Definite Logarithmic-Power Recurrences on [1,b]

From an Antiderivative Recurrence to a Definite Integral

An antiderivative recurrence expresses one integral in terms of a simpler one, usually with a lower power. For a definite integral, the same integration-by-parts step also produces endpoint terms. Those terms matter: they determine the recurrence’s exact form and its initial value.

Here the interval begins at $1$, where $\ln 1=0$. This makes the lower endpoint vanish for every positive power of the logarithm, but not for the zeroth power. Keeping that distinction clear is the central point of the method.

Definition: For $b\ge 1$ and an integer $n\ge 0$, define $$ I_n(b)=\int_1^b (\ln x)^n\,dx. $$ For $n=0$, the integrand is interpreted as $1$, so $I_0(b)=\int_1^b 1\,dx$.

The condition $b\ge 1$ keeps the interval in the positive domain of the natural logarithm and gives the stated orientation. If $b=1$, every integral is zero. The recurrence below is useful for $b>1$, and remains valid at $b=1$ as well.

Deriving the Definite Recurrence

For $n\ge 1$, apply integration by parts to $I_n(b)$, taking $u=(\ln x)^n$ and $dv=dx$. Then $du=n(\ln x)^{n-1}\frac{1}{x}\,dx$ and $v=x$. The definite integration-by-parts formula is $\int_a^b u\,dv=[uv]_a^b-\int_a^b v\,du$. Thus

I_n(b) = \left[x(\ln x)^n\right]_1^b - n\int_1^b (\ln x)^{n-1}\,dx.

At the upper endpoint, the boundary expression is $b(\ln b)^n$. At the lower endpoint it is $1(\ln 1)^n=0$, because $n\ge1$. The remaining integral is $I_{n-1}(b)$. Therefore:

Theorem: For $b\ge1$ and every integer $n\ge1$, $$ I_n(b)=b(\ln b)^n-nI_{n-1}(b). $$ The starting value is $$ I_0(b)=b-1. $$

The starting value follows directly from integrating $1$. It is not obtained by substituting $n=0$ into the positive-power recurrence. At $n=0$, the lower boundary term is $1(\ln 1)^0=1$, not zero: the zeroth power in the integrand is $1$. This is why the base case must be stated separately.

Why the Endpoint Term Takes This Form

The recurrence is not just an algebraic rule for changing a power. It records exactly what integration by parts does on the chosen interval. The factor $x$ from the antiderivative of $1$ remains in the boundary expression, and evaluating that expression at both endpoints gives

\left[x(\ln x)^n\right]_1^b = b(\ln b)^n-1(\ln 1)^n = b(\ln b)^n \qquad(n\ge1).

The lower endpoint contributes zero only for a positive logarithmic power. Meanwhile, $I_{n-1}(b)$ is the same kind of definite integral with its exponent lowered by one. These two facts give the recurrence. The sign before $nI_{n-1}(b)$ is negative because the integration-by-parts formula subtracts the integral of $v\,du$.

Choose the parts.
Set $u=(\ln x)^n$ and $dv=dx$, for $n\ge1$.

Differentiate and integrate.
Then $du=n(\ln x)^{n-1}\frac{1}{x}\,dx$ and $v=x$.

Evaluate the boundary term.
The value at $b$ is $b(\ln b)^n$, and the value at $1$ is zero.

Identify the lower-power integral.
The remaining integral is $I_{n-1}(b)$, multiplied by $-n$.

Worked Example: Evaluate $I_1(b)$ for General $b$

Use the recurrence with $n=1$, together with $I_0(b)=b-1$:

\begin{aligned} I_1(b) &=b\ln b-I_0(b)\\ &=b\ln b-(b-1)\\ &=b\ln b-b+1. \end{aligned}

To check this result independently, an antiderivative of $\ln x$ is $F(x)=x\ln x-x$, since

F'(x)=\ln x+1-1=\ln x.

Evaluating at the endpoints gives

F(b)-F(1) =(b\ln b-b)-(1\cdot 0-1) =b\ln b-b+1.

This agrees with the recurrence result. In particular, the $+1$ comes from subtracting the lower endpoint value $F(1)=-1$.

Worked Example: Compute $I_3(e)$

Set $b=e$, so $\ln e=1$. First compute the base value:

$$ I_0(e)=e-1. $$

For $n=1$, the recurrence gives

$$ I_1(e)=e(1)-(e-1)=1. $$

For $n=2$, it gives

$$ I_2(e)=e(1)^2-2I_1(e)=e-2. $$

Finally, for $n=3$,

$$ I_3(e)=e(1)^3-3I_2(e) =e-3(e-2) =e-3e+6 =6-2e. $$

For a check, define

F(x)=x\left((\ln x)^3-3(\ln x)^2+6\ln x-6\right).

Writing $L=\ln x$, differentiation gives

\begin{aligned} F'(x) &=L^3-3L^2+6L-6+3L^2-6L+6\\ &=L^3\\ &=(\ln x)^3. \end{aligned}

At $x=e$, $F(e)=e(1-3+6-6)=-2e$. At $x=1$, $F(1)=1(0-0+0-6)=-6$. Therefore

$$ F(e)-F(1)=-2e-(-6)=6-2e, $$

which matches the recurrence calculation.

Worked Example: Compute $I_2(2)$

Write $L=\ln 2$. Start with the base value and reduce the power one step at a time:

I_0(2)=2-1=1, \qquad I_1(2)=2L-I_0(2)=2L-1.

