How to tackle $\int_{0}^{\frac{\pi}{2}}y \ln (1+\cos y)\,d y$?

Question

I recently encountered an integral, $$ I:=\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y. $$ I tried to tackle $I$ using the double angle formula and the result of $$ \int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y $$

\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} y \ln \left(2 \cos ^{2} \frac{y}{2}\right) d y \\ &=\ln 2 \int_{0}^{\frac{\pi}{2}} y d y+2 \int_{0}^{\frac{\pi}{2}} y \ln \left(\cos \frac{y}{2}\right) d y \\ &=\frac{\pi^{2}}{8} \ln 2+8 \int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y \end{aligned}

By my post,$$\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y = \frac{\pi G}{8}-\frac{\pi^{2}}{32} \ln 2-\frac{21}{128} \zeta(3) $$

Now we can conclude that $$ \boxed{\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y = \pi G-\frac{21}{16} \zeta(3)-\frac{\pi^{2}}{8} \ln 2} $$

Suggestions for improvement and alternative methods are warmly welcome!

Answer 1

Another solution, obtained by one integration by parts is $$\int y \log (1+\cos (y))\,dy=$$ $$\frac{1}{2} y^2 \log (1+\cos (y))+2 i y \text{Li}_2\left(-e^{i y}\right)-2 \text{Li}_3\left(-e^{i y}\right)+\frac{i y^3}{6}-y^2 \log \left(1+e^{i y}\right)$$ giving for the definte integral $$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi C-2 \text{Li}_3(-i)-\frac{1}{4} \pi ^2 \log (1+i)-\frac{3 }{2} \zeta (3)$$ $$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi C-\frac{21 }{16}\zeta (3)-\frac{1}{8} \pi ^2 \log (2)$$

Just for the fun

Using the $1,400$ years old approximation $$\cos(y) \simeq\frac{\pi ^2-4y^2}{\pi ^2+y^2}\qquad \text{for} \qquad -\frac \pi 2 \leq y\leq\frac \pi 2$$ we should have $$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy\sim \pi ^2 \log \left(\frac{4}{5^{5/6}}\right) $$ (relative error equal to $0.07$%).

Answer 2

Using the series expansion of $ \log (\cos (y))$ an alternative method could be $$\int y \log (\cos (y))\,dy=\int\sum_{n=1}^\infty (-1)^n \,\frac{2^{2(n-1)}}{(2 n)!}\big[E_{2 n-1}(1)-E_{2 n-1}(0)\big]\, y^{2n+1}\,dy$$ where appear Euler polynomials.

So, $$\int_0^{\frac \pi 4} y \log (\cos (y))\,dy=\sum_{n=1}^\infty (-1)^n \,\frac{ \pi ^{2( n+1)}}{2^{2 n+7}\,(n+1)\, \Gamma (2 n+1)}\big[E_{2 n-1}(1)-E_{2 n-1}(0)\big]$$

Edit

We could do something similar using

$$\log (1+\cos (y))=\log(2)+\sum_{n=1}^\infty (-1)^n \,\Bigg[\frac{E_{2 n}(-1)-E_{2 n}(1)}{2 (2 n)!}+\frac{E_{2 n-1}(-1)+E_{2 n-1}(1)}{4 n (2 n-1)!}\Big]y^{2n}$$

Answer 3

$$I=\int_{0}^{\frac{\pi}{2}}x \ln (1+\cos x)\,d x$$

Applying $\color{red}{\cos(x)\to x}$,

$$I=-\int_{0}^{1} \frac{\arccos(x)\ln (1+x)}{\sqrt{1-x^2}}\,d x$$

Using inverse trig relation,

$$\color{red}{\arcsin(x)+\arccos(x)=\frac{\pi}{2}\implies -\arccos(x)=\arcsin(x)-\frac{\pi}{2}}$$

Using the following expansion,

$$\color{red}{\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n \choose n}}}$$

$$I=\int_{0}^{1} \frac{\arcsin(x)\ln (1+x)}{\sqrt{1-x^2}}\,d x-\frac{\pi}2\int_{0}^{1} \frac{\ln (1+x)}{\sqrt{1-x^2}}\,d x$$

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$$\int_{0}^{1} \frac{\ln (1+x)}{\sqrt{1-x^2}}\,d x=\int_0^\frac{\pi}{2} \ln(1 + \sin x) \ dx = \int_0^\frac{\pi}{2} \ln(\sec x + \tan x) \,dx+ \int_0^\frac{\pi}{2}\ln \cos x \ dx \\= -\frac{\pi \ln 2}{2} + \left[x \ln(\sec x + \tan x)\right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} x\sec x \ dx \\= - \frac{\pi \ln 2}{2} + \int_0^{\frac{\pi}{2}} \frac{x}{ \sin x} - \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \csc x \ dx \\= 2G - \frac{\pi \ln 2}{2} $$

$$\frac{\pi}{2}\int_{0}^{1} \frac{\ln (1+x)}{\sqrt{1-x^2}}\,d x=\pi G - \frac{\pi^2 \ln 2}{4} $$

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$$\int_{0}^{1} \frac{\arcsin(x)\ln (1+x)}{\sqrt{1-x^2}}\,d x=\frac12\sum_{n=1}^\infty\frac{(2)^{2n}}{n{2n \choose n}}\int_{0}^{1} x^{2n-1}\ln(1+x)\,d x$$

Using the below integral,

$$\int_0^1x^{2n-1}\ln(1+x)dx=\frac{H_{2n}-H_n}{2n}$$

$$\int_{0}^{1} \frac{\arcsin(x)\ln (1+x)}{\sqrt{1-x^2}}\,d x=\frac14\sum_{n=1}^\infty\frac{4^{n}H_{2n}}{n^2{2n \choose n}}-\frac14\sum_{n=1}^\infty\frac{4^{n}H_{n}}{n^2{2n \choose n}}=\frac{21}{16}\zeta(3)-\frac34\ln(2)\zeta(2)$$

