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Finding $\displaystyle \int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$

What I tried

Put $\displaystyle I =\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$

\begin{align*} I&=\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx\\ &=\int^{\frac{\pi}{4}}_{0}\ln\bigg(2\cos^2\frac{x}{2}\bigg)\mathrm dx\\ &=\frac{\pi}{4}\ln(2)+4\int^{\frac{\pi}{8}}_{0}\ln(\cos x)\mathrm dx\\ &=\frac{\pi}{4}\ln(2)+4\left[\left[x\ln(\cos x)\right]^{\frac{\pi}{8}}_{0}+\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx\right]\\ &=\frac{\pi}{8}\ln(2)+\frac{\pi}{2}\ln\left(\cos \frac{\pi}{8}\right)+4\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx \end{align*}

How do I solve it? Help me, please.

mrtaurho
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jacky
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  • This integral can not be expressed by the known elementary functions – Dr. Sonnhard Graubner Feb 13 '19 at 16:08
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    @ Dr. Sonnhard Graubner: if a function to integrate doesn't have an elementary antiderivative it doesn't mean that you can not obtain some close form for definite integrates involving this function. For example, $\int^{\frac{\pi}{4}}_{0}\ln(\tan x)dx$ have a closed form. But for the integral involved in this question i think the better you can do is to obtain an expression involving dilogarithms. – FDP Feb 13 '19 at 16:19

3 Answers3

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\begin{align}J=\int_0^{\frac{\pi}{8}}\ln\left(\tan x\right)\,dx\end{align}

Perform the change of variable $y=\left(\sqrt{2}+1\right)\tan x$,

\begin{align}J&=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big((\sqrt{2}-1)x\big) }{1+(\sqrt{2}-1)^2x^2}\,dx\\ &=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big(\sqrt{2}-1\big) }{1+(\sqrt{2}-1)^2x^2}\,dx+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\ &=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\ &=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(1-\sqrt{2}\big)x}\,dx+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(\sqrt{2}-1\big)x}\,dx\\ &=\boxed{\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}i\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)-\frac{1}{2}i\text{Li}_2\left(i\big(1-\sqrt{2}\big)\right)} \end{align}

NB:

For $0<|z|<1$ complex,

\begin{align}\int_0^1 \frac{\ln x}{1-zx}\,dx&=\int_0^1 \left(\sum_{n=0}^{\infty}(xz)^n\right)\ln x\,dx\\ &=\sum_{n=0}^{\infty} z^n\left(\int_0^1 x^n\ln x\,dx\right)\\ &=-\sum_{n=0}^{\infty} \frac{z^n}{(n+1)^2}\\ &=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)^2}\\ &=-\frac{1}{z}\text{Li}_2(z) \end{align}

PS:

if you prefer,

\begin{align}\boxed{J=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)-\Im{\left(\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)\right)}}\end{align}

FDP
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3

Using the Clausen Function $\operatorname{Cl}_2(z)$ we can directly write down an anti-derivative for your integral. Since we got that

$$\int_0^t \log(\tan x)\mathrm dx=-\frac12\operatorname{Cl}_2(2t)-\frac12\operatorname{Cl}_2(\pi-2t)\tag1$$

This formula can be shown by using the Fourier Series Expansions of $\log(\sin x)$ and $\log(\cos x)$ under the utilisation of the series representation of the Clausen Function. Using formula $(1)$ we may write your integral as

$$\int_0^{\pi/8} \log(\tan x)\mathrm dx=-\frac12\operatorname{Cl}_2\left(\frac\pi4\right)-\frac12\operatorname{Cl}_2\left(\frac{3\pi}4\right)$$

However, this does not help at all since there are no closed-form representations known for these values of the Clausen Function. By the Duplication Formula of the Clausen Function we can at least eliminate one of the unknown values.

$$\therefore~\int_0^{\pi/8} \log(\tan x)\mathrm dx~=~-\operatorname{Cl}_2\left(\frac\pi4\right)+\frac G4~=~-\operatorname{Cl}_2\left(\frac{3\pi}4\right)-\frac G4$$

I have doubts that we can get any closer to an actual closed-form. Anyway, these terms are still series in disguise. Note that the Clausen Function can be expressed as Dilogarithm aswell, precisely $\operatorname{Cl}_2(z)=\Im\operatorname{Li}_2(e^{iz})$, which moves this problem in the conetxt of Polylogarithms too.

mrtaurho
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0

Let’s find directly the original integral$$ \begin{aligned} \int_0^{\frac{\pi}{4}} \ln (1+\cos x) d x & =2\int_0^{\frac{\pi}{4}} \ln \left(2 \cos \frac{x}{2}\right) d x-\frac{\pi}{4} \ln 2 \\ & =2\int_0^{\frac{\pi}{4}} \ln \left(e^{\frac{x}{2} i}+e^{-\frac{x}{2} i}\right) d x-\frac{\pi}{4} \ln 2 \\ & =2\int_0^{\frac{\pi}{4}} \ln \left[e^{\frac{x}{2} i}\left(1+e^{-xi}\right)\right] d x-\frac{\pi}{4} \ln 2 \\ & =2\Re \int^{\frac{\pi}{4}} \ln \left(1+e^{-xi}\right) d x-\frac{\pi}{4} \ln 2 \end{aligned} $$ Putting $y=e^{-xi} $ yields $$ \begin{aligned} \int \ln \left(1+e^{-xi}\right) d x & =i \int \frac{\ln (1+y)}{y} d y \\ & =-i \operatorname {Li_2}(-y)+C\\&=- i \operatorname {Li_2}\left(-e^{-xi}\right)+C \end{aligned} $$ Plugging back the limits yields

$$ \begin{aligned} \int_0^{\frac{\pi}{4}} \ln \left(1+e^{-xi}\right) d x&=-2 i \operatorname {Li_2}\left(-e^{-\frac{\pi}{4} i}\right)+2 i \operatorname {Li_2}(-1) \\ & =-2 i \operatorname {Li_2}\left(-e^{-\frac{\pi}{4} i}\right)-\frac{\pi^2}{6} i \end{aligned} $$

Hence we can conclude that $$ \boxed{\int_0^{\frac{\pi}{4}} \ln (1+\cos x) d x =2\Im\left[\operatorname {L i_2}\left(-e^{-\frac{\pi}{4} i}\right)\right]-\frac{\pi}{4} \ln 2 \;} $$

Lai
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