Example 38: $\int\left[\log(\log x)+\frac{1}{(\log x)^2}\right]dx$

Basics you need

Solution

1. Where is the integrand defined?

   $\log(\log x)$ needs $\log x > 0$⁠, that is $x > 1$⁠. For such $x$⁠, $\log x \ne 0$⁠, so $\frac{1}{(\log x)^2}$ is defined too. The integrand is continuous on $(1, \infty)$⁠, and we work on that interval.

   A logarithm of a logarithm needs two conditions: $x > 0$ for the inner $\log x$⁠, and $\log x > 0$ for the outer one. The second is the stronger. (The term $\frac{1}{(\log x)^2}$ alone would also make sense for $0 < x < 1$⁠, but the sum needs both terms.)
   $\log x > 0 \iff x > 1$⁠, since $\log 1 = 0$ and $\log$ is increasing.

2. Split the integral

   $$I = \int \log(\log x)\,dx + \int \frac{dx}{(\log x)^2} .$$
   The integral of a sum is the sum of the integrals (Property III, Section 7.2.1). We will work on the first integral and leave the second untouched, waiting for the cancellation.

3. Parts on $\int \log(\log x)\,dx$⁠, with 1 as the second function

   Write $\log(\log x) = \log(\log x)\cdot 1$⁠. First function $f = \log(\log x)$⁠, second function $g = 1$⁠. Then $\int g\,dx = x$ and, by the chain rule,

   $$f'(x) = \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x} .$$

   Integration by parts gives

   $$\begin{aligned}\int\log(\log x)\,dx &= x\log(\log x) - \int\frac{1}{x\log x}\cdot x\,dx\\ &= x\log(\log x) - \int\frac{dx}{\log x} .\end{aligned}$$

   So

   $$I = x\log(\log x) - \int\frac{dx}{\log x} + \int\frac{dx}{(\log x)^2} \qquad (1)$$
   We do not know an integral of $\log(\log x)$⁠, but its derivative is simple. Taking 1 as the second function moves the work onto that derivative, and the factor $x$ from $\int 1\,dx$ cancels the $\frac1x$ in it.
   Chain rule: outer $\log u$ (derivative $\frac1u$⁠), inner $u = \log x$ (derivative $\frac1x$⁠). Lesson: integration by parts

4. Parts again on $\int\frac{dx}{\log x}$

   Write $\frac{1}{\log x} = \frac{1}{\log x}\cdot 1$⁠. First function $f = \frac{1}{\log x} = (\log x)^{-1}$⁠, second $g = 1$⁠. By the chain rule,

   $$f'(x) = -(\log x)^{-2}\cdot\frac1x = -\frac{1}{x(\log x)^2} .$$

   So

   $$\begin{aligned}\int\frac{dx}{\log x} &= \frac{x}{\log x} - \int x\left(-\frac{1}{x(\log x)^2}\right)dx\\ &= \frac{x}{\log x} + \int\frac{dx}{(\log x)^2} \qquad (2)\end{aligned}$$
   The same trick again: 1 as the second function. It produces the integral $\int\frac{dx}{(\log x)^2}$⁠, which is exactly the one already sitting in (1).
   $\frac{d}{dx}u^{-1} = -u^{-2}\,\frac{du}{dx}$ with $u = \log x$⁠.

5. Substitute (2) into (1): the hard integrals cancel

   $$\begin{aligned}I &= x\log(\log x) - \frac{x}{\log x} - \int\frac{dx}{(\log x)^2} + \int\frac{dx}{(\log x)^2}\\ &= x\log(\log x) - \frac{x}{\log x} + C .\end{aligned}$$
   The minus sign in front of $\int\frac{dx}{\log x}$ in (1) turns the $+\int\frac{dx}{(\log x)^2}$ of (2) into a minus, and it cancels the given second term. Each indefinite integral stands for a family of functions differing by constants, so "cancelling" them leaves an arbitrary constant: that is the $C$⁠. It can be shown that $\int\frac{dx}{\log x}$ has no formula in terms of the functions of this course, which is why the cancellation is the only way through.
   Basics: integration undoes differentiation

6. Check: $\frac{d}{dx}$ of our answer

   Product rule on $x\log(\log x)$⁠:

   $$\frac{d}{dx}\big[x\log(\log x)\big] = \log(\log x) + x\cdot\frac{1}{x\log x} = \log(\log x) + \frac{1}{\log x} .$$

   Quotient rule on $\frac{x}{\log x}$⁠:

   $$\frac{d}{dx}\left[\frac{x}{\log x}\right] = \frac{\log x - x\cdot\frac1x}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2} .$$

   Subtracting:

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

   This is the integrand.

   The two $\frac{1}{\log x}$ terms cancel in the derivative, mirroring the cancellation of the integrals in Step 5.

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