# Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$

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.

• 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 7:46
• @sos440: It's as simple as taylor expanding the log and evaluating the integral of powers of cosines. – Ron Gordon Nov 18 '13 at 7:56
• @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
• @RonGordon The other thread asks about a very specific approach, and only that approach is explained. – Random Variable Jun 9 '15 at 18:03

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

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

• Where does the evaluation of that sum come from Ron? (Second last line). – Bennett Gardiner Nov 18 '13 at 23:02
• @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 0:06
• @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 3:14

\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}


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}

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