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$$ I=\int_0^{\pi/2} \theta^2 \log ^4(2\cos \theta) d\theta =\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3). $$ Note $\zeta(3)$ is given by $$ \zeta(3)=\sum_{n=1}^\infty \frac{1}{n^3}. $$ I have a previous post related to this except the logarithm power is squared and not to the 4th power. If you are interested in seeing this result go here: Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$.. However, I am wondering how to calculate the result shown above. Thanks.

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3 Answers 3

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From Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik equation $3.631\ (9)$ we have

$$ \int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)} $$

Proof

Integrating $(1+z)^p z^q$, for $p,q\ge0$, in the $z=u+iv$ plane around the contour bounded by the $u$-axis from $-1$ to $1$ and the upper semicircle of unit radius yields $$ \int_{-1}^1(1+z)^p z^q\ dz=-i\int_0^\pi\left(1+e^{it}\right)^p e^{i(q+1)t}\ dt, $$ since $(1+z)^p z^q$ is holomorphic within and continuous on and within the given contour. The imaginary part of the RHS is $$ \Im\left[-i\int_0^\pi\left(1+e^{it}\right)^p e^{i(q+1)t}\ dt\right]=-\int_0^\pi\left(2\cos\frac t2\right)^p \cos bt\ dt, $$ where $b=q+\frac12p+1$. The LHS integral is equal to $$ \int_{0}^1(1+u)^p u^q\ du+e^{i\pi q}\int_{0}^1(1-u)^p u^q\ du $$ of which the imaginary part is \begin{align} \operatorname{B}\left(p+1,q+1\right)\sin\pi q&=-\frac{\Gamma\left(p+1\right)\Gamma\left(q+1\right)}{\Gamma\left(p+q+2\right)}\sin\pi q\\ &=-\frac{\Gamma\left(p+1\right)\Gamma\left(b-\frac12p\right)}{\Gamma\left(b+\frac12p+1\right)}\sin\pi \left(b-\frac12p\right)\\ &=-\frac{\pi\Gamma\left(p+1\right)}{\Gamma\left(1+\frac12p+b\right)\Gamma\left(1+\frac12p-b\right)}. \end{align} Final step is setting $t=2x$, $p=n-1$, and $a=2b$. $\qquad\color{blue}{\mathbb{Q.E.D.}}$


Thus \begin{align} \int_0^{\Large\frac\pi2}\theta^2\ln^4(2\cos \theta)\ d\theta&=\lim_{n\to5}\lim_{a\to0}\frac{\partial^6}{\partial n^4\partial a^2}\left[\int_0^{\Large\frac\pi2}(2\cos \theta)^{n-1}\cos a\theta\ d\theta\right]\\ &=\lim_{n\to5}\lim_{a\to0}\frac{\partial^6}{\partial n^4\partial a^2}\left[\frac{\pi}{2 n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}\right]\\ &=\large\color{blue}{\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3)}. \end{align}

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  • $\begingroup$ Differentiating the identity is indeed quite a nice way of evaluating the integral. (+1) $\endgroup$ Commented Sep 18, 2014 at 13:24
  • $\begingroup$ @SuperAbound But it's tedious. Thanks, +1 too. $\endgroup$
    – Tunk-Fey
    Commented Sep 18, 2014 at 13:31
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Note that $$\ln(1+e^{i2x})=\ln(2\cos{x})+ix$$ Raise both sides to the fourth power and extract the real part. This yields \begin{align} \Re\ln^4(1+e^{i2x})=\ln^4(2\cos{x})-6x^2\ln^2(2\cos{x})+x^4 \end{align} We first have to calculate $\displaystyle\int^{\pi/2}_0x^4\ln^2(2\cos{x}) \ {\rm d}x$. Squaring the above identity and extracting the real part gives $$\Re\ln^2(1+e^{i2x})=\ln^2(2\cos{x})-x^2$$ Then \begin{align} \int^{\pi/2}_0x^4\ln^2(2\cos{x}) \ {\rm d}x &=\Re\int^{\pi/2}_0x^4\ln^2(1+e^{i2x}) \ {\rm d}x+\frac{\pi^7}{896}\\ &=\frac{\pi^7}{896}+\frac{1}{32}\Im\int^{1}_0\frac{\ln^4{(-z)}\ln^2(1-z)}{z}{\rm d}z\\ &=\frac{\pi^7}{896}+\frac{\pi}{8}\int^1_0\frac{\ln^3{z}\ln^2(1-z)}{z}{\rm d}z-\frac{\pi^3}{8}\int^1_0\frac{\ln{z}\ln^2(1-z)}{z}{\rm d}z\\ &=\frac{\pi^7}{896}-\frac{\pi}{16}\sum^\infty_{n=1}H_n\int^1_0z^n\ln^4{z} \ {\rm d}z+\frac{\pi^3}{8}\sum^\infty_{n=1}H_n\int^1_0z^n\ln^2{z} \ {\rm d}z\\ &=\frac{\pi^7}{896}-\frac{3\pi}{2}\sum^\infty_{n=1}\frac{H_n}{(n+1)^5}+\frac{\pi^3}{4}\sum^\infty_{n=1}\frac{H_n}{(n+1)^3}\\ &=\frac{5\pi^7}{8064}+\frac{3\pi}{4}\zeta^2(3) \end{align} where I have used the result $$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)$$ which can be easily derived by integrating $\displaystyle f(z)=\frac{(\gamma+\psi(-z))^2}{z^q}$ along an infinitely large square.


