Logarithmic Integrals from Parameter Differentiation

Turning a Logarithmic Integral into a Parameter Problem

A logarithm in a denominator can make a direct antiderivative difficult to find. A useful strategy is to place the integrand in a family depending on a real parameter, differentiate that family under the integral sign, and solve the simpler resulting problem. For example, differentiating $x^a$ with respect to $a$ produces $x^a\ln x$; in a suitable quotient, this cancels a logarithmic denominator entirely.

All integrals in the main calculation are Lebesgue integrals on $((0,1),\mathcal{L},m)$, where $m$ is Lebesgue measure restricted to $(0,1)$. This is a finite measure space. The integrands below are real-valued at each $x\in(0,1)$. When a signed integral is finite, it means the integrand is integrable; for a nonpositive integrand whose absolute integral is infinite, the extended Lebesgue integral is $-\infty$.

Definition: For a real parameter $a$, define $q_a(x)=\dfrac{1-x^a}{-\ln x}$ for $0<x<1$. When $q_a$ is integrable, write $F(a)=\int_0^1 q_a(x)\,dm(x)$. The parameter integral will be finite exactly when $a>-1$.

For each fixed $x\in(0,1)$, the derivative of $q_a(x)$ with respect to $a$ is $x^a$: the derivative of $1-x^a$ is $-x^a\ln x$, and division by $-\ln x$ leaves $x^a$. The formal computation is short; its validity depends on justifying the passage of the derivative through the integral.

The Parameter Formula and Its Domain

Theorem: If $a>-1$, then $q_a$ is integrable on $(0,1)$, the function $F(a)=\int_0^1q_a\,dm$ is differentiable on $(-1,\infty)$, and $$ F(a)=\ln(1+a). $$ If $a\le -1$, then $q_a\le0$ on $(0,1)$ and $\int_0^1q_a\,dm=-\infty$.

Proof. First take $a>-1$. The substitution $x=e^{-t}$, with $dx=-e^{-t}dt$, gives the corresponding integral on $0<t<\infty$ with integrand $$ \frac{e^{-t}(1-e^{-at})}{t} =\frac{e^{-t}-e^{-(a+1)t}}{t}. $$ This change of variables is valid first on compact subintervals and then on the full interval by taking endpoint limits for the nonnegative or absolutely integrable parts. Near $t=0$, the mean value theorem applied to $s\mapsto e^{-as}$ gives $|1-e^{-at}|\le |a|t e^{|a|}$ for $0<t\le1$; after multiplication by $e^{-t}/t$, the absolute value is bounded by $|a|e^{|a|}$. At $t\ge1$, if $a\ge0$, then $|e^{-t}(1-e^{-at})|/t\le e^{-t}/t$. If $-1<a<0$, then \[ \frac{e^{-t}|1-e^{-at}|}{t} =\frac{e^{-(a+1)t}-e^{-t}}{t} \le \frac{e^{-(a+1)t}}{t}. \] The bounds at infinity are integrable because $a+1>0$. Thus $q_a$ is absolutely integrable.

Fix $a_0>-1$. Choose $\alpha$ with $-1<\alpha<a_0$, and restrict $h$ to be small enough that $a_0+h\ge\alpha$. For each $x\in(0,1)$, the mean value theorem in the parameter gives \[ \left|\frac{q_{a_0+h}(x)-q_{a_0}(x)}{h}\right|=x^s\le x^\alpha \] for some $s$ between $a_0$ and $a_0+h$; the inequality holds because $0<x<1$ and $s\ge\alpha$. The dominating function $x^\alpha$ is integrable, since $\int_0^1x^\alpha\,dm=1/(\alpha+1)<\infty$. The difference quotients converge pointwise to $x^{a_0}$. The Dominated Convergence Theorem therefore applies: the quotients are measurable, converge pointwise, and have the stated integrable bound. It gives \[ F'(a_0)=\int_0^1x^{a_0}\,dm=\frac{1}{a_0+1}, \] where the last equality follows by integrating $x^{a_0}$ and using $a_0>-1$. Also $F(0)=0$. The function $a\mapsto F(a)-\ln(1+a)$ has derivative zero on $(-1,\infty)$ and value zero at $a=0$. The mean value theorem on the interval between $0$ and any $a>-1$ implies $F(a)=\ln(1+a)$.

