Is $\int_0^{\pi/2}\Re\left\{\ln^a(1\pm e^{2ix})\right\}dx=0$ a new result?

Question

I managed to prove

$$\int_0^{\pi/2}\Re\left\{\ln^a(1\pm e^{2ix})\right\}dx=0,\qquad a=1,2,3,....$$

Is it a new result? and can be proved in a different way?

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Proof: We start with finding the derivative of the binomial coefficient: \begin{gather*} \frac{\partial}{\partial m}\binom{m+n-1}{n}=\frac{\partial}{\partial m}\frac{\Gamma(m+n)}{\Gamma(n+1)\Gamma(m)}\\ \left\{\text{use $\frac{d}{dx}\Gamma(x)=\Gamma(x)\psi(x)$}\right\}\\ =\frac{\Gamma(m)\Gamma(m+n)\psi(m+n)-\Gamma(m+n)\Gamma(m)\psi(m)}{\Gamma(n+1)\Gamma^2(m)}\\ =\frac{\Gamma(m+n)}{\Gamma(n+1)\Gamma(m)}\left(\psi(m+n)-\psi(m)\right)\\ =\binom{m+n-1}{n}\left(\psi(m+n)-\psi(m)\right). \end{gather*} By using $\displaystyle\psi(x)=H_{x-1}-\gamma=\sum_{k=1}^{x-1}\frac1k-\gamma$, we have \begin{gather} \psi(m+n)-\psi(m)=\sum_{k=1}^{m+n-1}\frac1k-\sum_{k=1}^{m-1}\frac1k\nonumber\\ \left\{\text{use $\sum_{k=1}^{a+b} f_k=\sum_{k=1}^{a-1}f_k+\sum_{k=a}^{a+b} f_k$ for the first sum}\right\}\nonumber\\ =\sum_{k=1}^{m-1}\frac1k+\sum_{k=m}^{m+n-1}\frac1k-\sum_{k=1}^{m-1}\frac1k\nonumber\\ =\sum_{k=m}^{m+n-1}\frac1k=\sum_{k=0}^{n-1}\frac1{k+m}.\label{psi(m+n)-psi(m)} \end{gather} Plugging this result back in, we get \begin{equation} \frac{\partial}{\partial m}\binom{m+n-1}{n}=\sum_{k=0}^{n-1}\frac{\binom{m+n-1}{n}}{k+m}.\label{rec12} \end{equation} Keep differentiating w.r.t $m$: \begin{equation*} \frac{\partial^2}{\partial m^2}\binom{m+n-1}{n}=-\sum_{k=0}^{n-1} \frac{\binom{m+n-1}{n}}{(k+m)^2}+\sum_{k=0}^{n-1} \frac{\frac{\partial}{\partial m}\binom{m+n-1}{n}}{k+m}, \end{equation*} \begin{gather*} \frac{\partial^3}{\partial m^3}\binom{m+n-1}{n}=2\sum_{k=0}^{n-1} \frac{\binom{m+n-1}{n}}{(k+m)^3}-2\sum_{k=0}^{n-1} \frac{\frac{\partial}{\partial m}\binom{m+n-1}{n}}{(k+m)^2}\\ +\sum_{k=0}^{n-1} \frac{\frac{\partial^2}{\partial m^2}\binom{m+n-1}{n}}{k+m}, \end{gather*} \begin{gather*} \frac{\partial^4}{\partial m^4}\binom{m+n-1}{n}=-6\sum_{k=0}^{n-1} \frac{\binom{m+n-1}{n}}{(k+m)^4}+6\sum_{k=0}^{n-1} \frac{\frac{\partial}{\partial m}\binom{m+n-1}{n}}{(k+m)^3}\\ -3\sum_{k=0}^{n-1} \frac{\frac{\partial^2}{\partial m^2}\binom{m+n-1}{n}}{(k+m)^2}+\sum_{k=0}^{n-1} \frac{\frac{\partial^3}{\partial m^3}\binom{m+n-1}{n}}{k+m}, \end{gather*} which can be generalized to \begin{equation*} \frac{\partial^a}{\partial m^a}\binom{m+n-1}{n}=-(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}\sum_{k=0}^{n-1} \frac{\frac{\partial^j}{\partial m^j}\binom{m+n-1}{n}}{(k+m)^{a-j}}. \end{equation*} Let $m\to1$ and define $\displaystyle\lim_{m\to1}\frac{\partial^a}{\partial m^a}\binom{m+n-1}{n}=f(a,n)$ observing that \begin{gather*} \lim_{m\to 1}\sum_{k=0}^{n-1} \frac{1}{(k+m)^{a-j}}=\sum_{k=0}^{n-1} \frac{1}{(k+1)^{a-j}}=\sum_{k=1}^{n} \frac{1}{k^{a-j}}=H_n^{(a-j)}; \\ f(0,n)=\lim_{m\to 1}\binom{m+n-1}{n}=\binom{n}{n}=1, \end{gather*} we get \begin{equation} f(a,n)=-(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}H_n^{(a-j)}f(j,n),\quad f(0,n)=1.\tag{1} \end{equation} On the other hand, we have \begin{equation*} \frac{1}{(1-x)^m}=(1-x)^{-m}=\sum_{n=0}^\infty \binom{m+n-1}{n}x^n. \end{equation*} Take the $a$-th derivative of both sides w.r.t $m$, \begin{equation*} (-1)^a\frac{\ln^a(1-x)}{(1-x)^m}=\sum_{n=0}^\infty \frac{\partial^a}{\partial m^a}\binom{m+n-1}{n} x^n. \end{equation*} Let $m\to 1$ and remember that $\displaystyle\lim_{m\to1}\frac{\partial^a}{\partial m^a}\binom{m+n-1}{n}=f(a,n)$, \begin{equation} (-1)^a\frac{\ln^a(1-x)}{1-x}=\sum_{n=0}^\infty f(a,n) x^n,\tag{2} \end{equation}

