}\frac{\log\big(u^2+q\big)}{u+1}{\rm d}u=\sqrt{q}\log q\int_{0}^{1/\sqrt{q}}\frac{{\rm d}t}{1+\sqrt{q}t}+\sqrt{q}\int_{0}^{1/\sqrt{q}}\frac{\log\big(t^2+1\big)}{1+t\sqrt{q}}{\rm d}t.\] Evaluating the first integral and performing integration by parts on the second yields \[\int_{0}^{1}\frac{\log\big(u^2+q\big)}{u+1}{\rm d}u=\log 2\log(q+1)-2\int_{0}^{1/\sqrt{q}}\frac{t\log\big(1+t\sqrt{q}\big)}{1+t^2}{\rm d}t.\]
Setting \(t=\frac{v}{\sqrt{q}}\), we obtain \[\int_{0}^{1}\frac{\log\big(u^2+q\big)}{u+1}{\rm d}u=\log 2\log(q+1)-\frac{2}{q}\int_{0}^{1}\frac{v\log(1+v)}{1+v^2/q}cv.\] The following result appears in [8]: \[\int_{0}^{1}\frac{x\log(1+x)}{1+ax^2}{\rm d}x=\frac{1}{2a}\arctan^2\big(\sqrt{a}\big)-\frac{1}{4a}\operatorname{Li}_2\left(\frac{a}{a+1}\right)\, (a>0).\] From this result with \(a=\frac{1}{q}\) it follows that \[\int_{0}^{1}\frac{\log\big(u^2+q\big)}{1+u}{\rm d}u=\log 2\log(1+q)-\arctan^2\left(\frac{1}{\sqrt{q}}\right)+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{q+1}\right).\] ◻
Lemma 3. Let \(q\) be a positive real number. Then we have \[\begin{align} \label{e10} \int_{0}^{1}\frac{\log \big(x^2+q\big)}{1+x^2}{\rm d}x=\frac{\pi}{2}\log\big(1+\sqrt{q}\big)+\operatorname{Ti}_2\left(\frac{\sqrt{q}-1}{\sqrt{q}+1}\right)-\operatorname{G}, \end{align}\qquad{(1)}\] where \(\operatorname{G}\) is the Catalan’s constant defined by \[\operatorname{G}=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}=0.915965...,\] and \(\operatorname{Ti}_2\) is the inverse tangent integral defined by \[\label{e10461} \operatorname{Ti}_2(x)=\int_{0}^{x}\frac{\tan^{-1}u}{u}{\rm d}u.\qquad{(2)}\]
Proof. We define \[g(q)=\int_{0}^{1}\frac{\log\big(x^2+q\big)}{x^2+1}{\rm d}x.\] Differentiation with respect to \(q\) gives \[\begin{align} g^\prime(q)&=\int_{0}^{1}\frac{{\rm d}x}{\big(x^2+q\big)\big(x^2+1\big)}\notag\\ &=\frac{1}{q-1}\int_{0}^1\frac{{\rm d}x}{x^2+1}-\frac{1}{q-1}\int_{0}^1\frac{{\rm d}x}{x^2+q}\notag\\ &=\frac{\pi}{4}\frac{1}{q-1}-\frac{1}{q-1}\int_{0}^1\frac{{\rm d}x}{x^2+q}. \end{align}\] It is very easy to see \[\begin{align} \int_{0}^1\frac{{\rm d}x}{x^2+q}=\frac{\arctan\left(\frac{1}{\sqrt{q}}\right)}{\sqrt{q}}. \end{align}\] Thus, we have \[\begin{align} g^\prime(q)&=\frac{\pi}{4}\frac{1}{q-1}-\frac{\arctan\left(\frac{1}{\sqrt{q}}\right)}{(q-1)\sqrt{q}}. \end{align}\] Integration gives \[\begin{align} g(q)&=\frac{\pi}{4}\log|q-1|-\int\frac{\arctan\left(\frac{1}{\sqrt{q}}\right)}{(q-1)\sqrt{q}}+C. \end{align}\] Making the change of variable \(u=\frac{1}{\sqrt{q}}\) we see that \[\begin{align} g(q)&=\frac{\pi}{4}\log|q-1|+2\int\frac{\arctan u}{1-u^2}{\rm d}u+C\\ &=\frac{\pi}{4}\log|q-1|+2\int_{0}^{u}\frac{\arctan y}{1-y^2}{\rm d}y+C\\ &=\frac{\pi}{4}\log|q-1|-\int_{0}^{u}\left(-\frac{1}{1-y}-\frac{1}{1+y}\right)\arctan y\thinspace{\rm d}y+C. \end{align}\] Applying integration by parts we obtain \[\begin{align} g(q)&=\frac{\pi}{4}\log|q-1|-\log\frac{|1-u|}{1+u}\arctan