Exercise 7.10 · Q10: $\int_0^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx$

Basics you need

Solution

1. Where the integrand is defined

   For $0 < x < \frac{\pi}{2}$⁠: $\sin x > 0$ and $0 < 2x < \pi$⁠, so $\sin 2x > 0$⁠. Both logarithms are defined and the integrand is continuous on the open interval. At the ends, $\sin 0 = 0$ and $\sin\left(2\cdot\frac{\pi}{2}\right) = \sin\pi = 0$⁠, whose logarithms do not exist.

   So, strictly, this is an improper integral at both ends: it means the limit of the integral over $[\varepsilon, \frac{\pi}{2} - \delta]$ as $\varepsilon, \delta \to 0^+$⁠. The book does not raise the point, but we need the integral to be a finite number before calculating with it. Step 3 settles this.
   $\sin\theta > 0$ for $0 < \theta < \pi$⁠. Basics: trigonometric identities

2. Simplify the integrand

   For $0 < x < \frac{\pi}{2}$⁠, all of 2, $\sin x$⁠, $\cos x$ are positive, so

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

   Hence

   $$\begin{aligned}&2\log\sin x - \log\sin 2x \\ &= \log\sin x - \log\cos x - \log 2 .\end{aligned}$$
   The product law of logarithms needs every factor positive, which holds on the open interval. One of the two copies of $\log\sin x$ is cancelled by the $\log\sin x$ inside $\log\sin 2x$⁠.
   $\log(pq) = \log p + \log q$ for $p, q > 0$⁠, used twice; $\sin 2x = 2\sin x\cos x$⁠. Basics: logarithms

3. Each piece has a finite integral

   Our worked solution of Example 34 (its Step 1; the book's own solution does not discuss this point) shows that $\log\sin x \le 0$ on $\left(0, \frac{\pi}{2}\right]$ and that $\int_0^{\pi/2}\log\sin x\,dx$ is finite. The same holds for $\log\cos x$ (Step 4 shows its integral is the same number). So we may split:

   $$\begin{aligned}I &= \int_0^{\frac{\pi}{2}}\log\sin x\,dx - \int_0^{\frac{\pi}{2}}\log\cos x\,dx \\ &\quad - \int_0^{\frac{\pi}{2}}\log 2\,dx .\end{aligned}$$
   Splitting an integral into a difference of integrals is only safe when each part is a finite number; "$\infty - \infty$" would mean nothing. That step squeezes $\log\sin x$ between $\log\frac{2x}{\pi}$ (whose integral over $\left[0, \frac{\pi}{2}\right]$ is $-\frac{\pi}{2}$⁠) and 0.
   Linearity: $\int(f - g - h) = \int f - \int g - \int h$ when all three are finite. Example 34

4. P4: the $\log\sin$ and $\log\cos$ integrals are equal

   Take $f(x) = \log\sin x$ and $a = \frac{\pi}{2}$⁠. Then

   $$f\left(\tfrac{\pi}{2} - x\right) = \log\sin\left(\tfrac{\pi}{2} - x\right) = \log\cos x ,$$

   so P4 gives

   $$\int_0^{\frac{\pi}{2}}\log\sin x\,dx = \int_0^{\frac{\pi}{2}}\log\cos x\,dx .$$

   The first two integrals in Step 3 cancel.

   P4 is proved by the substitution $t = \frac{\pi}{2} - x$⁠. Here we apply it on $\left[\varepsilon, \frac{\pi}{2}\right]$⁠, where $\log\sin x$ is continuous: it gives $\int_\varepsilon^{\pi/2}\log\sin x\,dx = \int_0^{\pi/2 - \varepsilon}\log\cos x\,dx$⁠. Letting $\varepsilon \to 0^+$ gives the equality of the two finite integrals. In pictures, $\log\cos x$ is the mirror image of $\log\sin x$ in $x = \frac{\pi}{4}$⁠.
   P4, Section 7.10; $\sin\left(\frac{\pi}{2} - x\right) = \cos x$⁠. Lesson: properties of definite integrals

5. Integrate the constant

   $$I = 0 - \log 2\cdot\frac{\pi}{2} = -\frac{\pi}{2}\log 2 \approx -1.0887930 .$$

   The book writes this as $\frac{\pi}{2}\log\frac12$⁠. They are equal, because $\log\frac12 = \log 1 - \log 2 = -\log 2$⁠.

   The constant $\log 2$ integrates to $\log 2$ times the interval length $\frac{\pi}{2}$⁠. The answer is negative because the integrand, $\log\frac{\tan x}{2}$⁠, is below the axis on most of the interval (it is negative wherever $\tan x < 2$⁠).
   $\int_a^b k\,dx = k(b - a)$⁠; $\log\frac{p}{q} = \log p - \log q$⁠, $\log 1 = 0$⁠. Basics: logarithms

6. The picture: the cancellation in Step 4

   The two cancelling terms combine into $\log\sin x - \log\cos x = \log\tan x$⁠. Replacing $x$ by $\frac{\pi}{2} - x$ gives $\log\cot x = -\log\tan x$⁠, so the graph of $\log\tan x$ is unchanged by a half turn about $\left(\frac{\pi}{4}, 0\right)$⁠, the way an odd function behaves about the origin.

   This is the same cancellation as P7(ii) for odd functions, centred at $x = \frac{\pi}{4}$ instead of 0. The equation $K = -K$⁠, with $K = \int_0^{\pi/2}\log\tan x\,dx$⁠, forces $K = 0$⁠.
   $\tan\left(\frac{\pi}{2} - x\right) = \cot x = \frac{1}{\tan x}$⁠, and $\log\frac1u = -\log u$⁠. Basics: symmetry and mirror images

7. A second route: Example 34 and P6

   Let $J = \int_0^{\pi/2}\log\sin x\,dx = -\frac{\pi}{2}\log 2$ (Example 34). For $\log\sin 2x$⁠, put $t = 2x$⁠, $dx = \frac12\,dt$⁠, limits $0$ to $\pi$⁠:

   $$\begin{aligned}\int_0^{\frac{\pi}{2}}\log\sin 2x\,dx &= \frac12\int_0^{\pi}\log\sin t\,dt \\ &= \frac12\cdot 2J = J ,\end{aligned}$$

   by P6 with $2a = \pi$⁠, since $\log\sin(\pi - t) = \log\sin t$⁠. So

   $$I = 2J - J = J = -\frac{\pi}{2}\log 2 .$$
   Two independent routes agree. The P6 condition $f(2a - t) = f(t)$ holds because $\sin(\pi - t) = \sin t$⁠: the graph of $\log\sin t$ on $[0, \pi]$ is symmetric about $t = \frac{\pi}{2}$⁠.
   Substitution in a definite integral (Section 7.9); P6, Section 7.10; Example 34.

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