Introduction
Some functions have antiderivatives that can be written using familiar elementary functions. For others, no such formula exists, even though the function is continuous on the interval of interest and its definite integral is a perfectly well-defined number.
The function \(1/\ln x\) is a standard example. It is called the logarithmic integral integrand. Its antiderivative is not elementary, but on any closed interval that stays within its domain, the definite integral exists. The distinction between these two facts is important: a limit on the form of an antiderivative does not prevent integration from being meaningful.
Domain, sign, and singularity
Here \(\ln x\) denotes the natural logarithm. The expression \(1/\ln x\) requires \(x>0\) and \(\ln x\ne 0\). Since \(\ln x=0\) exactly when \(x=1\), its domain is
The denominator is negative for \(0<x<1\) and positive for \(x>1\). Accordingly, the integrand is negative on \((0,1)\) and positive on \((1,\infty)\). At \(x=1\), it is undefined; this is a genuine singularity, not a removable gap in the formula.
The derivative gives useful information about the shape of the graph. Differentiating by the chain rule,
For every \(x>0\) with \(x\ne1\), the denominator \(x(\ln x)^2\) is positive, so \(f'(x)<0\). Thus \(f\) is strictly decreasing on each of the two domain intervals. This fact will let us bound integrals without finding an antiderivative.
Why the antiderivative is non-elementary
An elementary function is one obtained from algebraic functions, exponentials, logarithms, trigonometric functions and their inverses through finitely many arithmetic operations and compositions. A standard result about elementary antiderivatives is that \(1/\ln x\) has no elementary antiderivative on either \((0,1)\) or \((1,\infty)\). This is a non-elementarity result; the fact that a formula has not yet been found would not, by itself, prove that no elementary formula exists.
The antiderivative is commonly called a logarithmic integral. Because antiderivatives differ by constants, it is helpful to specify a base point when defining one on a particular interval. Choose \(a>1\), and for \(x>1\) set
The integrand is continuous on every closed interval between \(a\) and \(x\), provided both lie in \((1,\infty)\). The Fundamental Theorem of Calculus therefore gives
A corresponding definition works on \((0,1)\) by choosing a base point \(a\) in that interval and integrating only between points of \((0,1)\). Neither definition supplies an elementary formula for the resulting function. It supplies a precise function whose derivative is the integrand.
Why the definite integral remains meaningful
If \(0<a<b<1\), or if \(1<a<b\), then \(1/\ln x\) is continuous throughout \([a,b]\). The definite integral on that interval is therefore a proper Riemann integral: it exists and is finite. Its value is the signed accumulation of the function over the interval. In particular, it is negative when the interval lies in \((0,1)\) and positive when it lies in \((1,\infty)\).
The Fundamental Theorem also explains how the integral relates to the local antiderivative. For a fixed point \(a\) in either domain interval, define \(F_a(x)=\int_a^x 1/\ln t\,dt\) for \(x\) in that same interval. Continuity of the integrand on the interval between \(a\) and \(x\) is the condition needed to apply the theorem; it yields \(F_a'(x)=1/\ln x\). Consequently, for endpoints \(a\) and \(b\) in that interval,
since \(F_a(a)=0\). This gives an exact characterization of the integral even when \(F_a\) cannot be expressed using elementary functions.
Worked Example: Bounding an integral from 2 to 4
Worked Example: Bounding an integral from 2 to 4
Consider \(I=\int_2^4 dx/\ln x\). The interval \([2,4]\) lies in \((1,\infty)\), so the integrand is continuous and positive there. Since it is strictly decreasing, its endpoint values give bounds throughout the interval:
Integrating these inequalities over an interval of length \(4-2=2\), and using \(\ln4=2\ln2\), gives
The inequalities are strict because the integrand is not constant on \([2,4]\). Thus \(1/\ln2<I<2/\ln2\). As a check, both bounds are positive, as the integrand is positive throughout this interval; the interval length used in each bound is exactly \(2\).
Worked Example: Rewriting an integral with a substitution
Worked Example: Rewriting an integral with a substitution
Consider \(J=\int_e^{e^2} dx/\ln x\). This is a proper integral because \([e,e^2]\) stays in \((1,\infty)\). Use \(t=\ln x\), so \(x=e^t\) and \(dx=e^t\,dt\). The endpoints change from \(x=e\) and \(x=e^2\) to \(t=1\) and \(t=2\), respectively. Therefore
This substitution is valid because \(t=\ln x\) is differentiable and strictly increasing on the interval, and \(t\) does not vanish on \([1,2]\). It changes the representation, but it does not produce an elementary antiderivative. The transformed integrand is positive and continuous on \([1,2]\), so the transformed definite integral remains finite and positive.
To check the change of variables, substitute \(t=\ln x\) back: \(e^t=x\), \(dt=dx/x\), and \(e^t\,dt=x(dx/x)=dx\). Also, \(t=\ln x\), so \(e^t/t\,dt=dx/\ln x\), as required.
Worked Example: A negative integral below 1
Worked Example: A negative integral below 1
Let \(K=\int_{e^{-2}}^{e^{-1}} dx/\ln x\). Both endpoints are positive and less than \(1\), and the whole closed interval lies in \((0,1)\). The integrand is continuous there. Its endpoint values are
Since \(f\) is decreasing, \(-1\le f(x)\le-1/2\) on the interval. Its length is \(e^{-1}-e^{-2}>0\). Multiplying the bounds by that length and integrating gives
The inequalities are strict because \(f\) is strictly decreasing. In particular, \(K<0\), in agreement with the negative sign of the integrand throughout the interval. The interval length is positive, so the order of the two bounds is also consistent.
The singularity at 1 changes the question
A closed interval that includes \(1\) does not satisfy the continuity condition used above. In fact, the integrand behaves like \(1/(x-1)\) near \(1\). To see this, use the standard limit \(\ln x/(x-1)\to1\) as \(x\to1\); taking reciprocals gives
The comparison function \(1/(x-1)\) has a divergent integral at \(1\). Consequently, the ordinary improper integral of \(1/\ln x\) up to or across \(1\) does not converge: the one-sided integral diverges logarithmically. A Cauchy principal value, if defined by a specified limiting procedure, is a different notion and should not be confused with an ordinary convergent improper integral.
Why this example matters
The logarithmic integral shows why antiderivative formulas and definite integrals should not be treated as the same issue. An elementary antiderivative is a particularly convenient way to express an integral, but it is not a requirement for the integral to exist. On intervals where the integrand is continuous, the definite integral is a finite number defined independently of whether an elementary expression for it is available.
For this function, monotonicity can provide useful bounds, and a definite-integral definition gives an exact representation. The singularity at \(1\) still matters: intervals must be chosen so that the integral is being taken where the integrand is defined and continuous, unless a separate improper-integral or principal-value question is intended.
Check Your Understanding
Use the domain, monotonicity, and continuity conditions from this tutorial to answer the questions.
- What is the domain of \(1/\ln x\), and on which part of that domain is the function negative?
- Why does \(\int_a^b dx/\ln x\) exist as a proper integral when \([a,b]\) lies wholly in \((1,\infty)\)?
- What does the substitution \(t=\ln x\) change in an integral of \(1/\ln x\), and what does it not guarantee?
- Why is \(\int_{1/2}^{2} dx/\ln x\) not an ordinary convergent improper integral?
- How can a definite integral define an antiderivative even when that antiderivative is non-elementary?
