Problem. Evaluate the following integral. \begin{equation}\label{eqn:wts} \int_{0}^{1} \log (1-x) \log x \log (1+x) \; dx \end{equation}

We divide the solution into several steps.

1. Reduction to Euler series.

The key ingredient for the reduction is the following integral.

\begin{equation*} \int_{0}^{\frac{\pi}{2}} \sin^{p} \theta \cos^{q} \theta \log \sin\theta \log \cos\theta \; d\theta. \end{equation*} To find a closed form of it, we consider the beta function identity. \begin{equation*} 2 \int_{0}^{\frac{\pi}{2}} \sin^{2z-1} \theta \cos^{2w-1} \theta \; d\theta = \beta(z, w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)} \end{equation*} Differentiating with respect to $z$, we obtain \begin{equation*} 4 \int_{0}^{\frac{\pi}{2}} \sin^{2z-1} \theta \cos^{2w-1} \theta \log \sin\theta \; d\theta = \beta(z, w) \left\{ \psi_0 (z) - \psi_0 (z+w) \right\}, \end{equation*} where $\psi_k$ denotes the $k$-th polygamma function. Differentiating with respect to $w$, we finally obtain \begin{align} & 8 \int_{0}^{\frac{\pi}{2}} \sin^{2z-1} \theta \cos^{2w-1} \theta \log \sin\theta \log\cos\theta \; d\theta \nonumber \\ & \qquad = \beta(z, w) \left[ \left\{ \psi_0 (z) - \psi_0 (z+w) \right\}\left\{ \psi_0 (w) - \psi_0 (z+w) \right\} - \psi_1 (z+w) \right], \label{eqn:key_int} \end{align} the very closed form for \eqref{eqn:key_int}.

With this formula we want to reduce \eqref{eqn:wts} to a formula involving a certain type of infinite summations, namely the alternating Euler series. Our key formula \eqref{eqn:key_int} yields \begin{eqnarray} & & \int_{0}^{1} \log (1-x) \log x \log (1+x) \; dx \\ & = & \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^n \log (1-x) \log x \; dx \\ & = & \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{\frac{\pi}{2}} 8 \sin^{2n+1} \theta \cos \theta \log \sin\theta \log \cos\theta \; dx \qquad (x = \sin^2 \theta) \\ & = & \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \beta(n+1, 1) \left[ \{ \psi_0 (n+1) - \psi_0 (n+2) \} \{ \psi_0 (1) - \psi_0 (n+2) \} - \psi_1 (n+1) \right] \\ & = & \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)} \left[ \frac{H_{n+1}}{n+1} - \zeta(2) + H_{n+1}^{(2)} \right], \end{eqnarray} where \begin{equation*} H_n = \sum_{k=1}^{n} \frac{1}{k} \quad \text{and} \quad H_n^{(s)} = \sum_{k=1}^{n} \frac{1}{k^s} \end{equation*} are the $n$-th harmonic number and the $n$-th generalized harmonic number of order $s$, respectively, and we have exploited the following identities involving polygamma functions. \begin{equation*} \psi_0 (n+1) = -\gamma + H_n , \qquad \psi_1 (n+1) = \zeta(2) - H_n^{(2)}. \end{equation*}

Now consider the following family of alternating series \begin{equation*} A_{m}^{(s)} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} H_n^{(s)} }{n^m}, \end{equation*} namely the alternating Euler series. By some algebra, we obtain \begin{align*} \frac{(-1)^{n-1}}{n(n+1)} \left[ \frac{H_{n+1}}{n+1} - \zeta(2) + H_{n+1}^{(2)} \right] & = (-1)^{n-1} \Bigg[ \frac{H_n}{n} - \frac{H_{n+1}}{n+1} - \frac{H_{n+1}}{(n+1)^2} + \frac{H_n^{(2)}}{n} - \frac{H_{n+1}^{(2)}}{n+1} \\ & \qquad \qquad + \left( \frac{1}{n} - \frac{1}{n+1} \right) (2 - \zeta(2)) - \frac{1}{(n+1)^2} \Bigg], \end{align*} from which the following reduction formula follows. \begin{align} &\int_{0}^{1} \log (1-x) \log x \log (1+x) \; dx \nonumber \\ &\qquad = 2A_1^{(1)} + 2A_1^{(2)} + A_2^{(1)} + \zeta(2) \left( \frac{3}{2} - 2\log 2 \right) + 4 \log 2 - 6 \label{eqn:reduce} \end{align} Therefore it suffices to evaluate the following 3 series \begin{equation*} A_1^{(1)}, \quad A_2^{(1)}, \quad \text{and} \quad A_1^{(2)} \end{equation*} in the identity \eqref{eqn:reduce}.