Using $n=2$,

\begin{aligned} I_2(2) &=2L^2-2I_1(2)\\ &=2L^2-2(2L-1)\\ &=2L^2-4L+2. \end{aligned}

To check, an antiderivative of $(\ln x)^2$ is $G(x)=x((\ln x)^2-2\ln x+2)$. If $L_x=\ln x$, then

$$ G'(x)=L_x^2-2L_x+2+2L_x-2=L_x^2. $$

The endpoint values are $G(2)=2(L^2-2L+2)$ and $G(1)=2$. Their difference is

$$ G(2)-G(1)=2L^2-4L+4-2=2L^2-4L+2, $$

as required. Since $\ln x>0$ for $1<x\le2$, the integral is positive, consistent with this value.

Using the Recurrence Efficiently

To find $I_n(b)$, begin at $I_0(b)=b-1$, then apply the recurrence successively for $1,2,\ldots,n$. Each step uses the same endpoint parameter $b$; only the power and the multiplier $n$ change. Writing out the intermediate values reduces the chance of losing a sign or using the wrong lower-power integral.

The recurrence also gives a direct consistency check at each stage. Once $I_{n-1}(b)$ is known, the next value must satisfy

I_n(b)+nI_{n-1}(b)=b(\ln b)^n.

For $b>1$, the integrand $(\ln x)^n$ is nonnegative on $[1,b]$, so $I_n(b)\ge0$ for every $n\ge0$. This does not replace the calculation, but it is a useful check: a negative computed value signals an error. When $n$ is odd, the recurrence contains subtraction, so the sign of an individual term alone is not a reliable check; the integral’s nonnegativity is.

Common Pitfall: Applying the Formula at $n=0$

A tempting but incorrect step is to insert $n=0$ into the positive-power recurrence and conclude that $I_0(b)=b$. The mistake is using a boundary simplification that was valid only for $n\ge1$. At the lower endpoint, the zeroth-power integrand is $1$, not $0$.

Worked Example: Correct the Zeroth-Power Boundary Term

The incorrect substitution gives

I_0(b)\overset{\text{wrong}}{=}b(\ln b)^0-0I_{-1}(b)=b.

This is already inconsistent with a simple direct calculation: $\int_1^b 1\,dx=b-1$. The error is exposed by evaluating the boundary term for the zeroth power directly:

\left[x(\ln x)^0\right]_1^b =[x]_1^b=b-1.

Thus the correct base case is $I_0(b)=b-1$. The positive-power recurrence is derived for $n\ge1$, where $(\ln 1)^n=0$; it is not a substitute for the base case.

Why the Definite Form Is Useful

Definite recurrences let us evaluate a family of integrals on a fixed interval without repeatedly rebuilding an antiderivative from the beginning. The boundary contribution is explicit, and the base case anchors all later reductions. This makes the recurrence especially practical when several successive powers are needed or when a definite integral, rather than a general antiderivative, is the goal.

The interval matters. On $[1,b]$, the lower endpoint term disappears for positive powers because $\ln 1=0$. On a different interval, the same integration-by-parts formula would require evaluating the boundary expression at that interval’s actual endpoints. One should therefore derive or evaluate the endpoint terms for the interval in use rather than transferring this particular recurrence unchanged.

Key takeaway: For $n\ge1$, reduce $I_n(b)$ using $I_n(b)=b(\ln b)^n-nI_{n-1}(b)$, and start the calculation with $I_0(b)=b-1$. The separate base case accounts for the nonzero lower boundary contribution of the zeroth power.

Check Your Understanding

Use the recurrence and endpoint values to answer the following questions.

State the conditions on $b$ and $n$ for the positive-power recurrence.
Why does the lower endpoint contribute zero when $n\ge1$, but not when $n=0$?
Starting from $I_0(b)=b-1$, calculate $I_1(b)$.
Use the recurrence to express $I_2(b)$ in terms of $b$ and $\ln b$.
What antiderivative and endpoint substitution can be used to check a value of $I_2(2)$?

Definite Logarithmic-Power Recurrences on [1,b]

What You'll Learn

From an Antiderivative Recurrence to a Definite Integral

Deriving the Definite Recurrence

Why the Endpoint Term Takes This Form

Worked Example: Evaluate \(I_1(b)\) for General \(b\)

Worked Example: Evaluate \(I_1(b)\) for General \(b\)

Worked Example: Compute \(I_3(e)\)

Worked Example: Compute \(I_3(e)\)

Worked Example: Compute \(I_2(2)\)

Worked Example: Compute \(I_2(2)\)

Using the Recurrence Efficiently

Common Pitfall: Applying the Formula at \(n=0\)

Worked Example: Correct the Zeroth-Power Boundary Term

Why the Definite Form Is Useful

Check Your Understanding

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Vol. 2: Solutions Manual

Definite Logarithmic-Power Recurrences on [1,b]

What You'll Learn

From an Antiderivative Recurrence to a Definite Integral

Deriving the Definite Recurrence

Why the Endpoint Term Takes This Form

Worked Example: Evaluate \(I_1(b)\) for General \(b\)

Worked Example: Evaluate \(I_1(b)\) for General \(b\)

Worked Example: Compute \(I_3(e)\)

Worked Example: Compute \(I_3(e)\)

Worked Example: Compute \(I_2(2)\)

Worked Example: Compute \(I_2(2)\)

Using the Recurrence Efficiently

Common Pitfall: Applying the Formula at \(n=0\)

Worked Example: Correct the Zeroth-Power Boundary Term

Why the Definite Form Is Useful

Check Your Understanding

Test yourself: Calculus II final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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