Shifted Power Tails

Why a Finite Shift Matters

A power tail has a familiar convergence threshold: its integral over a half-line is finite when the power in the denominator is sufficiently large. A finite horizontal shift does not change that threshold, but it does change the value of a convergent integral. The distinction is important: convergence describes what happens far out along the half-line, while the value also depends on where the integration starts.

We will evaluate the integral directly by first replacing the infinite endpoint with a finite cutoff \(R\), then taking the limit as \(R\) tends to infinity. This also makes clear why the boundary case behaves differently from the convergent cases.

Definition and Domain

Let \(a,c\in\mathbb R\) and \(p\in\mathbb R\). To study \((x+c)^{-p}\) as a real-valued function throughout \([a,\infty)\), assume \(a+c>0\). Then \(x+c\) is positive on the entire interval, so real powers of \(x+c\) are defined there and the integrand has no singularity at the finite endpoint \(a\).

Definition: Under the condition \(a+c>0\), the shifted power tail is the improper integral \(\int_a^\infty (x+c)^{-p}\,dx\). It is defined by the cutoff limit \(\lim_{R\to\infty}\int_a^R (x+c)^{-p}\,dx\), if that limit exists as a finite real number. If the cutoff integrals grow without bound, the improper integral diverges.

The condition on \(a+c\) keeps this discussion focused on the infinite tail, rather than introducing a separate issue at the finite endpoint. The exponent \(p\) can be any real number; the outcome will depend on whether \(p\) is above, equal to, or below \(1\).

Deriving the Formula from a Finite Cutoff

For a finite \(R>a\), the integration is an ordinary definite integral. When \(p\ne 1\), an antiderivative is \((x+c)^{1-p}/(1-p)\), since differentiating it gives \((x+c)^{-p}\). When \(p=1\), the antiderivative is \(\ln(x+c)\), whose derivative is \(1/(x+c)\). Applying these antiderivatives first at finite \(R\) avoids treating infinity as an ordinary endpoint.

Use a finite endpoint.
Evaluate \(\int_a^R (x+c)^{-p}\,dx\) for \(R>a\), where the usual definite-integral formulas apply.

Keep the lower endpoint.
The value at \(a\) contains \(a+c\); it cannot be discarded when the cutoff is taken to infinity.

Take the limit.
Determine whether the resulting expression has a finite limit as \(R\to\infty\).

For \(p\ne1\), the finite-cutoff calculation is

\int_a^R (x+c)^{-p}\,dx = \left[\frac{(x+c)^{1-p}}{1-p}\right]_a^R = \frac{(R+c)^{1-p}-(a+c)^{1-p}}{1-p}.

If \(p>1\), then \(1-p<0\), so \((R+c)^{1-p}\) tends to \(0\) as \(R\to\infty\). If \(p<1\), then \(1-p>0\), so \((R+c)^{1-p}\) grows without bound. At \(p=1\), the cutoff integral is a logarithmic difference, which also grows without bound.

Theorem: If \(a+c>0\), then \(\int_a^\infty (x+c)^{-p}\,dx\) converges exactly when \(p>1\). In that case, \[ \int_a^\infty (x+c)^{-p}\,dx = \frac{(a+c)^{1-p}}{p-1}. \] For \(p=1\) or \(p<1\), the cutoff integrals tend to \(+\infty\), so the improper integral diverges.

Why the Convergence Threshold Does Not Shift

The exact cutoff expression gives a complete justification. For \(p>1\), the only term involving the growing endpoint is \((R+c)^{1-p}\), and this tends to zero because its exponent is negative. The remaining term depends on \(a+c\), so it determines the value but not whether the limit is finite.

For \(p<1\), the exponent \(1-p\) is positive. Thus \((R+c)^{1-p}\) tends to \(+\infty\), and the positive cutoff integral also tends to \(+\infty\). When \(p=1\), the cutoff expression is \(\ln(R+c)-\ln(a+c)\), which tends to \(+\infty\). Since a finite shift \(c\) leaves these limiting behaviors unchanged, it cannot move the convergence threshold from \(p=1\).

The formula also shows precisely how the shift affects the value when the integral converges: it appears through the positive quantity \(a+c\). Increasing that quantity changes the value, even though the threshold remains \(p>1\).

Worked Examples

Worked Example: A Convergent Tail with a Fractional Power

Evaluate \(\int_1^\infty (x+2)^{-3/2}\,dx\), and verify the result.

Here \(a=1\), \(c=2\), and \(p=3/2>1\). The domain condition holds because \(a+c=3>0\). First use a finite cutoff \(R>1\). Since \(1-p=-1/2\), an antiderivative is \(-2(x+2)^{-1/2}\): differentiating gives \((-2)(-1/2)(x+2)^{-3/2}=(x+2)^{-3/2}\).

\begin{aligned} \int_1^R (x+2)^{-3/2}\,dx &=\left[-2(x+2)^{-1/2}\right]_1^R\\ &=-\frac{2}{\sqrt{R+2}}+\frac{2}{\sqrt{3}}. \end{aligned}

As \(R\to\infty\), \(2/\sqrt{R+2}\to0\). Therefore the integral equals \(2/\sqrt{3}\). Checking with the general formula gives \((a+c)^{1-p}/(p-1)=3^{-1/2}/(1/2)=2/\sqrt{3}\), the same value. The derivative check above verifies the antiderivative used in the cutoff calculation.

