Why This Family of Integrals Matters
A function can become small as \(x\) grows and still contribute an unbounded total area over an infinite interval. The half-line \(p\)-integral provides a precise setting in which to see how the rate of decay controls that accumulated area.
The integrand \(x^{-p}\) is positive for \(x\ge 1\) and any real number \(p\). When \(p\) is positive, it also tends to zero as \(x\) tends to infinity. But that fact alone does not tell us whether the integral is finite. We will determine the exact threshold by evaluating finite-cutoff integrals and then taking a limit.
Definition by a Cutoff Limit
An integral with an infinite upper endpoint is defined using finite upper endpoints. For a real parameter \(p\), the half-line \(p\)-integral is
provided this limit exists as a finite real number. Here \(R\) is a finite real number with \(R\ge 1\). If the limit is \(+\infty\), the improper integral diverges; it also diverges if the limit does not exist. Because \(x^{-p}>0\) throughout the interval, the cutoff integrals increase as \(R\) increases.
Evaluating the Finite-Cutoff Integral
For \(p\ne 1\), the power rule gives an antiderivative
Indeed, differentiating the expression without the constant gives
Thus, when \(p\ne 1\), the fundamental theorem of calculus on the finite interval \([1,R]\) yields
The value \(p=1\) must be handled separately: the power-rule expression would have a zero denominator. Since \(x^{-1}=1/x\), its antiderivative is \(\ln x\), so
The Convergence Threshold
We can now evaluate the limit in each case. For \(p>1\), the exponent \(1-p\) is negative, so \(R^{1-p}\) tends to zero as \(R\) tends to infinity. For \(p<1\), the exponent \(1-p\) is positive, so \(R^{1-p}\) grows without bound. The remaining case \(p=1\) gives a logarithm that also grows without bound.
Here the notation \(+\infty\) describes the limit of the cutoff integrals, not a finite value of the improper integral. For \(p=1\), the cutoff integrals equal \(\ln R\), which tends to \(+\infty\). For \(p<1\), the power expression tends to \(+\infty\) as well. This includes \(p=0\) and negative values of \(p\).
Why the Threshold Follows
The key is the exponent \(1-p\) in the cutoff formula. The following calculation gives the result directly and shows why the boundary value \(p=1\) requires its own case.
For \(p\ne 1\), integrate on \([1,R]\) to obtain \(\dfrac{R^{1-p}-1}{1-p}\). This finite-interval calculation is valid for every such \(p\).
Then \(1-p<0\), so \(R^{1-p}=1/R^{p-1}\to 0\). The cutoff limit is \(\dfrac{-1}{1-p}=\dfrac{1}{p-1}\), which is finite.
Then \(1-p>0\), so \(R^{1-p}\to+\infty\). Since \(1-p\) is positive, \(\dfrac{R^{1-p}-1}{1-p}\to+\infty\), and the integral diverges.
The cutoff integral is \(\ln R\), which tends to \(+\infty\). Therefore it diverges, completing all cases.
This argument proves both directions of the convergence claim: when \(p>1\), the limit exists and is finite; when \(p\le 1\), the cutoff integrals grow without bound. It also gives the exact value whenever the integral converges.
Worked Examples
Worked Example: The Case \(p=2\)
Evaluate \(\displaystyle\int_1^\infty \frac{1}{x^2}\,dx\). Here \(p=2\), so \(p\ne 1\) and the finite-cutoff formula applies:
As \(R\to\infty\), \(1/R\to 0\). Therefore,
Check the calculation: differentiating \(-1/x\) gives \(1/x^2\), and the finite-cutoff value tends to \(1=1/(2-1)\), as the convergence formula predicts.
Worked Example: A Convergent Noninteger Power
Evaluate \(\displaystyle\int_1^\infty x^{-3/2}\,dx\). Since \(p=3/2>1\), the integral should converge. First calculate with a finite cutoff:
As \(R\to\infty\), \(1/\sqrt R\to 0\), and hence
For a check, the derivative of \(-2x^{-1/2}\) is \((-2)(-1/2)x^{-3/2}=x^{-3/2}\). The formula \(1/(p-1)\) gives \(1/(3/2-1)=1/(1/2)=2\), matching the cutoff calculation.
Worked Example: The Boundary Case \(p=1\)
Evaluate \(\displaystyle\int_1^\infty \frac{1}{x}\,dx\). Because \(p=1\), use the logarithmic antiderivative rather than the power formula:
Since \(\ln R\to+\infty\) as \(R\to\infty\), the improper integral diverges to \(+\infty\). The check is \(\dfrac{d}{dx}\ln x=1/x\) for \(x>0\), and the lower-endpoint contribution is \(\ln 1=0\).
Worked Example: A Positive \(p\) That Still Diverges
Consider \(\displaystyle\int_1^\infty x^{-1/2}\,dx\). Here \(p=1/2<1\), so the threshold predicts divergence. Directly,
As \(R\to\infty\), \(2\sqrt R-2\to+\infty\). Thus the improper integral diverges. Differentiating \(2\sqrt{x}\) gives \(2\cdot \frac{1}{2\sqrt{x}}=x^{-1/2}\), confirming the antiderivative. This example also shows that a positive exponent \(p\) alone is not enough for convergence.
A Common Mistake: Stopping at the Integrand
A tempting but incorrect argument is: “If \(p=1/2\), then \(x^{-1/2}\to 0\), so its integral must be finite.” The conclusion does not follow from the stated fact. The integrand does tend to zero, but the finite-cutoff areas are
The correction is to evaluate the improper integral by its cutoff limit, rather than infer convergence from the behavior of the integrand alone. Another common error is to substitute \(p=1\) into \(\frac{R^{1-p}-1}{1-p}\). That expression is not defined at \(p=1\); the correct calculation at the boundary uses \(\int_1^R 1/x\,dx=\ln R\), which diverges.
What the Result Tells Us
The parameter \(p\) controls how rapidly \(x^{-p}\) decreases, or, when \(p\le 0\), whether it decreases at all. The exact cutoff calculation distinguishes three behaviors: for \(p>1\), the remaining power \(R^{1-p}\) tends to zero and the accumulated area is finite; at \(p=1\), the accumulated area grows logarithmically; for \(p<1\), it grows as a positive power of \(R\).
The convergence threshold is therefore strict: \(p=1\) is on the divergent side, not the convergent side. When \(p>1\), the value \(1/(p-1)\) also shows how the total changes with \(p\). For example, as \(p\) approaches \(1\) from above, \(p-1\) approaches zero through positive values, so \(1/(p-1)\) becomes arbitrarily large. This is consistent with the divergence at the boundary.
Check Your Understanding
For each question, use a finite cutoff before deciding what happens as the upper endpoint tends to infinity.
- For which real values of \(p\) does \(\int_1^\infty x^{-p}\,dx\) converge?
- Evaluate \(\int_1^\infty x^{-4}\,dx\), showing the cutoff expression and its limit.
- Does \(\int_1^\infty x^{-2/3}\,dx\) converge? Compute its finite-cutoff integral to justify your answer.
- Why can the power-rule expression \(\frac{R^{1-p}-1}{1-p}\) not be used directly when \(p=1\)? What calculation replaces it?
- For \(p>1\), verify the stated value of the improper integral by taking the limit of the cutoff formula.
