Exercise 7.10 · Q16: $\int_0^{\pi}\log(1+\cos x)\,dx$

Basics you need

Solution

1. The trouble at $x = \pi$⁠, and why $I$ is finite

   For $0 \le x < \pi$⁠, $\cos x > -1$⁠, so $1 + \cos x > 0$ and $\log(1 + \cos x)$ is defined and continuous. At $x = \pi$⁠, $1 + \cos\pi = 0$ has no logarithm, and $\log(1 + \cos x) \to -\infty$ as $x \to \pi^-$⁠. So the integrand is not continuous on the closed interval: $I$ is an improper integral, meaning the limit of $\int_0^{b}$ as $b \to \pi^-$⁠. The book does not raise this point, but the method below needs $I$ to be a finite number.

   The half-angle formula gives, for $0 \le x < \pi$ (where $\cos\frac{x}{2} > 0$⁠):

   $$\begin{aligned}\log(1 + \cos x) &= \log\left(2\cos^2\tfrac{x}{2}\right) \\ &= \log 2 + 2\log\cos\tfrac{x}{2} .\end{aligned}$$

   With $t = \frac{x}{2}$ (so $dx = 2\,dt$⁠, and $t$ runs from 0 to $\frac{\pi}{2}$⁠), the second term integrates to $4\int_0^{\pi/2}\log\cos t\,dt$⁠, which is finite (shown in Step 1 of our Example 34 page; the book does not discuss it). So $I$ is finite. In the same way $\log(1 - \cos x)$⁠, which equals $\log 2 + 2\log\sin\frac{x}{2}$⁠, has a finite integral over $[0, \pi]$⁠, although it tends to $-\infty$ at $x = 0$⁠.

   In Step 3 we will solve an equation for $I$⁠; that only makes sense if $I$ is a finite number. As in Example 34 (Step 1 there), the properties P4 and P6 remain true for these convergent improper integrals. For instance, for P4: the substitution $x \to \pi - x$ turns $\int_0^{\pi-\varepsilon}\log(1 + \cos x)\,dx$ into $\int_{\varepsilon}^{\pi}\log(1 - \cos x)\,dx$⁠, where both integrands are continuous; letting $\varepsilon \to 0^+$ gives the improper version.
   $\cos x = 2\cos^2\frac{x}{2} - 1$⁠, the double-angle formula read backwards. Basics: trigonometric identities

2. P4 with $a = \pi$⁠: $\cos x$ becomes $-\cos x$

   P4 replaces $x$ by $\pi - x$⁠. Since $\cos(\pi - x) = -\cos x$⁠:

   $$\begin{aligned}I &= \int_0^{\pi}\log\big(1 + \cos(\pi - x)\big)\,dx \\ &= \int_0^{\pi}\log(1 - \cos x)\,dx .\end{aligned}$$
   P4 has no condition beyond the integrals existing (Step 1). Reading $[0, \pi]$ backwards turns the graph of $\log(1 + \cos x)$ into its mirror image in $x = \frac{\pi}{2}$⁠, which is the graph of $\log(1 - \cos x)$ (dashed in the figure).
   $\cos(\pi - x) = -\cos x$⁠. Basics: symmetry and mirror images

3. Add the two forms: $\sin^2 x$ appears

   For $0 < x < \pi$⁠, both $1 + \cos x$ and $1 - \cos x$ are positive and $\sin x > 0$⁠, so

   $$\begin{aligned}&\log(1 + \cos x) + \log(1 - \cos x) \\ &= \log(1 - \cos^2 x) = \log\sin^2 x \\ &= 2\log\sin x .\end{aligned}$$

   Adding the original $I$ and the form from Step 2:

   $$2I = 2\int_0^{\pi}\log\sin x\,dx, \quad\text{so}\quad I = \int_0^{\pi}\log\sin x\,dx .$$
   The product law $\log p + \log q = \log pq$ needs $p, q > 0$⁠, which holds strictly inside the interval; the endpoints are single points and do not affect an integral. We want $\sin x$ because Example 34 already knows the integral of $\log\sin x$⁠.
   $(1 + c)(1 - c) = 1 - c^2$⁠; $\sin^2 x + \cos^2 x = 1$⁠; $\log p^2 = 2\log p$ for $p > 0$⁠. Basics: logarithms

4. P6 with $2a = \pi$⁠: fold $[0, \pi]$ in half

   Take $g(x) = \log\sin x$ and $a = \frac{\pi}{2}$⁠. The condition of P6 holds:

   $$\begin{aligned}g(\pi - x) &= \log\sin(\pi - x) \\ &= \log\sin x = g(x) .\end{aligned}$$

   So the first case of P6 gives

   $$I = \int_0^{\pi}\log\sin x\,dx = 2\int_0^{\pi/2}\log\sin x\,dx .$$
   $\sin(\pi - x) = \sin x$⁠: the graph of $\log\sin x$ on $[0, \pi]$ is symmetric in the line $x = \frac{\pi}{2}$⁠, so the two halves contribute equally.
   P6, first case, Section 7.10. Lesson: properties

5. Use Example 34

   Example 34 found $\int_0^{\pi/2}\log\sin x\,dx = -\frac{\pi}{2}\log 2$⁠. Therefore

   $$I = 2\left(-\frac{\pi}{2}\log 2\right) = -\pi\log 2 \approx -2.177586 .$$
   The answer is negative: the region below the axis near $x = \pi$⁠, where the integrand plunges to $-\infty$⁠, outweighs the small region above the axis near $x = 0$⁠, where the integrand is at most $\log 2 \approx 0.693147$⁠.
   Basics: area as a sum of thin strips (signed area)

6. Check: a second route through the half-angle formula

   From Step 1, for $0 \le x < \pi$⁠, $\log(1 + \cos x)$ $= \log 2 + 2\log\cos\frac{x}{2}$⁠. Integrate term by term, putting $t = \frac{x}{2}$ in the second term:

   $$\begin{aligned}I &= \pi\log 2 + 4\int_0^{\pi/2}\log\cos t\,dt \\ &= \pi\log 2 + 4\left(-\frac{\pi}{2}\log 2\right) \\ &= \pi\log 2 - 2\pi\log 2 = -\pi\log 2 .\end{aligned}$$
   This route uses the $\cos$ form of Example 34 (equal to the $\sin$ form by P4) instead of the adding trick, and it gives the same value, so the two routes confirm each other.
   Substitution $t = \frac{x}{2}$⁠: $dx = 2\,dt$⁠; $x = 0 \Rightarrow t = 0$⁠, $x = \pi \Rightarrow t = \frac{\pi}{2}$⁠. Basics: substitution

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