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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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    It is a special case of a more general integral. though I suspect that it can also be attacked by a more elementary method. – Sangchul Lee Nov 18 '13 at 07:46
  • @sos440: It's as simple as taylor expanding the log and evaluating the integral of powers of cosines. – Ron Gordon Nov 18 '13 at 07:56
  • @RonGordon, Amazing! I wish I were able to tackle this problem. I'm plowing through a swamp of homework... :( – Sangchul Lee Nov 18 '13 at 10:15
  • 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. – Lucian Nov 18 '13 at 17:39
  • @Lucian: did you post this comment before or after you read my solution? – Ron Gordon Nov 18 '13 at 19:00
  • It's redundant, I know, but I always tell people the same thing when dealing with logs of fractions and/or powers: Simplify your expressions ! Also, this is one of those very rare instances in which the Taylor series is too hard to miss. – Lucian Nov 18 '13 at 19:19
  • This question is not a duplicate of the linked-to question. While the integral is the same, the requirement of the answer was "The book says that I should use the method of "differentiating parameters" by using the provided Leibniz Formula (?)" The only answer given used that method. On the other hand, there are varied solutions here. Moreover, this question is 19 months old and has an accepted answer. What on earth is the point? Yes, I marked an identical question to this one as a duplicate, but that was because it truly was. And it was new. – Ron Gordon Jun 09 '15 at 21:45
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3 Answers3

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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

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

Ron Gordon
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  • Where does the evaluation of that sum come from Ron? (Second last line). – Bennett Gardiner Nov 18 '13 at 23:02
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    @BennettGardiner: that is a binomial expansion. Apply a Taylor expansion to the square root and you can easily see where the sum comes from. – Ron Gordon Nov 19 '13 at 00:06
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    @Bennett

    $$\frac{1}{(1-a^{2})^{1/2}} = \sum_{k=0}^{\infty} \binom{-1/2}{k} (-a^{2})^{k} = \sum_{k=0}^{\infty} \frac{(-1/2)(-1/2-1) \cdots (-1/2-k+1)}{k!} (-1)^{k} a^{2k}$$

    $$ = \sum_{k=0}^{\infty} (-1)^{k} \frac{(1/2+k-1)\ldots (1/2+1) (1/2)}{k!} (-1)^{k} a^{2k} = \sum_{k=0}^{\infty} \frac{\Gamma(k+1/2)}{\Gamma(k+1)\Gamma(1/2)} a^{2k}$$

    $$ = \sum_{k=0}^{\infty} \frac{\Gamma(2k) \Gamma(1/2)}{2^{2k-1} \Gamma(k)\Gamma(k+1)\Gamma(1/2)} a^{2k} \frac{k}{k}=\sum_{k=0}^{\infty}\frac{\Gamma(2k+1)}{\Gamma^{2}(k+1)} \frac{a^{2k}}{2^{2k}} = \sum_{k=0}^{\infty} \binom{2k}{k} \frac{a^{2k}}{2^{2k}}$$

    – Random Variable Nov 19 '13 at 03:14
10

$$\begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1+ b\cos x}\right)\frac{1}{\cos x}dx &= \int_{0}^{\pi/2} \int_{b}^{a} \frac{1}{1+t \cos x} \ dt \ dx \\ &= \int_{b}^{a} \int_{0}^{\pi/2}\frac{1}{1+t \cos x} \ dx \ dt \end{align}$$

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

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

5

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

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}

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

Felix Marin
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