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How can I evaluate the definite integral $$\int_{0}^{\pi}\log(2+\cos x)dx$$ without using integral with parameters? This question is not difficult , if we use the acknowledge of integral with parameters.Actually, I want find a series of skillful substitutions to deal with it concisely.

Applying the integral with parameters,my answer is $ \pi\log(1+\frac{\sqrt{3}}{2})$.

Quanto
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Elliot
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    Why don't you wish to try the Feynman technique by introducing a parameter? I don't think the solution is possible by using some sort of elementary anti-derivative. – Paramanand Singh May 20 '20 at 08:46
  • Please add details as to what techniques you have tried or what type of answer you are expecting. In this site, your question will most likely be answered if you provide your workings and experimentation. – Mathsisfun May 20 '20 at 09:03
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    Ok you may add in your question that you know the Feynman approach and seek alternative methods. – Paramanand Singh May 20 '20 at 09:11
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    If this can help,$$I-\pi\log2=\int_0^\pi\log\left(1+\frac{\cos x}2\right),dx=\sum_{k=0}^\infty\frac1{k,2^{2k+1}}\int_0^\pi\cos^{2k}(x),dx \=\sum_{k=0}^\infty\frac1{k,2^{2k+1}}\frac{(2k-1)!!}{(2k)!!}\pi.$$ –  May 20 '20 at 09:33

2 Answers2

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Here is to integrate with just substitutions. Let $a=2-\sqrt3$ and rewrite the integral as

$$\int_{0}^{\pi}\ln(2+\cos x)dx=2\pi \ln\frac{1+\sqrt3}2 + \int_{0}^{\pi}\ln(1+a^2-2a\cos \gamma)d\gamma\tag 1 $$ Substitute $c^2= 1+a^2-2a\cos \gamma$ while recognizing that $c$, $a$ and $1$ represent the sides of a triangle with the angles $\gamma = \pi-\beta(\gamma) - \alpha (\gamma) $. Then

\begin{align} & \frac12\int_{0}^{\pi}\ln(1+a^2-2a\cos \gamma)d\gamma = -\int_{0}^{\pi}\ln c(\gamma) \>( d\beta(\gamma)+ d\alpha(\gamma) )\\ = &-\int_{0}^{\pi}\ln \frac{1\cdot \sin\gamma}{\sin\beta(\gamma) }\>d\beta -\int_{0}^{\pi}\ln \frac{a\sin\gamma}{\sin\alpha(\gamma) }\>d\alpha\\ = & \int_{0}^{\pi}\ln \sin\gamma\>d\gamma +\int_{0}^{\pi}\ln \sin\beta(\gamma) \>\overset{ t= \beta(\gamma) }{d\beta(\gamma)}+ \int_{0}^{\pi}\ln \frac{\sin\alpha(\gamma) }a \>\overset{ s= \alpha(\gamma)} {d\alpha(\gamma)}\\ = &\int_{0}^{\pi}\ln \sin\gamma\>d\gamma +\int_{\pi}^{0}\ln \sin tdt+ \int_{0}^{0}\ln \frac{\sin s}ads =0 \\ \end{align}

Thus, the integral (1) is $$\int_{0}^{\pi}\ln(2+\cos x)dx=2\pi \ln\frac{1+\sqrt3}2 $$

Quanto
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8

In General

For $|\alpha|\lt1$, $$ \begin{align} \log\left(1+\alpha^2+2\alpha\cos(\theta)\right) &=\log\left(1+\alpha e^{i\theta}\right)+\log\left(1+\alpha e^{-i\theta}\right)\tag1\\[9pt] &=\sum_{k=1}^\infty(-1)^{k-1}\frac{\alpha^k}ke^{ik\theta}+\sum_{k=1}^\infty(-1)^{k-1}\frac{\alpha^k}ke^{-ik\theta}\tag2\\ &=\sum_{k=1}^\infty(-1)^{k-1}2\frac{\alpha^k}k\cos(k\theta)\tag3 \end{align} $$ Explanation:
$(1)$: $1+\alpha^2+2\alpha\cos(\theta)=\left(1+\alpha e^{i\theta}\right)\left(1+\alpha e^{-i\theta}\right)$
$(2)$: power series for $\log(1+x)$
$(3)$: $e^{ik\theta}+e^{-ik\theta}=2\cos(k\theta)$

Thus, integrating $(3)$ over $[0,\pi]$ gives $$ \begin{align} 0 &=\int_0^\pi\log\left(1+\alpha^2+2\alpha\cos(\theta)\right)\,\mathrm{d}\theta\tag4\\ &=\pi\log(2\alpha)+\int_0^\pi\log\left(\frac{1+\alpha^2}{2\alpha}+\cos(\theta)\right)\,\mathrm{d}\theta\tag5\\ &=\pi\log\left(\frac2{\beta+\sqrt{\beta^2-1}}\right)+\int_0^\pi\log\left(\beta+\cos(\theta)\right)\,\mathrm{d}\theta\tag6 \end{align} $$ Explanation:
$(4)$: $(3)$ and $\int_0^\pi\cos(k\theta)\,\mathrm{d}\theta=0$
$(5)$: pull $\pi\log(2\alpha)$ out of the integral
$(6)$: $\alpha=\frac1{\beta+\sqrt{\beta^2-1}}$ is the root of $\frac{1+\alpha^2}{2\alpha}=\beta$ with $|\alpha|\lt1$

Therefore, $$ \int_0^\pi\log\left(\beta+\cos(\theta)\right)\,\mathrm{d}\theta =\pi\log\left(\frac{\beta+\sqrt{\beta^2-1}}2\right)\tag7 $$ Substitute $\beta\mapsto\frac\beta\gamma$ and add $\pi\log(\gamma)$: $$ \int_0^\pi\log\left(\beta+\gamma\cos(\theta)\right)\,\mathrm{d}\theta =\pi\log\left(\frac{\beta+\sqrt{\beta^2-\gamma^2}}2\right)\tag8 $$

Applied to this problem $$ \int_0^\pi\log(2+\cos(x))\,\mathrm{d}x=\pi\log\left(\frac{2+\sqrt3}2\right)\tag9 $$

robjohn
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