u+\int_{0}^{u}\frac{1}{y^2+1}\log\bigg|\frac{1-y}{1+y}\bigg|{\rm d}y+C. \end{align}\] Letting \(q\to 1\) yields \[\begin{align} \label{e11} \int_{0}^{1}\frac{\log\big(x^2+1\big)}{x^2+1}{\rm d}x=&\lim\limits_{q\to 1}\bigg[\frac{\pi}{4}\log|q-1|-\log\frac{|1-u|}{1+u}\arctan u\bigg]\notag\\ &+\int_{0}^{1}\frac{1}{y^2+1}\log\bigg|\frac{1-y}{1+y}\bigg|{\rm d}y+C. \end{align}\tag{6}\] By [7] (see also [9]) we have \[\begin{align} \label{e12} \int_{0}^{1}\frac{\log\big(x^2+1\big)}{x^2+1}{\rm d}x=\frac{\pi}{2}\log 2-\operatorname{G}. \end{align}\tag{7}\] We can also easily evaluate that \[\begin{align} \label{e13} \lim\limits_{q\to 1}\bigg[\frac{\pi}{4}\log|q-1|-\log\frac{|1-u|}{1+u}\arctan u\bigg]=\frac{\pi}{4}\log 2 \end{align}\tag{8}\] and \[\begin{align} \label{e14} \int_{0}^{1}\frac{1}{y^2+1}\log\bigg|\frac{1-y}{1+y}\bigg|{\rm d}y=-\operatorname{G}. \end{align}\tag{9}\] Combining 6 , 7 , 8 and 9 we conclude that \(C=0\). Thus we have
\[\begin{align} \label{e15} g(q)&=\frac{\pi}{4}\log|q-1|-\log\frac{|1-u|}{1+u}\arctan u+\int_{0}^{u}\frac{1}{1+y^2}\log\bigg|\frac{1-y}{1+y}\bigg|{\rm d}y. \end{align}\tag{10}\] We recall the following two integral formulas from [8]: \[\begin{align} \int_{0}^{a}\frac{\log (1-y)}{1+y^2}{\rm d}y=&\arctan a\log(1-a)+\frac{\pi}{4}\tanh^{-1}a -\frac{1}{2}\operatorname{G}-\frac{1}{2}\operatorname{Ti}_2(a)\\ &+\frac{1}{2}\operatorname{Ti}_2\left(\frac{1-a}{1+a}\right)+\frac{1}{4}\operatorname{Ti}_2\left(\frac{2a}{1-a^2}\right)\quad |a|<1 \end{align}\] and \[\begin{align} \int_{0}^{a}\frac{\log (1+y)}{1+y^2}{\rm d}y=&\arctan a\log(1+a)-\frac{\pi}{4}\tanh^{-1}a +\frac{1}{2}\operatorname{G}-\frac{1}{2}\operatorname{Ti}_2(a)\\ &-\frac{1}{2}\operatorname{Ti}_2\left(\frac{1-a}{1+a}\right)+\frac{1}{4}\operatorname{Ti}_2\left(\frac{2a}{1-a^2}\right)\quad |a|<1 \end{align}\] Substituting the values of these two integrals in 10 , after replacing \(a\) by \(1/\sqrt{q}\), we find that \[\begin{align} \int_{0}^{1}\frac{\log \big(x^2+q\big)}{1+x^2}{\rm d}x=\frac{\pi}{2}\log\big(1+\sqrt{q}\big)+\operatorname{Ti}_2\left(\frac{\sqrt{q}-1}{\sqrt{q}+1}\right)-\operatorname{G}. \end{align}\] ◻
Remark 1. We would very much like to express the integral in ?? in terms of some well-known special functions. However, since it is not known at present whether the inverse tangent integral \(\operatorname{Ti}_2\) can be represented in terms of such functions, we are compelled to leave it in this form. Although \(\operatorname{Ti}_2\) is not well known as the others, it has been studied by Ramanujan in [10] and many of its properties has been discovered. It is worth noting that Entry 4.295.22 in [11] provides a related generalisation, from which the following integral formula can be obtained: \[\int_{0}^{\infty}\frac{\log\big(x^2+q\big)}{x^2+1}{\rm d}x=\pi\log\big(1+\sqrt{q}\big).\]
We begin this section by presenting a solution to the problem proposed in [7].
