Deriving the Logarithmic-Power Recurrence

Why a Recurrence Helps

A power of a logarithm is not a product of two different types of functions, so a familiar strategy is to treat the logarithm itself as the factor to differentiate and $1$ as the factor to integrate. This choice has a useful effect: differentiating $(\ln x)^n$ lowers its power, while integrating $1$ leaves the simple factor $x$.

The resulting relation expresses an antiderivative with logarithmic power $n$ using one with power $n-1$. Such a relation is called a reduction formula or recurrence. To state it precisely, we must specify the domain, the allowed values of $n$, and how we handle the arbitrary constants in indefinite integration.

Domain, Index, and Base Case

Throughout, $x$ is a real variable. The real logarithm $\ln x$ is defined only for $x>0$, so the integrand $(\ln x)^n$ and the recurrence below are considered on the interval $(0,\infty)$. We take $n$ to be a nonnegative integer. The finite reduction proceeds from a positive integer $n$ to $n-1$, and eventually reaches $n=0$.

Definition: For each nonnegative integer $n$, let $I_n(x)$ denote a chosen antiderivative of $(\ln x)^n$ on $(0,\infty)$. We select the base representative $I_0(x)=x$. For $n\geq 1$, the recurrence will define the representative $I_n$. The full family of antiderivatives is $I_n(x)+C$, where $C$ is any real constant.

This convention separates two related ideas. The indefinite integral $\int(\ln x)^n\,dx$ represents all antiderivatives, which differ by constants on the interval. A recurrence between chosen representatives can be written as an exact equality once their constants are fixed. The base case is immediate because $(\ln x)^0=1$: an antiderivative of $1$ is $x$.

I_0(x)=\int 1\,dx=x

Deriving the Recurrence

Integration by parts states that $\int u\,dv=uv-\int v\,du$, provided $u$ is differentiable and $v$ is an antiderivative of $dv$ on the interval in question. For this integrand, choose $u=(\ln x)^n$ and $dv=dx$. Since $x>0$, the chain rule gives the derivative of $u$, and $v=x$.

Choose the parts.
Set $u=(\ln x)^n$ and $dv=dx$. Then $v=x$.

Differentiate the logarithmic power.
By the chain rule, $du=n(\ln x)^{n-1}\frac{1}{x}\,dx$.

Substitute in integration by parts.
The factor $x$ in $uv$ cancels the $1/x$ in $du$, leaving an integral with logarithmic power $n-1$.

\begin{aligned} \int(\ln x)^n\,dx &=x(\ln x)^n-\int x\left(n(\ln x)^{n-1}\frac{1}{x}\right)\,dx\\ &=x(\ln x)^n-n\int(\ln x)^{n-1}\,dx. \end{aligned}

Thus, with the chosen representatives, the recurrence is

Recurrence: For every integer $n\geq 1$ and every $x>0$, define $I_0(x)=x$ and $$ I_n(x)=x(\ln x)^n-nI_{n-1}(x). $$ Then $I_n'(x)=(\ln x)^n$. Every antiderivative of $(\ln x)^n$ is $I_n(x)+C$.

Why the Relation Works

The cancellation in the integration-by-parts calculation explains the shape of the recurrence. The derivative of $(\ln x)^n$ contributes a factor $1/x$; the integrated factor $v=x$ cancels it. The remaining logarithmic power is exactly one lower, and the coefficient $n$ comes from differentiating a power.

We can also verify the recurrence directly, which confirms that it produces an antiderivative. Suppose $I_{n-1}'(x)=(\ln x)^{n-1}$. Differentiate the right-hand side of the recurrence using the product rule and the chain rule:

\begin{aligned} \frac{d}{dx}\left[x(\ln x)^n-nI_{n-1}(x)\right] &=(\ln x)^n +x\left(n(\ln x)^{n-1}\frac{1}{x}\right) -n(\ln x)^{n-1}\\ &=(\ln x)^n+n(\ln x)^{n-1}-n(\ln x)^{n-1}\\ &=(\ln x)^n. \end{aligned}

The base case has derivative $I_0'(x)=1=(\ln x)^0$. Therefore, by induction on the nonnegative integer $n$, each recursively defined $I_n$ is indeed an antiderivative of $(\ln x)^n$ throughout $(0,\infty)$.

Worked Examples

Worked Example: The First Reduction Step

Use integration by parts to obtain the recurrence instance for $n=1$, then check the result by differentiation.

With $u=\ln x$ and $dv=dx$, we have $du=\frac{1}{x}\,dx$ and $v=x$. Therefore,

\begin{aligned} I_1(x) &=x\ln x-\int x\frac{1}{x}\,dx\\ &=x\ln x-\int 1\,dx\\ &=x\ln x-x, \end{aligned}

where the chosen base representative is $I_0(x)=x$. To check, differentiate the result:

\frac{d}{dx}(x\ln x-x) =\ln x+1-1 =\ln x.