2. Evaluation of $A_m^{(s)}$.

For the evaluation, we make the following observation.

Observation. Suppose the power series $\sum a_n x^n$ converges with the radius of convergence $\geq 1$. If we write $s_n = a_0 + \cdots + a_n$ for the partial sum, then for all $|x| < 1$ we have \begin{equation*} \frac{1}{1-x} \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} s_n x^n. \end{equation*}

Consider $a_n = \frac{1}{n}$ with $a_0 = 0$. Then \begin{equation*} -\frac{\log(1-x)}{1-x} = \sum_{n=1}^{\infty} H_n x^n. \end{equation*} Integrating this identity, \begin{align*} \sum_{n=1}^{\infty} \frac{H_n}{n} x^n & = - \int_{0}^{x} \frac{\log(1-t)}{t(1-t)} \; dt = - \int_{0}^{x} \left( \frac{\log(1-t)}{t} + \frac{\log(1-t)}{1-t} \right) \; dt \\ & = \left[ \mathrm{Li}_2 (t) + \frac{1}{2} \log^2 (1-t) \right]_{0}^{x} = \mathrm{Li}_2 (x) + \frac{1}{2} \log^2 (1-x). \end{align*} Finally, plugging $x = -1$ yields \begin{equation} A_{1}^{(1)} = \frac{1}{2}\zeta(2) - \frac{1}{2}\log^2 2. \label{eqn:iden_1} \end{equation}

Let's review on the idea in the calculation of \eqref{eqn:iden_1}. We first identified the series as a value of a power series evaluated at some point, and then derived the general formula for the power series. This small triumph seems to suggest that our strategy would also work for other two summations. However, \begin{align} \sum_{n=1}^{\infty} \frac{H_n}{n^2} x^n & = \int_{0}^{x} \frac{1}{t} \sum_{n=1}^{\infty} \frac{H_n}{n} t^n \; dt = \int_{0}^{x} \frac{1}{t} \left( \mathrm{Li}_2 (t) + \frac{1}{2} \log^2 (1-t) \right) \; dt \nonumber \\ & = \mathrm{Li}_3 (x) + \frac{1}{2} \int_{0}^{x} \frac{\log^2 (1-t)}{t} \; dt \label{eqn:iden_2} \end{align} and also \begin{align} \sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n} x^n & = \int_{0}^{x} \frac{1}{t} \sum_{n=1}^{\infty} H_n^{(2)} t^n \; dt = \int_{0}^{x} \frac{1}{t} \left( \frac{\mathrm{Li}_2 (t)}{1-t} \right) \; dt \nonumber \\ & = \int_{0}^{x} \left( \frac{\mathrm{Li}_2 (t)}{t} + \frac{\mathrm{Li}_2 (t)}{1-t} \right) \; dt = \mathrm{Li}_3 (x) + \int_{0}^{x} \frac{\mathrm{Li}_2 (t)}{1-t} \; dt \nonumber \\ & = \mathrm{Li}_3 (x) + \left[ - \mathrm{Li}_2 (t) \log (1 - t) \right]_{0}^{x} - \int_{0}^{x} \frac{\log^2 (1 - t)}{t} \; dt \nonumber \\ & = \mathrm{Li}_3 (x) - \mathrm{Li}_2 (x) \log (1 - x) - \int_{0}^{x} \frac{\log^2 (1 - t)}{t} \; dt \label{eqn:iden_3} \end{align} shows that this idea is thwarted by the same factor \begin{equation}\label{eqn:int_in_q} I(x) = \frac{1}{2} \int_{0}^{x} \frac{\log^2 (1-t)}{t} \; dt. \end{equation} Therefore it is essential for our calculation to calculate the value of \eqref{eqn:int_in_q} at $x = -1$.