Integrate both sides of (2),

\begin{equation} \frac{(-1)^{a+1}}{a+1}\ln^{a+1}(1-x)=\sum_{n=0}^\infty f(a,n)\frac{x^{n+1}}{n+1}=\sum_{n=1}^\infty f(a,n-1)\frac{x^{n}}{n}\tag{3} \end{equation}

Replace $x=-e^{2ix}$ in (3) and consider the real parts then integrate from $x=0\to \pi/2$

\begin{equation} \frac{(-1)^{a+1}}{a+1}\int_0^{\pi/2}\Re\left\{\ln^{a+1}(1+e^{2ix})\right\}dx=\sum_{n=1}^\infty \frac{f(a,n-1)(-1)^n}{n}\int_0^{\pi/2}\Re\left\{e^{2inx}\right\}dx\\ =\sum_{n=1}^\infty \frac{f(a,n-1)(-1)^n}{n}\int_0^{\pi/2}\cos(2nx)dx\\ =\sum_{n=1}^\infty \frac{f(a,n-1)(-1)^n}{n}.\frac{\sin(n\pi)}{2n}=0. \end{equation}

and if we replace $x=e^{2ix}$, we get

\begin{equation} \int_0^{\pi/2}\Re\left\{\ln^{a+1}(1-e^{2ix})\right\}dx=0. \end{equation}

Thus $$\int_0^{\pi/2}\Re\left\{\ln^a(1\pm e^{2ix})\right\}dx=0.$$

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Examples of (2):

\begin{gather} -\frac{\ln(1-x)}{1-x}=\sum_{n=1}^\infty H_n x^n;\\ \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n};\\ -\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n. \end{gather}

Answer 1

The function $f(z) = \left(\ln(1+z)\right)^a$ is holomorphic in the unit disk. Here we choose the principal branch of the logarithm, i.e. $$ \ln(1+z) = \ln |1+z| + i \arg(1+z) \, , \, -\pi < \arg(1+z) < \pi \, . $$ It follows that its real part $$ u(z) = \operatorname{Re} \left(\left(\ln(1+z)\right)^a\right) $$ is harmonic in the unit disk. Using the mean-value property we get for $0 < r < 1$ $$ 0 = u(0) = \frac{1}{2\pi}\int_0^{2\pi} u(re^{it}) \, dt = \frac{1}{2\pi}\int_0^{\pi} \left(u(re^{it} + u(re^{-it})\right) \, dt \\ = \frac{1}{\pi}\int_0^{\pi} u(re^{it}) \, dt\, . $$ The last equality follows from the fact that the Taylor series of $f$ has only real coefficients, so that $f(\bar z) = \overline{f(z)}$ and consequently $u(\bar z) = u(z)$.

It remains to show that we can take the limit $r\to 1$, and that is true because of the dominated convergence theorem, since $$ |u(re^{it})| \le \left| \ln(1+re^{it})\right|^a \le \left( \ln |1+re^{it}| + \pi \right)^a = \left( \ln |\cos \frac t2| + \ln 2 + \pi \right)^a $$ is an integrable majorant.

Answer 2

We can also use the power series for the logarithm, namely: $$\ln\left(1\pm e^{2ix}\right)=\sum_{n=1}^\infty \frac{(\pm1)^n}{n}e^{2in x}$$ Multiplying this $a$ times gives: $$\ln^a\left(1\pm e^{2ix}\right)=\sum_{n_1=1}^\infty\frac{(\pm1)^{n_1}}{n_1}\sum_{n_2=1}^\infty\frac{(\pm1)^{n_2}}{n_2}... \sum_{n_a=1}^\infty \frac{(\pm1)^{n_a}}{n_a} e^{2ix(n_1+n_2+..+n_a)} $$ And integrating over $\left(0,\frac{\pi}{2}\right)$ will always give $0$ since: $$\Re \int_0^\frac{\pi}{2}e^{2ikx}dx=\int_0^\frac{\pi}{2}\cos(2kx)dx=0,\ k\in N$$

Answer 3

Utilize the Gauss Mean Value Theorem to extablish

$$ \frac1{2\pi}\int_0^{2\pi} \ln^n(1\pm e^{i x})dx =\ln^n(1)=0$$

and note that \begin{align} &\int_{0}^{2\pi} \ln^n \left ( 1\pm e^{i x} \right )\overset{x\to 2x}{dx } = 2\int_{0}^{\pi} \ln^n \left ( 1\pm e^{i 2x} \right )dx\\ = &\ 2\int_{0}^{\pi/2} \ln^n \left ( 1\pm e^{i 2x} \right )dx + 2\int^{\pi}_{\pi/2} \ln^n \left ( 1\pm e^{i 2x} \right ) \overset{x\to\pi-x}{dx} \\ = &\ 2\int_{0}^{\pi/2} \ln^n \left ( 1\pm e^{i 2x} \right ) + \ln^n \left ( 1\pm e^{-i 2x} \right ) \ {dx} \\ =&\ 4\int_0^{\pi/2}\Re\ln^n(1\pm e^{i2x})dx \end{align}