Exercise 7.10 · Q8: $\int_0^{\pi/4}\log(1+\tan x)\,dx$

Basics you need

Solution

1. The integrand is defined and continuous on $\left[0, \frac{\pi}{4}\right]$

   On $\left[0, \frac{\pi}{4}\right]$⁠, $\tan x$ is continuous and increases from 0 to 1, so

   $$1 \le 1 + \tan x \le 2 .$$

   The logarithm of a number between 1 and 2 is defined, and $\log(1 + \tan x)$ is continuous, with values from 0 to $\log 2$⁠. Label

   $$I = \int_0^{\frac{\pi}{4}}\log(1 + \tan x)\,dx \qquad (1)$$
   We need the integrand continuous on the closed interval for P4 and so that $I$ is a finite number (Step 5 subtracts it). We also need $1 + \tan x > 0$⁠, so that the logarithm exists and the log law in Step 4 is allowed.
   $\tan$ is continuous on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, which contains $\left[0, \frac{\pi}{4}\right]$⁠; $\log$ is continuous on $(0, \infty)$⁠. Basics: logarithms

2. Apply P4 with $a = \frac{\pi}{4}$

   $$I = \int_0^{\frac{\pi}{4}}\log\left(1 + \tan\left(\tfrac{\pi}{4} - x\right)\right)dx \qquad (2)$$
   P4 holds for every function continuous on $[0, a]$ (Step 1). As $x$ runs over $\left[0, \frac{\pi}{4}\right]$⁠, so does $\frac{\pi}{4} - x$⁠, so $\tan\left(\frac{\pi}{4} - x\right)$ is defined and lies in $[0, 1]$⁠. The point is what this new integrand simplifies to.
   P4, Section 7.10. Lesson: properties of definite integrals

3. Simplify $1 + \tan\left(\frac{\pi}{4} - x\right)$

   By the difference formula with $\tan\frac{\pi}{4} = 1$⁠:

   $$\tan\left(\tfrac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} .$$

   So

   $$\begin{aligned}&1 + \tan\left(\tfrac{\pi}{4} - x\right) \\ &= \frac{(1 + \tan x) + (1 - \tan x)}{1 + \tan x} \\ &= \frac{2}{1 + \tan x} .\end{aligned}$$
   This is the heart of the problem: after P4, the expression inside the logarithm becomes 2 divided by the original expression. The formula is valid because its denominator $1 + \tan x \ge 1$ is never 0 here.
   $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$ with $A = \frac{\pi}{4}$⁠, $B = x$⁠. Basics: trigonometric identities

4. Split the logarithm

   Since $2 > 0$ and $1 + \tan x > 0$⁠:

   $$\log\frac{2}{1 + \tan x} = \log 2 - \log(1 + \tan x) .$$

   Putting this into (2):

   $$\begin{aligned}I &= \int_0^{\frac{\pi}{4}}\big(\log 2 - \log(1 + \tan x)\big)dx \\ &= \log 2\int_0^{\frac{\pi}{4}}dx - \int_0^{\frac{\pi}{4}}\log(1 + \tan x)\,dx \\ &= \frac{\pi}{4}\log 2 - I .\end{aligned}$$
   The quotient law of logarithms needs both numbers positive, which holds. The constant $\log 2$ integrates to $\log 2$ times the length $\frac{\pi}{4}$ of the interval, and the remaining integral is exactly (1), our $I$⁠.
   $\log\frac{p}{q} = \log p - \log q$ for $p, q > 0$⁠. Basics: logarithms

5. Solve for $I$

   $$\begin{aligned}2I &= \frac{\pi}{4}\log 2, \\ I &= \frac{\pi}{8}\log 2 \approx 0.2721983 .\end{aligned}$$
   Adding $I$ to both sides is allowed because $I$ is a finite number (Step 1). The value is positive and less than $\frac{\pi}{4}\log 2 \approx 0.5443965$⁠, as it must be: the integrand lies between 0 and $\log 2$ on an interval of length $\frac{\pi}{4}$⁠.
   $\log 2 \approx 0.6931472$ and $\frac{\pi}{8} \approx 0.3926991$⁠.

6. The picture

   The curve passes through $\left(\frac{\pi}{8}, \frac12\log 2\right)$ and is symmetric under a half turn about that point, so it cuts the rectangle into two equal halves. That is the equation $2I = \frac{\pi}{4}\log 2$ in picture form.
   $\tan\frac{\pi}{8} = \sqrt2 - 1$⁠, so $1 + \tan\frac{\pi}{8} = \sqrt2$ and $\log\sqrt2 = \frac12\log 2$⁠. Basics: symmetry and mirror images

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