Theorem 4. Let \(m\) be any positive odd integer. Then \[\begin{align} \label{e1} J(m)=\frac{m}{2}\log^22-\frac{(m^2-1)\pi^2}{24m}+\frac{1}{2}\sum_{k=0}^{\frac{m-3}{2}}\operatorname{Li}_2\bigg(\cos^2\left(\frac{(2k+1)\pi}{2m}\right)\bigg). \end{align}\qquad{(3)}\] Also, in view of 3 , we have \[\begin{align} J\left(\frac{1}{m}\right)=\frac{2-m}{2}\log^22+\frac{(m^2-1)\pi^2}{24m}-\frac{1}{2}\sum_{k=0}^{\frac{m-3}{2}}\operatorname{Li}_2\bigg(\cos^2\left(\frac{(2k+1)\pi}{2m}\right)\bigg), \end{align}\] where \(\varphi_k=\cos\left(\frac{(2k+1)\pi}{m}\right)\), and \(\operatorname{Li}_2\) is the dilogarithm function defined by \[\operatorname{Li}_2(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^2}\quad (|z|\leq 1).\]
Proof. We make the substitution \(x=\frac{1-u}{1+u}\) and \({\rm d}x=\frac{-2}{(1+u)^2}{\rm d}u\). It follows that \[\begin{align} J(m)=\int_{0}^{1}\frac{1}{u+1}\log\left[\left(\frac{1-u}{1+u}\right)^m+1\right]{\rm d}u. \end{align}\] By Lemma 1 we get \[\begin{align} J(m)=&\frac{1}{2}\log^22+\log2\sum_{k=0}^{\frac{m-3}{2}}\log(1+\varphi_k)+\sum_{k=0}^{\frac{m-3}{2}}\int_{0}^{1}\frac{1}{u+1}\log\bigg[u^2+\frac{1-\varphi_k}{1+\varphi_k}\bigg]{\rm d}u. \end{align}\] Applying Lemma 2 it follows that \[\begin{align} J(m)=&\frac{m}{2}\log^22+\log2\sum_{k=0}^{\frac{m-3}{2}}\log(1+\varphi_k)+\sum_{k=0}^{\frac{m-3}{2}}\bigg[\log2\log\bigg(\frac{2}{1+\varphi_k}\bigg)\\ &-\arctan^2\bigg(\sqrt{\frac{1+\varphi_k}{1-\varphi_k}}\bigg)+\frac{1}{2}\operatorname{Li}_2\bigg(\frac{1+\varphi_k}{2}\bigg)\bigg]. \end{align}\] Simplifying this identity, we get \[\begin{align} \label{rfjhnoqe} J(m)=\frac{m}{2}\log^22-\sum_{k=0}^{\frac{m-3}{2}}\arctan^2\bigg(\sqrt{\frac{1+\varphi_k}{1-\varphi_k}}\thinspace\bigg)+\frac{1}{2}\sum_{k=0}^{\frac{m-3}{2}}\operatorname{Li}_2\bigg(\frac{1+\varphi_k}{2}\bigg). \end{align}\tag{11}\] Since \(1+\cos x=2\cos^2\frac{x}{2}\) and as can be easily shown that \[\arctan\left(\sqrt{\frac{1+\varphi_k}{1-\varphi_k}}\thinspace\right)=\frac{\pi}{4}+\frac{1}{2}\arcsin(\varphi_k)\] and \[\arcsin(\varphi_k)=\frac{\pi}{2}-\frac{(2k+1)\pi}{m},\] we arrive at \[\arctan\left(\sqrt{\frac{1+\varphi_k}{1-\varphi_k}}\thinspace\right)=\frac{(2k+1)\pi}{2m}.\] Replacing this in 2 , we get the result. ◻
Theorem 5. Let \(m\) be any positive odd integer. Then \[\begin{align} &\int_{0}^{1}\frac{\log\big(x^m+1\big)}{x^2+1}{\rm d}x=\frac{m\pi}{8}\log2-\frac{(m-1)\operatorname{G}}{2}\notag\\ &+\frac{\pi}{4}\sum_{k=0}^{\frac{m-3}{2}}\log\left(1+\sin\frac{\pi(2k+1)}{m}\right) +\sum_{k=0}^{\frac{m-3}{2}}\operatorname{Ti}_2\bigg(\frac{1}{\varphi_k}\left(\sqrt{1-\varphi_k^2}-1\right)\bigg), \end{align}\] where \(\operatorname{Ti}_2\) is the inverse tangent integral defined in ?? .
