Exercise 7.2 · Q14: $\int\frac{dx}{x(\log x)^m}$⁠, $m \ne 1$

Basics you need

Solution

1. Spot a function together with its derivative

   Write the integrand as a product:

   $$\frac{1}{x(\log x)^m} = (\log x)^{-m}\cdot\frac{1}{x}.$$

   The factor $\frac1x$ is the derivative of $\log x$⁠, and the rest is a function (a power) of $\log x$⁠.

   This is the pattern $f(g(x))\,g'(x)$ with inner function $g(x) = \log x$ and outer function $f(t) = t^{-m}$⁠. Seeing that pattern is the whole skill in substitution: the derivative of the inner function must be present (up to a constant factor) so that it can be absorbed into $dt$⁠.
   $\frac{1}{a^m} = a^{-m}$ for $a \ne 0$ (laws of indices). Basics: powers

2. Substitute $t = \log x$

   Let $t = \log x$⁠. Differentiating, $\frac{dt}{dx} = \frac1x$⁠, so

   $$dt = \frac{dx}{x}.$$

   Replace $\log x$ by $t$ and $\frac{dx}{x}$ by $dt$⁠:

   $$\int (\log x)^{-m}\,\frac{dx}{x} = \int t^{-m}\,dt.$$
   Every piece of the old integral has been replaced: there is no $x$ left, which is the sign that the substitution is complete. The substitution is legitimate because $x > 0$⁠, so $\log x$ is defined and differentiable there (Section 7.3.1, change of variable).
   Section 7.3.1: $\int f(x)\,dx$ can be changed into an integral in $t$ by a substitution, with $dx$ rewritten through the derivative. Lesson: integration by substitution

3. Integrate the power of $t$

   Use the power formula with $n = -m$⁠. It needs $n \ne -1$⁠, that is, $-m \ne -1$⁠, that is, $m \ne 1$⁠, which the question gives us:

   $$\begin{aligned} \int t^{-m}\,dt &= \frac{t^{-m+1}}{-m+1} + C \\ &= \frac{t^{1-m}}{1-m} + C. \end{aligned}$$
   This is why the question says $m \ne 1$⁠. For $m = 1$ the formula would divide by $1 - m = 0$⁠, and the integral is really $\int\frac{dt}{t} = \log|t| + C = \log|\log x| + C$⁠, a different kind of answer (that is Question 3's pattern).
   Standard formula (i), Section 7.2: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$⁠, $n \ne -1$⁠. Basics: the power rule

4. Return to $x$

   Put back $t = \log x$⁠:

   $$\int \frac{dx}{x(\log x)^m} = \frac{(\log x)^{1-m}}{1-m} + C.$$
   The question is about a function of $x$⁠, so the answer must be in $x$⁠. Leaving $t$ in the answer is a common slip.

5. Check: differentiate the answer back

   Let $F(x) = \dfrac{(\log x)^{1-m}}{1-m}$⁠. By the chain rule (outer: the power $u^{1-m}$⁠; inner: $u = \log x$⁠):

   $$\begin{aligned} F'(x) &= \frac{(1-m)(\log x)^{-m}}{1-m}\cdot\frac{1}{x} \\ &= (\log x)^{-m}\cdot\frac{1}{x} \\ &= \frac{1}{x(\log x)^m}. \end{aligned}$$

   This is the integrand, so the answer is correct.

   The factor $1 - m$ brought down by the power rule cancels the $1 - m$ we divided by. That cancellation is exactly what the integration step was designed to produce.
   Chain rule, Theorem 4, Section 5.3.1; $\frac{d}{du}u^k = k\,u^{k-1}$⁠. Basics: integration undoes differentiation

6. Where the answer is valid

   The integrand needs $x > 0$ (for $\log x$⁠), and the power $(\log x)^m$ must make sense and be nonzero:

   • For every $m \ne 1$⁠, the answer is valid for $x > 1$⁠, where $\log x > 0$⁠, so any real power of $\log x$ is defined.
   • When $m$ is an integer, negative values of $\log x$ are also allowed, so the answer also holds on $0 < x < 1$⁠. (For $m = 2, 3, \dots$ the point $x = 1$ is excluded, since $\log 1 = 0$ is in a denominator; for $m = 0, -1, -2, \dots$ there is no such denominator and the answer holds for every $x > 0$⁠.)
   • For a fraction such as $m = \frac12$⁠, $(\log x)^{1/2}$ needs $\log x \ge 0$⁠, so only $x > 1$ works.
   The book states only $x > 0$⁠, following its Note in Section 7.2 that intervals are normally not mentioned. A careful answer adds that real powers of a negative number are not always defined. Where the domain falls into two pieces, $(0,1)$ and $(1,\infty)$⁠, the constant $C$ may even be chosen differently on each piece.

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