Now suppose $a\le-1$. Since $x^a\ge1$ for $0<x<1$, the numerator $1-x^a$ is nonpositive and $-\ln x>0$, so $q_a\le0$. Under the same substitution, its integral is represented by $(e^{-t}-e^{-(a+1)t})/t$. If $a=-1$, this is $(e^{-t}-1)/t$; for $t\ge1$, it is at most $-(1-e^{-1})/t$, whose integral diverges to $-\infty$. If $a<-1$, put $c=-(a+1)>0$. The expression is $(e^{-t}-e^{ct})/t$. For all sufficiently large $t$, $e^{-t}\le e^{ct}/2$, so it is at most $-e^{ct}/(2t)$, whose integral also diverges to $-\infty$. Because the integrand is nonpositive, this proves that its extended Lebesgue integral is $-\infty$. $\square$

The sign in the last case matters: the divergence is not to positive infinity. For $-1<a<0$, the integral is finite but negative, agreeing with $\ln(1+a)<0$. At $a=0$, the integrand is identically zero.

Worked Evaluations

Worked Example: The Integral with a Linear Numerator

Evaluate $\int_0^1\dfrac{1-x}{-\ln x}\,dm(x)$.

This is the theorem with $a=1$, which lies in the finite range. Differentiating the parameter integral gives $F'(a)=1/(1+a)$, and the theorem’s normalization is $F(0)=0$. Thus the value at $a=1$ is $\ln(1+1)=\ln2$. We can check it independently after setting $x=e^{-t}$: \[ \int_0^1\frac{1-x}{-\ln x}\,dm(x) =\int_0^\infty\frac{e^{-t}-e^{-2t}}{t}\,dt. \] For $t>0$, $\int_1^2e^{-ut}\,du=(e^{-t}-e^{-2t})/t$. The integrand in this double integral is nonnegative, so Tonelli’s theorem applies and gives \[ \int_1^2\int_0^\infty e^{-ut}\,dt\,du =\int_1^2\frac1u\,du =\ln2. \] Check: substituting $a=1$ in the parameter formula gives $\ln(1+1)=\ln2$, the same value.

Worked Example: A Negative Parameter

Evaluate $\int_0^1\dfrac{1-x^{-1/2}}{-\ln x}\,dm(x)$.

Here $a=-1/2>-1$, so the integral is finite and the parameter formula gives \[ F(-1/2)=\ln(1-1/2)=\ln(1/2)=-\ln2. \] To verify both the sign and the value, set $x=e^{-t}$. The transformed integrand is \[ \frac{e^{-t}-e^{-t/2}}{t} =-\int_{1/2}^{1}e^{-ut}\,du, \] which is nonpositive. Tonelli’s theorem applied to its nonnegative magnitude gives \[ -\int_{1/2}^{1}\int_0^\infty e^{-ut}\,dt\,du =-\int_{1/2}^{1}\frac1u\,du =-\bigl(\ln1-\ln(1/2)\bigr) =-\ln2. \] Check: at $x=1/4$, the numerator is $1-2=-1$, while $-\ln(1/4)=\ln4>0$; the integrand is indeed negative, consistent with the computed value.

Worked Example: A Higher Power

Evaluate $\int_0^1\dfrac{1-x^2}{-\ln x}\,dm(x)$.

Since $a=2>-1$, the theorem gives $F(2)=\ln(1+2)=\ln3$. For an independent check, the substitution $x=e^{-t}$ yields \[ \int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,dt. \] For $t>0$, $\int_1^3 e^{-ut}\,du=(e^{-t}-e^{-3t})/t$. The integrand is nonnegative, so Tonelli applies: \[ \int_1^3\int_0^\infty e^{-ut}\,dt\,du =\int_1^3\frac1u\,du =\ln3. \] Check: the parameter substitution $a=2$ gives $\ln(1+2)=\ln3$, matching the independent calculation.