Proof. The proof is based on the same arguments used in the proof of Theorem 4. We define \[I(m):=\int_{0}^{1}\frac{\log\big(x^m+1\big)}{x^2+1}{\rm d}x.\] Making the change of variable \(x=\frac{1-u}{1+u}\), so that, \({\rm d}x=-\frac{2}{(1+u)^2}{\rm d}u\), we get \[\begin{align} I(m)=\int_{0}^{1}\log\bigg(1+\left(\frac{1-u}{1+u}\right)^m\bigg)\frac{{\rm d}u}{u^2+1}. \end{align}\] Applying Lemma 1 we obtain
\[\begin{align} I(m)=&\int_{0}^{1}\frac{1}{u^2+1}\bigg[\frac{m+1}{2}\log 2+\sum_{k=0}^{\frac{m-3}{2}}\log(1+\varphi_k)-m\log(1+u)\\ &+\sum_{k=0}^{\frac{m-3}{2}}\log\left(u^2+\frac{1-\varphi_k}{1+\varphi_k}\right)\bigg]{\rm d}u \end{align}\] or
\[\begin{align} I(m)=&\frac{(m+1)\pi}{8}\log 2+\frac{\pi}{4}\sum_{k=0}^{\frac{m-3}{2}}\log((1+\varphi_k)-m\int_{0}^{1}\frac{\log(1+u)}{1+u^2}{\rm d}u\\ &+\sum_{k=0}^{\frac{m-3}{2}}\int_{0}^{1}\frac{\log\big(u^2+q\big)}{u^2+1}{\rm d}u, \end{align}\] where \(q=\frac{1-\varphi_k}{1+\varphi_k}\). Since \[\int_0^1 \frac{\log (u+1)}{u^2+1} \, du=\frac{ \pi}{8}\log 2\] we get \[\begin{align} I(m)=\frac{\pi}{8}\log 2+\frac{\pi}{4}\sum_{k=0}^{\frac{m-3}{2}}\log((1+\varphi_k)+\sum_{k=0}^{\frac{m-3}{2}}\int_{0}^{1}\frac{\log\big(u^2+q\big)}{u^2+1}{\rm d}u, \end{align}\] Applying Lemma 3 with \(q=\frac{1-\varphi_k}{1+\varphi_k}\), we get \[\begin{align} I(m)=&\frac{\pi}{8}\log2-\frac{(m-1)\operatorname{G}}{2}+\frac{\pi}{4}\sum_{k=0}^{\frac{m-3}{2}}\log(1+\varphi_k)\\ &+\frac{\pi}{2}\sum_{k=0}^{\frac{m-3}{2}}\log\bigg(1+\sqrt{\frac{1-\varphi_k}{1+\varphi_k}}\bigg)+\sum_{k=0}^{\frac{m-3}{2}}\operatorname{Ti}_2\bigg(\frac{\sqrt{1-\varphi_k^2}-1}{\varphi_k}\bigg). \end{align}\] Simplifying this expression we arrive at the desired result stated in Theorem 5. ◻
