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How can I evaluate the following integral?

$$ \int_0^{\pi/2} \log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\, \frac{1}{\cos\left(x\right)}\,{\rm d}x\,, \qquad\left\vert\,a\,\right\vert \le 1$$

I tried differentiating under the integral with respect to the parameter $a$, and I also tried expanding the log term in a Taylor series and then switching the order of integration and summation. I ran into difficulties with both approaches.

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    $\begingroup$ It is a special case of a more general integral. though I suspect that it can also be attacked by a more elementary method. $\endgroup$ – Sangchul Lee Nov 18 '13 at 7:46
  • $\begingroup$ @sos440: It's as simple as taylor expanding the log and evaluating the integral of powers of cosines. $\endgroup$ – Ron Gordon Nov 18 '13 at 7:56
  • $\begingroup$ @RonGordon, Amazing! I wish I were able to tackle this problem. I'm plowing through a swamp of homework... :( $\endgroup$ – Sangchul Lee Nov 18 '13 at 10:15
  • $\begingroup$ Use $\ln\frac ab=\ln a-\ln b$, then expand each new term according to the well-known Taylor series for the natural logarithm, $\ln(1-x)=\sum_{n=1}^\infty\frac{x^n}n$, and use the fact that the integral of a sum is the same as the sum of integrals. $\endgroup$ – Lucian Nov 18 '13 at 17:39
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    $\begingroup$ @RonGordon The other thread asks about a very specific approach, and only that approach is explained. $\endgroup$ – Random Variable Jun 9 '15 at 18:03
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Use the expansion for $|z| < 1$

$$\log{\left ( \frac{1+z}{1-z}\right )} = 2 \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$

Then the integral is equal to

$$2 \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \int_0^{\pi/2} dx \, \cos^{2 k}{x}$$

It is straightforward to show that

$$\int_0^{\pi/2} dx \, \cos^{2 k}{x} = \frac{1}{2^{2 k}} \binom{2 k}{k} \frac{\pi}{2}$$

Thus the integral $I(a)$ is

$$I(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \frac{1}{2^{2 k}} \binom{2 k}{k}$$

We may evaluate this sum by considering

$$I'(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k}}{2^{2 k}} \binom{2 k}{k} = \pi \left (1-a^2\right)^{-1/2}$$

Integrating with respect to $a$ and noting that $I(0)=0$, we find that

$$I(a) = \pi \arcsin{a}$$

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  • $\begingroup$ Where does the evaluation of that sum come from Ron? (Second last line). $\endgroup$ – Bennett Gardiner Nov 18 '13 at 23:02
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    $\begingroup$ @BennettGardiner: that is a binomial expansion. Apply a Taylor expansion to the square root and you can easily see where the sum comes from. $\endgroup$ – Ron Gordon Nov 19 '13 at 0:06
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    $\begingroup$ @Bennett $$\frac{1}{(1-a^{2})^{1/2}} = \sum_{k=0}^{\infty} \binom{-1/2}{k} (-a^{2})^{k} = \sum_{k=0}^{\infty} \frac{(-1/2)(-1/2-1) \cdots (-1/2-k+1)}{k!} (-1)^{k} a^{2k}$$ $$ = \sum_{k=0}^{\infty} (-1)^{k} \frac{(1/2+k-1)\ldots (1/2+1) (1/2)}{k!} (-1)^{k} a^{2k} = \sum_{k=0}^{\infty} \frac{\Gamma(k+1/2)}{\Gamma(k+1)\Gamma(1/2)} a^{2k}$$ $$ = \sum_{k=0}^{\infty} \frac{\Gamma(2k) \Gamma(1/2)}{2^{2k-1} \Gamma(k)\Gamma(k+1)\Gamma(1/2)} a^{2k} \frac{k}{k}=\sum_{k=0}^{\infty}\frac{\Gamma(2k+1)}{\Gamma^{2}(k+1)} \frac{a^{2k}}{2^{2k}} = \sum_{k=0}^{\infty} \binom{2k}{k} \frac{a^{2k}}{2^{2k}}$$ $\endgroup$ – Random Variable Nov 19 '13 at 3:14
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$$\begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1+ b\cos x}\right)\frac{1}{\cos x}dx &= \int_{0}^{\pi/2} \int_{b}^{a} \frac{1}{1+t \cos x} \ dt \ dx \\ &= \int_{b}^{a} \int_{0}^{\pi/2}\frac{1}{1+t \cos x} \ dx \ dt \end{align}$$

