Example 34: $\int_0^{\pi/2}\log\sin x\,dx$

Basics you need

Solution

1. The trouble at $x = 0$⁠, and why $I$ is still a finite number

   On $0 < x \le \frac{\pi}{2}$ we have $0 < \sin x \le 1$⁠, so $\log \sin x$ is defined, continuous and $\le 0$⁠. But at $x = 0$⁠, $\sin 0 = 0$ has no logarithm, and as $x \to 0^+$⁠, $\log\sin x \to -\infty$⁠. So the integrand is not continuous on the closed interval $[0, \frac{\pi}{2}]$⁠, and Theorem 2 of Section 7.8.3 does not apply directly (Remark (iv) there asks for a function continuous on $[a, b]$⁠). Such an integral is called improper: it means the limit of $\int_\varepsilon^{\pi/2}\log\sin x\,dx$ as $\varepsilon \to 0^+$⁠. The textbook does not raise this point, but the method below needs $I$ to be a finite number, so we check it.

   The graph of $\sin$ is concave on $[0, \frac{\pi}{2}]$⁠, so it lies above its chord from $(0, 0)$ to $(\frac{\pi}{2}, 1)$⁠:

   $$\sin x \ge \frac{2x}{\pi} \qquad \left(0 \le x \le \frac{\pi}{2}\right).$$

   Since $\log$ is increasing, for $0 < x \le \frac{\pi}{2}$⁠:

   $$\log\frac{2x}{\pi} \le \log\sin x \le 0 .$$

   With $u = \frac{2x}{\pi}$ (so $dx = \frac{\pi}{2}\,du$⁠) and $\int \log u\,du = u\log u - u$⁠:

   $$\int_0^{\pi/2}\log\frac{2x}{\pi}\,dx = \frac{\pi}{2}\Big[u\log u - u\Big]_0^1 = -\frac{\pi}{2},$$

   using $u\log u \to 0$ as $u \to 0^+$⁠. So the negative function $\log \sin x$ is squeezed between $\log\frac{2x}{\pi}$⁠, whose integral is finite, and 0. Its integral therefore converges, and

   $$-\frac{\pi}{2} \le I \le 0 .$$
   This matters in Step 6, where we subtract $I$ from both sides of an equation. That is only allowed for a finite number: "$\infty = \infty - c$" tells us nothing. The same bound (reflected or rescaled) shows that every other integral below, of $\log\cos x$⁠, $\log\sin 2x$ and $\log\sin t$ on $[0, \pi]$⁠, is also finite. The properties P0 to P7 and the substitution rule then hold for them too, because each one can be applied on a slightly shorter closed interval, where everything is continuous, before letting the cut-off shrink to 0.
   Concave means $\frac{d^2}{dx^2}\sin x = -\sin x \le 0$ on $[0, \frac{\pi}{2}]$⁠: the curve bends downwards, so it stays above any chord. The antiderivative $u\log u - u$ of $\log u$ is found by parts in Section 7.6 (check: its derivative is $\log u + 1 - 1$⁠).

2. P4: the same integral with $\cos$ in place of $\sin$

   P4 with upper limit $a = \frac{\pi}{2}$ replaces $x$ by $\frac{\pi}{2} - x$⁠:

   $$I = \int_0^{\pi/2}\log\sin\left(\frac{\pi}{2}-x\right)dx = \int_0^{\pi/2}\log\cos x\,dx .$$
   Replacing $x$ by $\frac{\pi}{2} - x$ reads the interval $[0, \frac{\pi}{2}]$ backwards, so the area does not change. On this interval it turns $\sin$ into $\cos$⁠: the graph of $\log\cos x$ is the mirror image of the graph of $\log\sin x$ in the line $x = \frac{\pi}{4}$ (see the figure in Step 6).
   $\sin\left(\frac{\pi}{2} - x\right) = \cos x$⁠. Basics: symmetry and mirror images

3. Add the two forms of $I$

   $$2I = \int_0^{\pi/2}\big(\log\sin x + \log\cos x\big)\,dx = \int_0^{\pi/2}\log(\sin x\cos x)\,dx .$$