Theorem 6. Let the parameters \(a\), \(b\), \(\alpha\), \(\beta\), and the functions \(\phi\) and \(\varphi\) are such that the following integrals are convergent. \[\mathcal{F}(x;\alpha,\beta)=\alpha\int_{0}^{1}\phi^\prime(\alpha t)\phi(\beta t^x){\rm d}t\] and \[\mathcal{G}(x;a,b)=a\int_{0}^{\infty}\varphi^\prime(at)\varphi(bt^x){\rm d}t.\] Then we have \[\label{e461463} \mathcal{F}(x;\alpha,\beta)+\mathcal{F}(1/x;\beta, \alpha)=\phi(\alpha)\phi(\beta)-\phi^2(0)\qquad{(4)}\] and \[\label{e461464} \mathcal{G}(x;a,b)+\mathcal{G}(1/x;b, a)=\lim\limits_{t\to\infty}\varphi(at)\varphi(bt)-\varphi^2(0).\qquad{(5)}\]
Proof. By partial integration we have
\[\begin{align} \mathcal{F}(x;\alpha,\beta)=\phi(\alpha t)\phi(\beta t^x)\big|_{t=0}^{t=1}-\beta\int_{0}^{1}\phi(\alpha t)xt^{x-1}\phi^\prime(\beta t^x){\rm d}t. \end{align}\] Making the change of variable \(u=t^x\) it follows that \[\begin{align} \mathcal{F}(x;\alpha,\beta)&=\phi(\alpha)\phi(\beta)-\phi^2(0)-\beta\int_{0}^{1}\phi^\prime(\beta u)\phi(\alpha u^{1/x}){\rm d}u\\ &=\phi(\alpha)\phi(\beta)-\phi^2(0)-\mathcal{F}(1/x;\beta, \alpha), \end{align}\] which proves ?? . The proof of ?? can be achieved similarly. ◻
Example 1. Letting \(m=3\) in Theorem 5 we obtain \[\label{e3461} \int_{0}^{1}\frac{\log\big(x^3+1\big)}{x^2+1}{\rm d}x=-\frac{\pi}{8}\log2-G+\frac{\pi}{2}\log\big(1+\sqrt{3}\big) -\operatorname{Ti}_2\big(2-\sqrt{3}\big).\tag{12}\]
Remark 2. Combining the formulas 5 and 12 we get \[\begin{align} \label{e3462} \operatorname{Ti}_2\big(2-\sqrt{3}\big)&=\frac{2}{3}\operatorname{G}-\frac{\pi}{12}\log \big(2+\sqrt{3}\big). \end{align}\tag{13}\] The inverse tangent integral satisfies \(\operatorname{Ti}_2(x)-\operatorname{Ti}_2\big(\frac{1}{x}\big)=\frac{\pi}{2}\log x\). Setting \(x=2-\sqrt{3}\) here gives \[\begin{align} \label{e3463} \operatorname{Ti}_2\big(2+\sqrt{3}\big)=\frac{2}{3}\operatorname{G}-\frac{\pi}{12} \log \left(362-209 \sqrt{3}\right). \end{align}\tag{14}\] Identity 13 appears in [12] and 14 recovers the formula given in [13].