In the calculation \eqref{eqn:iden_3}, we actually derived the following identity \begin{equation*} \int_{0}^{x} \frac{\mathrm{Li}_2 (t)}{1-t} \; dt \ = \ - \mathrm{Li}_2 (x) \log (1 - x) - 2I(x). \end{equation*} Now we are going to evaluate the left-hand side of this identity in another way. Using Euler's reflection formula for the dilogarithm, \begin{align*} \int_{0}^{x} \frac{\mathrm{Li}_2 (t)}{1-t} \; dt & = \int_{0}^{x} \frac{1}{1-t} \left[ \zeta(2) - \log t \log (1-t) - \mathrm{Li}_2 (1-t) \right] \; dt \\ & = \int_{0}^{x} \frac{1}{1-t} \left[ \zeta(2) - \mathrm{Li}_2 (1-t) \right] \; dt - \int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt \\ & = \int_{1-x}^{1} \frac{\zeta(2) - \mathrm{Li}_2 (u)}{u} \; du - \int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt \qquad (u = 1-t) \\ & = - \zeta(2) \log (1-x) - \zeta(3) + \mathrm{Li}_3 (1-x) - \int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt \\ & = - \zeta(2) \log (1-x) - \zeta(3) + \mathrm{Li}_3 (1-x) + \left[ \frac{1}{2} \log^2 (1-t) \log t \right]_{0}^{x} - I(x) \\ & = - \zeta(2) \log (1-x) - \zeta(3) + \mathrm{Li}_3 (1-x) + \frac{1}{2} \log^2 (1-x) \log x - I(x). \end{align*} From this calculation, we obtain a closed for for \eqref{eqn:int_in_q} as follows: \begin{equation}\label{eqn:int_in_q2} I(x) \ = \ \zeta(3) - \mathrm{Li}_3 (1-x) + [\zeta(2) - \mathrm{Li}_2 (x)] \log (1 - x) - \frac{1}{2} \log^2 (1-x) \log x. \end{equation} This identity, however, does not immediately yield the value for \eqref{eqn:iden_2} and \eqref{eqn:iden_3}. We have to evaluate the value of $I(x)$ at $x = -1$, but there are some terms in \eqref{eqn:int_in_q2} which are undefined at $x = -1$. This means that, in order to utilize this formula, we have to analyze the behavior of each terms of \eqref{eqn:int_in_q2} more carefully.

3. Principal Value of the Polylogarithm.

At the core of the problem lies the branch cut, namely $(-\infty, 0)$, of $\mathrm{Li}_3 (1-x)$ and $\log x$. To circumvent this problem, we consider $I(x)$ as a function on a region in the complex plane. Since the branch cut of the integrand is $[1, \infty)$ and $I(1)$ is defined, \eqref{eqn:int_in_q} defines a holomorphic function on $\mathbb{C}-(1,\infty)$. By analytic continuation, the identity \eqref{eqn:int_in_q2} remains valid whenever both sides are defined. Therefore, instead of plugging $x = -1$ to \eqref{eqn:int_in_q2} directly, we can evaluate $I(-1)$ by taking limit to \eqref{eqn:int_in_q2} along a certain path.

Here, we choose the path as $x = -1 - ih$ as $h \to 0^+$. From Euler's reflection formula again, \begin{align*} \mathrm{Li}_2 (2+0^+ i) & = \mathrm{Li}_2 (1 - (-1 - 0^+ i)) \\ & = \zeta(2) - \log (-1 - 0^+ i) \log (1 - (-1 - 0^+ i)) - \mathrm{Li}_2 (-1 - 0^+ i) \\ & = \zeta(2) - (-i \pi) \log 2 - \text{Li}_2 (-1) \\ & = \frac{3}{2} \zeta(2) + i \pi \log 2 . \end{align*} Also, plugging $x = -1-0^+ i$ and $x = \frac{1}{2}$ respectively to the following trilogarithm identity \begin{equation*} \mathrm{Li}_3 (x) + \mathrm{Li}_3 (1-x) + \mathrm{Li}_3 \big( 1 - \tfrac{1}{x} \big) = \zeta(3) + \zeta(2) \log x + \tfrac{1}{6}\log^3 x - \tfrac{1}{2}\log^2 x \log(1-x), \end{equation*} we obtain \begin{equation*} \text{Li}_3 (2+0^+ i) \ = \ \frac{7}{8}\zeta(3) + \frac{3}{2}\zeta(2) + \frac{i\pi}{2} \log^2 2. \end{equation*} Therefore, plugging $x = -1-0^+ i$ to \eqref{eqn:int_in_q2} and simplifying, we obtain \begin{equation*} I(-1) = \frac{1}{8} \zeta(3) \end{equation*} Therefore, from \eqref{eqn:iden_2} and \eqref{eqn:iden_3}, we obtain \begin{equation*} A_2^{(1)} \ = \ \frac{5}{8} \zeta(3) , \qquad A_1^{(2)} \ = \ \zeta(3) - \frac{1}{2}\zeta(2)\log 2. \end{equation*}