Moving on to the actual integral, \begin{align} \int^{\pi/2}_0x^2\ln^4(2\cos{x}) \ {\rm d}x &=\frac{\pi^7}{384}+\frac{9\pi}{2}\zeta^2(3)+\Re\int^{\pi/2}_0x^2\ln^4(1+e^{i2x}) \ {\rm d}x\\ &=\frac{\pi^7}{384}+\frac{9\pi}{2}\zeta^2(3)-\frac{1}{8}\Im\int^{1}_0\frac{\ln^2(-z)\ln^4(1-z)}{z}\ {\rm d}z\\ &=\frac{\pi^7}{384}+\frac{9\pi}{2}\zeta^2(3)-\frac{\pi}{4}\int^{1}_0\frac{\ln^4{z}\ln(1-z)}{1-z}\ {\rm d}z\\ &=\frac{\pi^7}{384}+\frac{9\pi}{2}\zeta^2(3)+\frac{\pi}{4}\sum^\infty_{n=1}H_n\int^1_0z^n\ln^4{z} \ {\rm d}z\\ &=\frac{\pi^7}{384}+\frac{9\pi}{2}\zeta^2(3)+6\pi\sum^\infty_{n=1}\frac{H_n}{(n+1)^5}\\ &=\frac{\pi^7}{384}+\frac{9\pi}{2}\zeta^2(3)+\frac{\pi^7}{210}-3\pi\zeta^2(3)\\ &=\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3) \end{align} as was to be shown.


Just for fun, here is a derivation of $$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)$$ Consider $\displaystyle f(z)=\frac{(\gamma+\psi(-z))^2}{z^q}$. At the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{z^q(z-n)^2}+\frac{2H_n}{z^q(z-n)}\right]\\ &=2\sum^\infty_{n=1}\frac{H_n}{n^q}-q\sum^\infty_{n=1}\frac{1}{n^{q+1}}\\ &=2\sum^\infty_{n=1}\frac{H_n}{n^q}-q\zeta(q+1)\\ \end{align} At zero, \begin{align} {\rm Res}(f,0) &=[z^{q-1}]\left(\frac{1}{z}-\sum^\infty_{k=1}\zeta(k+1)z^k\right)^2\\ &=[z^{q-1}]\left(-2\sum^\infty_{k=1}\zeta(k+1)z^{k-1}+\sum^\infty_{k=1}\sum^k_{j=1}\zeta(j+1)\zeta(k-j+2)z^{k+1}\right)\\ &=-2\zeta(q+1)+\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j) \end{align} where I have removed the first (irrelevant) term and used the Cauchy product in the second equality. Since the sum of the residues is zero, we conclude $$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)$$