Thus $I_1$ has the required derivative for every $x>0$. The full family of antiderivatives is $x\ln x-x+C$.

Worked Example: Applying the Recurrence at $n=2$

Use the recurrence to express $I_2$ in terms of the base case, and verify the resulting antiderivative.

Set $n=2$ in $I_n=x(\ln x)^n-nI_{n-1}$. This gives

\begin{aligned} I_2(x) &=x(\ln x)^2-2I_1(x)\\ &=x(\ln x)^2-2(x\ln x-x)\\ &=x(\ln x)^2-2x\ln x+2x. \end{aligned}

Differentiate each term. The product rule gives $\frac{d}{dx}[x(\ln x)^2]=(\ln x)^2+2\ln x$, and $\frac{d}{dx}[-2x\ln x]=-2\ln x-2$. Hence

\begin{aligned} I_2'(x) &=(\ln x)^2+2\ln x-2\ln x-2+2\\ &=(\ln x)^2. \end{aligned}

The derivative matches the integrand, so the recurrence has produced the correct representative on $x>0$.

Worked Example: Reducing the Third Power

Apply the recurrence once more, now with $n=3$, and check the resulting expression.

First substitute $I_2(x)=x(\ln x)^2-2x\ln x+2x$ into $I_3=x(\ln x)^3-3I_2$:

\begin{aligned} I_3(x) &=x(\ln x)^3-3\left[x(\ln x)^2-2x\ln x+2x\right]\\ &=x(\ln x)^3-3x(\ln x)^2+6x\ln x-6x. \end{aligned}

Differentiate the four terms individually:

\begin{aligned} I_3'(x) &=\left[(\ln x)^3+3(\ln x)^2\right] +\left[-3(\ln x)^2-6\ln x\right]\\ &\quad+\left[6\ln x+6\right]-6\\ &=(\ln x)^3. \end{aligned}

The intermediate terms cancel, leaving the required integrand. This is also the pattern in the general verification: the extra derivative term from $x(\ln x)^n$ cancels the derivative of the lower-index term.

A Common Mistake: Losing the Chain-Rule Factor

When differentiating a logarithmic power, it is not enough to differentiate the outer power. The inner function $\ln x$ has derivative $1/x$. Omitting this factor changes the integral produced by parts.

For instance, an incorrect calculation might claim $d(\ln x)=dx$, and then write

\int\ln x\,dx \overset{\text{wrong}}{=}x\ln x-\int x\,dx =x\ln x-\frac{x^2}{2}.

This answer fails the derivative check:

\frac{d}{dx}\left(x\ln x-\frac{x^2}{2}\right) =\ln x+1-x,

which is not $\ln x$ for all $x>0$. The correction is $d(\ln x)=\frac{1}{x}\,dx$. Then the integration-by-parts remainder is $\int x\frac{1}{x}\,dx=\int 1\,dx$, giving $x\ln x-x$, whose derivative is $\ln x$.

Takeaway: The factor $1/x$ from differentiating $\ln x$ is essential. It cancels the $x$ from $v=x$; it does not disappear before that cancellation is shown.

What the Recurrence Does—and Does Not—Say

For each positive integer $n$, the recurrence lowers the logarithmic power by one. Repeating it reaches the base case $I_0=x$, so it provides a finite procedure for constructing a chosen antiderivative. The recurrence is an identity for $x>0$, where the real logarithm and all the derivatives used in the argument are defined.

The integer condition matters for this finite reduction: the index decreases through $n,n-1,\ldots,0$. The derivation does not assert a recurrence for arbitrary non-integer indices, nor does it define real logarithms for $x\leq 0$. Also, because this is an indefinite integral, there are no endpoint terms. The arbitrary constant belongs to the general antiderivative and is not a boundary contribution.

The main value of the formula is structural. It turns the antiderivative problem at index $n$ into a simpler one at index $n-1$, with an explicit term $x(\ln x)^n$. The base case and the derivative check ensure that this reduction is anchored correctly rather than being only a formal manipulation.

Check Your Understanding

Use the domain and recurrence stated above when answering these questions.

Why must the real-variable recurrence be restricted to $x>0$?
What are $u$, $dv$, $du$, and $v$ in the integration-by-parts derivation?
State the recurrence for $I_n$ and its base case.
Why does the factor $x$ cancel the $1/x$ in the derivative of $(\ln x)^n$?
What goes wrong if the chain-rule factor in $d(\ln x)$ is omitted?

Deriving the Logarithmic-Power Recurrence

What You'll Learn

Why a Recurrence Helps

Domain, Index, and Base Case

Deriving the Recurrence

Why the Relation Works