Where,

$$\frac14\sum_{n=1}^\infty\frac{4^{n}H_{2n}}{n^2{2n \choose n}}=\frac{3}{4}\ln(2)\zeta(2)+\frac{35}{16}\zeta(3)$$

Partial proof is below

$$\frac14\sum_{n=1}^\infty\frac{4^{n}H_{n}}{n^2{2n \choose n}}=\frac32\ln(2)\zeta(2)+\frac{14}{16}\zeta(3)$$

Proof to above sum

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$$\therefore I=\pi G-\frac{21}{16}\zeta(3)-\frac{\pi^2 }{8}\ln 2$$

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Partial proof to $\sum_{n=1}^\infty \frac{4^{n}H_{2n}}{n^2{2n \choose n}}$ is below,

$$ \frac{\arcsin(x)}{\sqrt{1 - x^2}} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{4^n}{n\binom{2n}{n}} {x^{2n-1}} $$

$$ \sum_{n=1}^\infty \frac{4^{n-1}H_{2n}}{n^2{2n \choose n}} = \int_0^1 \frac{\arcsin (x) \ln(1 - x)}{\sqrt{1 - x^2}}\, dx $$

Make the change $x \to \sin x$ and applying Fourier series expansion of $\ln(\sin x)$ and then standard Zeta function definition, we can finish the whole proof.

Answer 4

Use the following Fourier series, shown in this answer $$\ln(1+\cos x)=-\ln 2+2\sum_{k=1}^\infty\frac{(-1)^{k-1}\cos(kx)}{k}$$ Substitute the sum for the expression and integrate term-by-term $$I=\int_0^{\pi/2}x\ln(1+\cos x)dx$$$$=-\frac{\pi^2}{8}\ln2+2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\int_0^{\pi/2}x\cos(kx)dx$$ $$=-\frac{\pi^2}{8}\ln2+2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\frac{kx\sin(kx)+\cos(kx)}{k^2}\Bigg|_{x=0}^{x=\pi/2}$$ $$=-\frac{\pi^2}{8}\ln2+2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\frac{k\pi/2\sin(\pi k/2)+\cos(\pi k/2)-1}{k^2}$$ $$=-\frac{\pi^2}{8}\ln2+\pi\sum_{k=1}^\infty\frac{(-1)^{k-1}\sin(\pi k/2)}{k^2}+2\sum_{k=1}^\infty\frac{(-1)^{k-1}\cos(\pi k/2)}{k^3}-2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^3}$$ $$=-\frac{\pi^2}{8}\ln2+\pi\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(2k-1)^2}+2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(2k)^3}-2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^3}$$$$=-\frac{\pi^2}8\ln2+\pi G-\frac{21\zeta(3)}{16}$$ Therefore, $$I=\int_0^{\pi/2}x\ln(1+\cos x)dx=-\frac{\pi^2}8\ln2+\pi G-\frac{21\zeta(3)}{16}$$

Answer 5

By integration by parts, $I=\int_0^{\frac\pi2}y\ln(1+\cos y)=4\int_0^{\frac\pi4}x^2\tan xdx.$ Let $J=4\int_0^{\frac\pi4}x^2\cot xdx$. Then, by robjohn's answer $$I+J=\int_0^{\frac\pi2}\frac{x^2}{\sin x}dx=2\pi G-\frac72\zeta(3)$$ and by using integration by parts first and then the Foruier series $\ln\sin x=-\ln2-\sum_{n=1}^\infty\frac{\cos 2nx}{n}$, we have $$J-I=\int_0^{\frac\pi2} x^2\cot x dx=\frac{\pi^2}4\ln2-\frac78\zeta(3).$$ Hence the result in OP follows.

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Some notable zeta series show up in computations. Two of them are: $$\sum_{n=1}^\infty\frac{\zeta(2n)}{4^n(n+1)}=\frac12-\ln 2+\frac7{2\pi^2}\zeta(3)$$ $$\sum_{n=1}^\infty\frac{\zeta(2n)}{16^n(n+1)}=\frac{35}{4\pi^2}\zeta(3)-\frac{\ln2}2-\frac{4G}\pi$$

Answer 6

\begin{align} I=&\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y\overset{y=2x}= 8\int_{0}^{\frac{\pi}{4}}x\ln(\sqrt2\cos x)dx\\ =& \ 4\int_0^{\frac\pi4 }x\ln(\sin 2x)\overset{2x\to x}{dx }- 4\int_0^{\frac\pi4 }x\ln(\tan x) \overset{x\to \frac\pi2-x}{dx }\\ =& \ \frac12 \int_0^{\frac\pi2 }x\ln\frac{\sin 2x}2 {dx } -\frac32\int_0^{\frac\pi2}x\ln(\tan x) dx -\pi\int_0^{\frac\pi4 }\ln(\tan x)dx \end{align} where $\int_0^{\frac\pi2}x\ln(\tan x) dx=\frac78\zeta(3)$, $\int_0^{\frac\pi4 }\ln(\tan x)dx=-G$ \begin{align} \int_0^{\frac\pi2 }x\ln\frac{\sin 2x}2 \overset{2x\to x}{dx } =\frac14\int_0^{\pi} \underset{=0}{x\ln(2\sin x)}\overset{x\to\frac\pi2-x}{dx}-\frac{\pi^2}{4}\ln2 \end{align} As a result $$I= \pi G-\frac{\pi^{2}}{8} \ln 2 -\frac{21}{16} \zeta(3) $$