Using the Same Method for Logarithmic Moments

Parameter differentiation also computes integrals in which the logarithm appears in the numerator. For $a>-1$, define $J(a)=\int_0^1x^a\,dm(x)$. Direct integration gives $J(a)=1/(a+1)$. Near any fixed $a_0>-1$, choose $-1<\alpha<a_0$; for sufficiently small parameter increments, the difference quotients for $x^a$ are bounded in absolute value by $x^\alpha|\ln x|$. This bound is integrable: substituting $x=e^{-t}$ gives $\int_0^\infty t e^{-(\alpha+1)t}\,dt<\infty$, evaluated by integration by parts. The Dominated Convergence Theorem therefore justifies differentiating $J$, and the derivative of the integrand is $x^a\ln x$.

Corollary: For $a>-1$, $$ \int_0^1x^a(-\ln x)\,dm(x)=\frac{1}{(a+1)^2}. $$

Proof. The justified derivative calculation gives $J'(a)=\int_0^1x^a\ln x\,dm(x)$. Differentiating $J(a)=1/(a+1)$ gives $J'(a)=-1/(a+1)^2$. Multiplying the integral identity by $-1$ proves the claim. $\square$

Worked Example: A Logarithmic Moment

Compute $\int_0^1x^2(-\ln x)\,dm(x)$.

The corollary applies with $a=2$, so the value is $1/(2+1)^2=1/9$. To check by direct integration by parts, take $u=-\ln x$ and $dv=x^2\,dx$ on $[\varepsilon,1]$, where $0<\varepsilon<1$. Then $du=-dx/x$ and $v=x^3/3$, giving \[ \int_\varepsilon^1x^2(-\ln x)\,dx =\left[-\frac{x^3\ln x}{3}\right]_\varepsilon^1 +\frac13\int_\varepsilon^1x^2\,dx. \] The upper boundary term is zero, and $\varepsilon^3\ln\varepsilon\to0$ as $\varepsilon\downarrow0$. The remaining integral tends to $\frac13\cdot\frac13=1/9$. Check: the parameter formula gives $1/(2+1)^2=1/9$, as does the direct computation.

A Sign Pitfall and a Practical Check

A common error is to assume $1-x^a\ge0$ for every real $a$. That is true for $a\ge0$, but false for $a<0$: because $0<x<1$, a negative power satisfies $x^a>1$. For instance, at $a=-1/2$ and $x=1/4$, $x^a=2$, so $1-x^a=-1$, not a positive number. Since $-\ln(1/4)=\ln4>0$, the quotient at that point is negative.

The corrected calculation is $F(-1/2)=\ln(1/2)=-\ln2$; the integral is finite because $-1/2>-1$. At and below the boundary $a=-1$, the same quotient remains nonpositive but its integral is $-\infty$, as proved above. In particular, it is incorrect to describe the $a\le-1$ case as a nonnegative integrand with positive infinite integral.

For this family, a reliable workflow is to check the sign first, identify whether the parameter lies above the integrability threshold $-1$, and only then apply the differentiated formula. Differentiation under the integral sign is justified locally by an integrable dominating function; it is not licensed merely by a pointwise derivative calculation.

Check Your Understanding

For each question, state the parameter range and sign where relevant, and identify the justification for any differentiation or limit interchange.

Use the parameter formula to evaluate $\int_0^1\dfrac{1-x^{3}}{-\ln x}\,dm(x)$, then check the parameter substitution in the answer.
For $a=-1/3$, determine whether the integral is finite and whether its value is positive, negative, or zero.
Explain why $x^\alpha$ is an integrable dominating function for the difference quotients defining $F'(a_0)$ when $-1<\alpha<a_0$.
What is the extended Lebesgue integral of $q_{-1}$, and what sign does its integrand have on $(0,1)$?
Use the logarithmic-moment corollary to compute $\int_0^1x(-\ln x)\,dm(x)$.