   Now the book adds and subtracts $\log 2$ and asks "Why?". The reason is the double-angle formula: $2\sin x\cos x = \sin 2x$⁠. So

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

   Integrating each term (the constant $\log 2$ integrates to $\frac{\pi}{2}\log 2$ over an interval of length $\frac{\pi}{2}$⁠):

   $$2I = \int_0^{\pi/2}\log\sin 2x\,dx - \frac{\pi}{2}\log 2 .$$
   For $0 < x < \frac{\pi}{2}$ both $\sin x$ and $\cos x$ are positive, so $\log\sin x + \log\cos x = \log(\sin x\cos x)$ is a correct use of the product law. We want a single sine, because a single sine can be turned back into the original integral; the factor $2$ is exactly what $\sin x\cos x$ is missing to become $\sin 2x$⁠.
   $\log(pq) = \log p + \log q$ for $p, q > 0$ (here $p = 2$⁠, $q = \sin x\cos x$⁠). Basics: logarithms

4. Substitute $t = 2x$ in the first integral

   Put $t = 2x$⁠. Then $dt = 2\,dx$⁠, so $dx = \frac{1}{2}\,dt$⁠. When $x = 0$⁠, $t = 0$⁠; when $x = \frac{\pi}{2}$⁠, $t = \pi$⁠. Hence

   $$\int_0^{\pi/2}\log\sin 2x\,dx = \frac{1}{2}\int_0^{\pi}\log\sin t\,dt ,$$

   and

   $$2I = \frac{1}{2}\int_0^{\pi}\log\sin t\,dt - \frac{\pi}{2}\log 2 .$$
   The substitution makes the argument of $\sin$ a single variable again, at the price of doubling the interval to $[0, \pi]$⁠. The map $x \mapsto 2x$ is increasing and takes $[0, \frac{\pi}{2}]$ exactly onto $[0, \pi]$⁠, so it is a legitimate change of variable on the whole interval. On $0 < t < \pi$⁠, $\sin t > 0$⁠, so $\log\sin t$ is defined there.
   Section 7.9: change the limits along with the variable, then there is no need to substitute back. Basics: substitution

5. P6 folds $[0, \pi]$ back onto $[0, \frac{\pi}{2}]$

   Take $f(t) = \log\sin t$ and $2a = \pi$⁠, so $a = \frac{\pi}{2}$⁠. The condition of P6 holds:

   $$f(\pi - t) = \log\sin(\pi - t) = \log\sin t = f(t) .$$

   So P6 gives

   $$\int_0^{\pi}\log\sin t\,dt = 2\int_0^{\pi/2}\log\sin t\,dt = 2I ,$$

   where the last equality is just P0: the name of the variable does not matter. Therefore

   $$2I = \frac{1}{2}\cdot 2I - \frac{\pi}{2}\log 2 = I - \frac{\pi}{2}\log 2 .$$
   The graph of $\sin t$ on $[0, \pi]$ is symmetric about $t = \frac{\pi}{2}$⁠, so the two halves $[0, \frac{\pi}{2}]$ and $[\frac{\pi}{2}, \pi]$ contribute equally. The left half is our original integral. This is the moment the integral "reproduces itself".
   $\sin(\pi - t) = \sin t$ (supplementary angles). P0: $\int_a^b f(t)\,dt = \int_a^b f(x)\,dx$⁠.

6. Solve for $I$

   Subtract $I$ from both sides (allowed because $I$ is a finite number, Step 1):

   $$I = -\frac{\pi}{2}\log 2 \approx -1.088793 .$$

   This lies between $-\frac{\pi}{2} \approx -1.570796$ and 0, as Step 1 promised. It is negative because $\log\sin x \le 0$ on the whole interval: the region lies below the $x$-axis.

   The final answer can also be read as: the average value of $\log\sin x$ over $[0, \frac{\pi}{2}]$ is $\frac{I}{\pi/2} = -\log 2$⁠, the same as $\log\frac{1}{2}$⁠.
   Basics: area as a sum of thin strips (signed area)

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