18
$\begingroup$

$$ I=\int_0^{\pi/2} \theta^2 \log ^4(2\cos \theta) d\theta =\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3). $$ 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.

$\endgroup$
10
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Differentiating the identity is indeed quite a nice way of evaluating the integral. (+1) $\endgroup$ – SuperAbound Sep 18 '14 at 13:24
  • $\begingroup$ @SuperAbound But it's tedious. Thanks, +1 too. $\endgroup$ – Tunk-Fey Sep 18 '14 at 13:31
8
$\begingroup$

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 $$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)$$

$\endgroup$
  • $\begingroup$ unique and well explained. $\endgroup$ – Ali Shather May 11 at 21:51
5
$\begingroup$

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

\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: $$\color{#66f}{\large% \int_{0}^{\pi/2}\theta^{2}\ln^{4}\pars{2\cos\pars{\theta}}\,\dd\theta ={33\pi^{7} \over 4480} + {3\pi \over 2}\,\zeta^{2}\pars{3}} \approx {\tt 29.0568} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.