# Evaluating $\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x$, $b$ is a parameter

Evaluating $$\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x$$ where $$b$$ is a parameter

I've tried Integration by parts which yield $$\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x=\pi\ln(1-b)+b\int_{0}^{\pi}{x\sin x\over 1+b\cos x}\text{d}x$$ I cannot figure out what do next.

I also tried using Leibniz integral rule by putting $$I(b)=\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x$$ to form a differential equation.

$${\text{d}I(b)\over \text{d}b}=\int_{0}^{\pi}{\cos x\over 1+b\cos x}\text{d}x$$ but I'm not able to solve the integral on right.

I've looked similar questions like this one Evaluating $\int_{0}^{\pi}\ln (1+\cos x)\, dx$ to no avail. Also I'm high school student so I don't understand advanced calculus stuff yet.

• I think that $0<b < 1$ Commented Sep 22, 2020 at 19:47
• Have you tried the Weierstrass substitution? i.e. $t = \tan\left(\frac{x}{2}\right)$. My hunch is that this should work (for $b$ values in the correct range, of course), although I haven't done the calculation yet myself. Commented Sep 22, 2020 at 20:01
• If $-1 \leq b \leq 1$, then the integral is $$\int_0^\pi \ln |1 + b \cos x| \,d x = \int_0^\pi (\ln |b/2| + \ln |1 - x_1 e^{i x}| + \ln |1 - x_2 e^{i x}|) \, dx,$$ where $x_{1, 2}$ are the roots of $b x^2 + 2 x + b$. The integral of $\ln |1- a e^{i x}|$ can be found as here. Commented Sep 23, 2020 at 10:04