Let $ \displaystyle u = \tan \frac{x}{2}$.

$$\begin{align} &= \int_{b}^{a} \int_{0}^{1} \frac{1}{1+ t \left(\frac{1-u^{2}}{1+u^{2}} \right)} \frac{2}{1+u^{2}} \ du \ dt \\ &= 2 \int_{b}^{a} \int_{0}^{1} \frac{1}{1+t} \frac{1}{1+ \frac{1-t}{1+t} u^{2}} du \ dt \end{align}$$

Let $\displaystyle w = \sqrt{\frac{1-t}{1+t}} u $.

$$ \begin{align} &= 2 \int_{b}^{a} \int_{0}^\sqrt{\frac{1-t}{1+t}} \frac{1}{\sqrt{1-t^{2}}} \frac{1}{1+w^{2}} \ dw \ dt \\ &= 2 \int_{b}^{a} \frac{1}{\sqrt{1-t^{2}}} \arctan \sqrt{\frac{1-t}{1+t}}\ dt \\ &= \int_{b}^{a} \frac{\arccos t}{\sqrt{1-t^{2}}} \ dt \\ &= \frac{1}{2} \Big(\arccos^{2} (b)- \arccos^{2} (a)\Big) \end{align}$$

Then

$$ \begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1-a \cos x}\right)\frac{1}{\cos x}dx &= \frac{1}{2} \Big(\arccos^{2} (-a)- \arccos^{2} (a)\Big) \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} - \arcsin (-a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} + \arcsin (a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big(2 \pi \arcsin a\Big) = \pi \arcsin a \end{align}$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_0^{\pi/2}\ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\, {1 \over \cos\pars{x}}\,\dd x:\ {\large ?}\,.\qquad\qquad\verts{a}\ <\ 1}$.

The general idea is to derivate respect of $\ds{\quad a\quad}$ in order " to kill " the " $\ds{\cos\pars{x}}$ term " in the denominator:

\begin{align}&\color{#c00000}{\partiald{}{a}\bracks{\int_0^{\pi/2} \ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\,{\dd x \over \cos\pars{x}}}} \\[3mm]&=\int_0^{\pi/2}\bracks{{\cos\pars{x} \over 1 + a\cos\pars{x}} -{-\cos\pars{x} \over 1 - a\cos\pars{x}}}\,{\dd x \over \cos\pars{x}} =2\int_{0}^{\pi/2}{\dd x \over 1 - a^{2}\cos^{2}\pars{x}} \\[3mm]&=2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over \sec^{2}\pars{x} - a^{2}} =2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over \tan^{2}\pars{x} + 1 - a^{2}} =2\int_{0}^{\infty}{\dd x \over x^{2} + 1 - a^{2}} \\[3mm]&={2 \over \root{1 - a^{2}}}\ \overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\ =\ \color{#c00000}{\pi \over \root{1 - a^{2}}} \end{align}

$$\color{#66f}{\large% \int_0^{\pi/2}\ln\pars{{1 + a\cos\pars{x} \over 1 - a\cos\pars{x}}}\, {1 \over \cos\pars{x}}\,\dd x} =\int_{0}^{a}{\pi\,\dd t \over \root{1 - t^{2}}} =\color{#66f}{\large\pi\ \arcsin\pars{a}} $$

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