4. Conclusion.

Summarizing all our efforts, we obtain \begin{align*} \int_{0}^{1} \log (1-x) \log x \log (1+x) \; dx & = \frac{21}{8} \zeta(3) + \frac{1}{2} \zeta(2) (5 - 6 \log 2) + 4 \log 2 - \log^2 2 - 6. \end{align*}

References

  1. Integrals in "AoPS Forum"
  2. An Interesting Integral in "AoPS Forum"
  3. Philippe Flajolet and Bruno Salvy, "Euler Sums and Contour Integral Representations".
Posted by aficionado

댓글을 달아 주세요

  1. jackal_anu 2010.05.20 22:59 신고  댓글주소  수정/삭제  댓글쓰기

    궁금한건 저런 적분 문제는 도데체 어디서 구하는 걸까요 _-;;

  2. ㅁㄴㅇㄹ 2010.05.21 06:01  댓글주소  수정/삭제  댓글쓰기

    전 이래서 수학과를 존경합니다..

  3. 프리뱅 2010.05.21 12:49  댓글주소  수정/삭제  댓글쓰기

    아저씨, 편지 잘 갔냐고요. (...)

  4. 지나가던 수학과생 2010.05.24 00:20  댓글주소  수정/삭제  댓글쓰기

    오홍 ㅎㅎ 전수학과 학부생1학년이에요 물론 관악산(?)속에있는 수리과학부는아니구용 ㅋㅋ;

    전 포항 남구 산31번지에 속해있는대학이에용 ㅇㅇ..

    여기 블로그?? 뭐라부르죠?? 티스토리?? 명칭은 뭔지 모르겟고 앞으로 자주 들락날락해봐야겟네용 ㅎㅎ 나름 재밌는계산이나 토픽등이 매력적이군요 ㅎㅎ

  5. 지나가며 늅늅하는 인문계 고1 2011.05.29 00:31  댓글주소  수정/삭제  댓글쓰기

    부산촌놈이라 관악산 네이버두드리니 서울대
    포항 남구 산31번지 두드리니 포항공대
    대체 어떻게 하셨길래...ㅠ
    전 적어도 이 글을 하루만에 이해할 수 있지만요

    • aficionado 2011.05.29 03:56 신고  댓글주소  수정/삭제

      감사합니다. 댓글 올려주신 덕분에 이 글에 대한 기억이 떠올라서, 시험삼아 검산해봤는데, 맨 마지막에 -6을 빼먹었다는 걸 깨닫고 얼른 고쳤네요 -ㅅ-;; 근데 그럼 그 전까지 사람들은 아무도 글을 안 읽어봤다는 건가 OTL

    • 지나가며 늅늅하는 인문계 고1 2011.05.29 17:41  댓글주소  수정/삭제

      그리고 중간중간에 수식이 짤려있네요
      ("그것도 아무도 못봤단 건가"가 아니라 스프링노트에 제대로 되어있네요 문제는 댓글0)
      그것보단 이해는 할 수 있겠는데 중간중간에
      단순계산생략이 많아서 조금 읽기 힘드네요..ㅠ
      교대급수 압박

    • aficionado 2011.05.29 19:23 신고  댓글주소  수정/삭제

      수식이 너무 길어서 짤립니다... 스프링노트에서는 잘 보이는 것 같은데 말이죠 =ㅁ=;;