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  • $\begingroup$ unique and well explained. $\endgroup$ Commented May 11, 2019 at 21:51
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\pi/2}\theta^{2}\ln^{4}\pars{2\cos\pars{\theta}}\,\dd\theta ={33\pi^{7} \over 4480} + {3\pi \over 2}\,\zeta^{2}\pars{3}:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\int_{0}^{\pi/2}% \theta^{2}\ln^{4}\pars{2\cos\pars{\theta}}\,\dd\theta} =\int_{0}^{\pi/2}\theta^{2}\ln^{4}\pars{2\root{1 + \cos\pars{2\theta} \over 2}} \,\dd\theta \\[3mm]&={1 \over 8}\int_{0}^{\pi} \theta^{2}\ln^{4}\pars{2^{1/2}\root{1 + \cos\pars{\theta}}}\,\dd\theta ={1 \over 16}\int_{-\pi}^{\pi} \theta^{2}\ln^{4}\pars{2^{1/2}\root{1 + \cos\pars{\theta}}}\,\dd\theta \\[3mm]&={1 \over 16} \oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} \bracks{-\ic\ln\pars{z}}^{2}\ln^{4}\pars{2^{1/2}\root{1 + {z^{2} + 1 \over 2z}}} \,{\dd z \over \ic z} \\[3mm]&={\ic \over 16} \oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} \ln^{2}\pars{z}\ln^{4}\pars{z + 1 \over z^{1/2}}\,{\dd z \over z} \\[3mm]&={\ic \over 16}\lim_{\mu\ \to\ 0 \atop \nu\ \to\ 0}\partiald[2]{}{\mu}\partiald[4]{}{\nu} \oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} z^{\mu}\pars{z + 1 \over z^{1/2}}^{\nu}\,{\dd z \over z} \end{align}

$$ \color{#c00000}{\int_{0}^{\pi/2}% \theta^{2}\ln^{4}\pars{2\cos\pars{\theta}}\,\dd\theta} ={\ic \over 16}\lim_{\mu\ \to\ 0 \atop \nu\ \to\ 0} \partiald[2]{}{\mu}\partiald[4]{}{\nu}\color{#00f}{% \oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} z^{\mu - \nu/2 - 1}\pars{z + 1}^{\nu}\,\dd z} $$

\begin{align}&\color{#00f}{% \oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} z^{\mu - \nu/2 - 1}\pars{z + 1}^{\nu}\,\dd z} \\[3mm]&=-\int_{-1}^{0}\pars{-x}^{\mu - \nu/2 - 1} \exp\pars{\bracks{\mu - {\nu \over 2} - 1}\pi\ic}\pars{x + 1}^{\nu}\,\dd x \\[3mm]&\phantom{=\!}-\int_{0}^{-1}\pars{-x}^{\mu - \nu/2 - 1} \exp\pars{-\bracks{\mu - {\nu \over 2} - 1}\pi\ic}\pars{x + 1}^{\nu}\,\dd x \\[1cm]&=\exp\pars{\bracks{\mu - {\nu \over 2}}\pi\ic} \int_{0}^{1}x^{\mu - \nu/2 - 1}\pars{1 - x}^{\nu}\,\dd x \\[3mm]&-\exp\pars{-\bracks{\mu - {\nu \over 2}}\pi\ic} \int_{0}^{1}x^{\mu - \nu/2 - 1}\pars{1 - x}^{\nu}\,\dd x \\[1cm]&=2\ic\sin\pars{\bracks{\mu - {\nu \over 2}}\pi} \int_{0}^{1}x^{\mu - \nu/2 - 1}\pars{1 - x}^{\nu}\,\dd x \\[3mm]&=2\ic\sin\pars{\bracks{\mu - {\nu \over 2}}\pi}\, {\Gamma\pars{\mu - \nu/2}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu/2 + 1}} =2\pi\ic\, {\Gamma\pars{\nu + 1} \over \Gamma\pars{1 - \mu + \nu/2}\Gamma\pars{1 + \mu + \nu/2 }} \end{align}

$$ \color{#c00000}{\int_{0}^{\pi/2}% \theta^{2}\ln^{4}\pars{2\cos\pars{\theta}}\,\dd\theta} =-\,{\pi \over 8}\lim_{\mu\ \to\ 0 \atop \nu\ \to\ 0} \partiald[2]{}{\mu}\partiald[4]{}{\nu}{\nu \choose \mu + \nu/2} $$ The derivatives and the limits are a formidable task which we can evaluate with a CAS: $$\color{#66f}{\large% \int_{0}^{\pi/2}\theta^{2}\ln^{4}\pars{2\cos\pars{\theta}}\,\dd\theta ={33\pi^{7} \over 4480} + {3\pi \over 2}\,\zeta^{2}\pars{3}} \approx {\tt 29.0568} $$

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