Worked Example: A Shift Changes the Value

Evaluate \(\int_0^\infty (x+4)^{-2}\,dx\), and compare the result with the unshifted case.

Here \(a=0\), \(c=4\), and \(p=2>1\), with \(a+c=4>0\). For \(R>0\), an antiderivative is \(-(x+4)^{-1}\), because its derivative is \((x+4)^{-2}\). Thus

\begin{aligned} \int_0^R (x+4)^{-2}\,dx &=\left[-\frac{1}{x+4}\right]_0^R\\ &=-\frac{1}{R+4}+\frac14. \end{aligned}

Taking \(R\to\infty\) gives \(1/4\). The general formula confirms this: \((a+c)^{1-p}/(p-1)=4^{-1}/1=1/4\). For comparison, the unshifted integral \(\int_0^\infty x^{-2}\,dx\) is not covered by the same endpoint condition at \(a=0\), because it has a singularity there. The safe comparison is to start both tails at a positive point: for example, \(\int_1^\infty x^{-2}\,dx=1\), while \(\int_1^\infty (x+4)^{-2}\,dx=1/5\). The shift changes the value, not the threshold \(p>1\).

Worked Example: The Boundary Exponent

Determine whether \(\int_2^\infty (x+1)^{-1}\,dx\) converges.

Here \(p=1\), so use the logarithmic antiderivative. For \(R>2\),

\begin{aligned} \int_2^R \frac{1}{x+1}\,dx &=\left[\ln(x+1)\right]_2^R\\ &=\ln(R+1)-\ln 3. \end{aligned}

As \(R\to\infty\), \(\ln(R+1)\to+\infty\), while \(\ln 3\) is a fixed finite number. Therefore the cutoff integrals tend to \(+\infty\) and the improper integral diverges. The check is immediate from differentiating: \(\frac{d}{dx}\ln(x+1)=1/(x+1)\), as required. A finite shift does not make the boundary case converge.

Worked Example: A Power Below the Threshold

Determine whether \(\int_0^\infty (x+2)^{-1/2}\,dx\) converges.

Here \(a=0\), \(c=2\), and \(p=1/2<1\), with \(a+c=2>0\). An antiderivative is \(2(x+2)^{1/2}\), because its derivative is \(2(1/2)(x+2)^{-1/2}=(x+2)^{-1/2}\). The cutoff calculation is

\begin{aligned} \int_0^R (x+2)^{-1/2}\,dx &=\left[2\sqrt{x+2}\right]_0^R\\ &=2\sqrt{R+2}-2\sqrt{2}. \end{aligned}

As \(R\to\infty\), the first term grows without bound, so the integral diverges. The derivative check verifies the antiderivative, and the cutoff expression verifies the divergence directly. This illustrates that a shift cannot rescue an exponent below the threshold.

A Common Error: Dropping the Lower-Endpoint Contribution

A frequent mistake is to use the unshifted half-line formula as though every convergent power tail had value \(1/(p-1)\), regardless of its starting point or shift. For example, applying that expression to \(\int_0^\infty (x+2)^{-2}\,dx\) would give \(1/(2-1)=1\). That is incorrect: the formula would have discarded the dependence on \(a+c\).

The cutoff calculation gives the correction. For \(R>0\),

\int_0^R (x+2)^{-2}\,dx = \left[-\frac{1}{x+2}\right]_0^R = -\frac{1}{R+2}+\frac12.

Taking \(R\to\infty\) gives \(1/2\), not \(1\). Checking against the convergent-case formula gives \((a+c)^{1-p}/(p-1)=2^{-1}/1=1/2\). The exponent determines whether the tail converges; the lower endpoint and shift determine its numerical value.

What the Result Tells Us

The convergence test and the value formula answer different questions. First check whether \(p>1\). If not, the integral diverges, so there is no finite value to calculate. If it does converge, the quantity \(a+c\) supplies the finite-endpoint contribution in the formula.

This separation is useful when comparing tails. Changing \(c\) by a finite amount does not change the convergence classification, but it can substantially change the integral. Likewise, changing the finite lower endpoint affects the value. Neither change alters the threshold, provided the integrand remains defined and positive throughout the interval.

Key takeaway: For \(a+c>0\), the shifted power tail converges precisely when \(p>1\). Its value in that case is \((a+c)^{1-p}/(p-1)\). The finite shift changes the value through \(a+c\), but not the convergence threshold.

Check Your Understanding

For each question, distinguish the convergence decision from the calculation of a finite value.

For \(a=1\), \(c=3\), and \(p=2\), does the shifted power tail converge? If so, what is its value?
For \(a=0\), \(c=5\), and \(p=1\), write the finite-cutoff integral and determine its limit.
For \(a=2\), \(c=-1\), and \(p=1/2\), does the domain condition hold, and does the tail converge?
Explain why replacing \(a+c\) by \(1\) in the convergent-case value formula can produce an incorrect answer.
What feature of the cutoff expression establishes that a finite shift does not change the threshold?

What You'll Learn