Example 2. Letting \(m=5\) in Theorem 5 we get: \[\begin{align} &\int_{0}^{1}\frac{\log\big(x^5+1\big)}{x^2+1}{\rm d}x=-\frac{3 \pi}{8} \log 2-2\operatorname{G}+\operatorname{Ti}_2\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\thinspace\right)\\ &+\operatorname{Ti}_2\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)+\frac{\pi}{2} \log \left(1+\sqrt{5}+\sqrt{10+2 \sqrt{5}}\thinspace\right). \end{align}\] Here we used \[\cos \frac{\pi}{5}=\frac{1+\sqrt{5}}{4} \quad and \quad \cos \frac{3\pi}{5}=\frac{1-\sqrt{5}}{4}.\]
Example 3. Setting \(m=3\) in ?? we get \[\begin{align} \int_{0}^{1}\frac{\log\big(x^3+1\big)}{x+1}{\rm d}x=\frac{1}{2}\operatorname{Li}_2\left(\frac{3}{4}\right)+\frac{3}{2}\log^22-\frac{\pi^2}{9}. \end{align}\]
Example 4. Setting \(m=5\) in ?? we get \[\begin{align} &\int_{0}^{1}\frac{\log\big(x^5+1\big)}{x+1}{\rm d}x=\frac{5}{2}\log ^22-\frac{\pi^2}{5}+\frac{1}{2}\operatorname{Li}_2\left(\frac{5+\sqrt{5}}{8}\right)+\frac{1}{2}\operatorname{Li}_2\left(\frac{5-\sqrt{5}}{8}\right). \end{align}\]
Example 5. Example 1 together with [7] (with m=3) lead to the following series evaluation:
\[\begin{align} \sum_{n=1}^{\infty}\frac{2^n\sum_{k=1}^{n}\frac{(3/2)^k}{k}}{(2n+1)\binom{2n}{n}}=\frac{\pi}{2}\log 2+\frac{10}{3}\operatorname{G}-\frac{2 \pi }{3} \log \left(2+\sqrt{3}\right). \end{align}\]
Example 6. Letting \(\alpha=-v\), \(\beta=-u\) and \(\phi(t)=\log(1+t)\) in ?? , we have \[\int_{0}^{1}\frac{\log(1-ut^x)}{v^{-1}-t}{\rm d}t+ \int_{0}^{1}\frac{\log(1-ut^{1/x})}{v^{-1}-t}{\rm d}t=-\log(1-u)\log(1-v).\]
Remark 3. This gives a new, very short and elementary proof of [3].
Example 7. For \(\alpha=\beta=1\) and \(\phi(t)=\log(1+t)\), we get from in ?? \[\int_{0}^{1}\frac{\log(1+t^x)}{1+t}{\rm d}t+\int_{0}^{1}\frac{\log(1+t^{1/x})}{1+t}{\rm d}t=\log^22.\]
Example 8. For \(\alpha=\beta=1\) and \(\phi(t)=\tan^{-1}t\), we get from in ?? \[\int_{0}^{1}\frac{\tan^{-1}\big(t^x\big)}{1+t^2}{\rm d}t+ \int_{0}^{1}\frac{\tan^{-1}\big(t^{1/x}\big)}{1+t^2}{\rm d}t=\frac{\pi^2}{16}.\]
Example 9. Setting \(\alpha=\beta=1\) and \(\phi(t)=\sin^{-1}t\) in ?? , we have \[\int_{0}^{1}\frac{\sin^{-1}\big(t^x\big)}{\sqrt{1-t^2}}{\rm d}t+ \int_{0}^{1}\frac{\sin^{-1}\big(t^{1/x}\big)}{\sqrt{1-t^2}}{\rm d}t=\frac{\pi^2}{4}.\]
Example 10. Setting in ?? , we have \(\alpha=\beta=1\) and \(\phi(t)=\sinh^{-1}t\) in ?? , we have \[\int_{0}^{1}\frac{\sinh^{-1}\big(t^x\big)}{\sqrt{1+t^2}}{\rm d}t+ \int_{0}^{1}\frac{\sinh^{-1}\big(t^{1/x}\big)}{\sqrt{1+t^2}}{\rm d}t=-\log^2\big(1+\sqrt{2}\big).\]
Example 11. For \(\alpha=\beta=1\) and \(\phi(t)=\operatorname{Li}_2(t)\), we get from ?? \[\int_{0}^{1}\frac{\log(1-t)\operatorname{Li}_2\big(t^x\big)}{t}{\rm d}t+ \int_{0}^{1}\frac{\log(1-t)\operatorname{Li}_2\big(t^{1/x}\big)}{t}{\rm d}t=-\frac{\pi^4}{36}.\]
Example 12. Setting \(a=b=1\) and \(\varphi(t)=\tan^{-1}(t)\) in ?? , we get \[\int_{0}^{\infty}\frac{\tan^{-1}(t^x)}{t^2+1}{\rm d}t+\int_{0}^{\infty}\frac{\tan^{-1}(t^{1/x})}{t^2+1}{\rm d}t=\frac{\pi^2}{4}.\]
Acknowledgements
We would like to thank the editor and the reviewer for their very careful reading and numerous helpful suggestions.
Author contributions Both authors have contributed equally in all aspects of the preparation of this submission.
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