\begin{align*} \frac{\mathrm{d} I(b)}{\mathrm{d} b}=\int_{0}^{\pi}{\cos x\over 1+b\cos x}\; \mathrm{d}x &= \frac{1}{b}\int_0^{\pi} \frac{ 1+b \cos{x}-1}{1+b\cos{x}} \; \mathrm{d}x \\ &= \frac{\pi}{b}-\frac{1}{b} \int_0^{\pi} \frac{1}{1+b \cos{x}} \; \mathrm{d}x\\ &= \frac{\pi}{b}-\frac{2}{b} \int_0^{\infty} \frac{1}{(t^2+1)+b(1-t^2)} \; \mathrm{d}t \tag{1}\\ &= \frac{\pi}{b}-\frac{2}{b} \int_0^{\infty} \frac{1}{(1-b)t^2+(1+b)} \; \mathrm{d}t\\ &= \frac{\pi}{b}-\frac{2}{b} \left(\frac{\pi}{2\sqrt{1-b^2}}\right) \\ &= \frac{\pi}{b}- \frac{\pi}{b\sqrt{1-b^2}} \\ I(b) &= \int \frac{\pi}{b}- \frac{\pi}{b\sqrt{1-b^2}} \; \mathrm{d}b \\ &= \pi \ln|b| + \pi \operatorname{artanh}{\left(\sqrt{1-b^2}\right)}+C \\ I(1)&=-\pi \ln{2} \implies C=-\pi \ln{2}\\ I(b) &= \pi \ln|b| + \pi \operatorname{artanh}{\left(\sqrt{1-b^2}\right)}-\pi \ln{2} \\ &= \pi \ln\bigg|\frac{b}{2}\bigg| -\frac{\pi}{2} \ln\left(1-\sqrt{1-b^2}\right)+\frac{\pi}{2}\ln \left(1+\sqrt{1-b^2}\right) \\ &= \boxed{\pi \ln\left(\frac{1+\sqrt{1-b^2}}{2}\right)} \end{align*}
Additionally, note that $$-1.
$$(1):$$ Weierstrass substitution
$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\bbox[5px,#ffd]{}}$$
\begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\pi} \ln\pars{1 + b\cos\pars{x}}\,\dd x \,\right\vert_{\ b\ \in\ \pars{-1,1}}} \\[5mm] = &\ \int_{0}^{\pi/2}\ln\pars{1 + b\cos\pars{x}}\,\dd x \\[2mm] + &\ \int_{-\pi/2}^{0}\ln\pars{1 - b\cos\pars{x}}\,\dd x \\[5mm] = &\ \int_{0}^{\pi/2}\ln\pars{1 - b^{2}\cos^{2}\pars{x}}\,\dd x \\[5mm] = &\ \int_{0}^{\pi/2}\int_{0}^{b^{2}} {-\cos^{2}\pars{x} \over 1 - y\cos^{2}\pars{x}}\,\dd y\,\dd x \\[5mm] = &\ \int_{0}^{b^{2}}\int_{0}^{\pi/2}\bracks{% 1 - {1 \over 1 - y\cos^{2}\pars{x}}}\dd x\,{\dd y \over y} \\[5mm] = &\ \int_{0}^{b^{2}}\bracks{% {\pi \over 2} - \int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \sec^{2}\pars{x} - y}\,\dd x}{\dd y \over y} \\[5mm] = &\ \int_{0}^{b^{2}}\bracks{% {\pi \over 2} - \int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \tan^{2}\pars{x} + 1 - y}\,\dd x}{\dd y \over y} \\[5mm] = &\ \int_{0}^{b^{2}}\left\{% {\pi \over 2}\right. \\ & \left.- {1 \over \root{1 - y}}\int_{0}^{\pi/2}\!\!\!\!\!\!\! {\sec^{2}\pars{x}/\root{1 - y} \over \bracks{\tan\pars{x}/\root{1 - y}}^{2} + 1}\,\dd x\right\} {\dd y \over y} \\[5mm] = &\ {\pi \over 2} \int_{0}^{b^{2}} \pars{{1 \over y} - {1 \over y\root{1 - y}}} \dd y \\[5mm] & \stackrel{y\ =\ 1 - t^{2}}{=}\,\,\, \pi\int_{1}^{\root{1 - b^{2}}} {\dd t \over t + 1} \\[5mm] = &\ \bbx{\pi\ln\pars{1 + \root{1 - b^{2}} \over 2}} \\ & \end{align}
Since $$\mathcal{I}=\int_0^{\pi}\ln(1+b\cos x)dx =\int_0^{\frac{\pi}{2}}\ln(1-b^2\cos^2 x)dx$$ See Flexin Marin and for all $$b\in\mathbb(0,1)$$ we notice $$-1< b\cos x <1$$ we use the series for $$\ln(1-x)$$, giving us. $$\mathcal{I}=-\sum_{p=1}^{\infty}\frac{1}{p}\int_0^{\frac{\pi}{\color{red}{2}}}b^{2p}\cos^{2p}x dx$$Latter integral we have Wallis integral which further reduces to $$\mathcal{I}=-\frac{\pi}{2}\sum_{p=1}^{\infty}\frac{ b^{2p}}{2^{2p}p}{2n\choose n}$$ Since the generating function of central binomial coefficients is given as $$\sum_{p=0}^{\infty}{2p\choose p}x^p=\frac{1}{\sqrt{1-4x}}, \; \; |p|< 1/4$$ Divide by $$x$$ and hence on integrating from $$0$$ to $$\frac{b^2}{4}$$ we have $$\sum_{p=1}^{\infty}\frac{b^{2p}}{2^{2p}}{2p\choose p}=\int_0^{-\frac{b^2}{4}}\left(\frac{1}{x\sqrt{1-4x}}-\frac{1}{x}\right)dx=-2\log\left(1+\sqrt{1-4x}\right)\Bigg|_0^{\frac{b^2}{4}}=2\left(\log 2-\log(1+\sqrt{1-b^2})\right)$$ hence $$\mathcal{I}={\pi}\log\left(\frac{1+\sqrt{1-b^2}}{2}\right)$$