Why Antiderivatives Reveal What Happens in the Tail
A rational function with a quadratic denominator can have a tail whose integral converges, or one whose integral grows without bound. The integrand’s value tending to zero does not settle the question: what matters is the limit of the accumulated integrals over longer and longer finite intervals.
For quadratic denominators that stay positive, completing the square leads to two familiar antiderivatives. A constant numerator produces an arctangent term, which remains bounded as the upper endpoint tends to infinity. A linear numerator may also produce a logarithmic term, which grows without bound. Working out the antiderivative makes this distinction precise.
On each finite interval \([a,R]\), the Fundamental Theorem of Calculus lets us evaluate the integral using an antiderivative \(F\): \(\int_a^R f(x)\,dx=F(R)-F(a)\). Thus, for these rational tails, the central task is to find \(F\) and examine its limit as \(R\to\infty\).
The Antiderivatives for a Quadratic Denominator
Consider a denominator of the form \(x^2+px+q\). Completing the square gives \[ x^2+px+q=(x+p/2)^2+c,\qquad c=q-\frac{p^2}{4}. \] When \(c>0\), this denominator is positive for every real \(x\). A linear numerator \(Ax+B\) can be split into a multiple of the derivative of the denominator plus a constant: \[ Ax+B=\frac{A}{2}(2x+p)+\left(B-\frac{Ap}{2}\right). \] The first part integrates to a logarithm, while completing the square in the second part gives an arctangent.
The formula is useful only if its coefficients are correct. The factor in front of each term follows from differentiating it, as the derivation below shows. On a half-line, the arctangent has a finite limit as \(x\to\infty\), whereas \(\ln(x^2+px+q)\) grows without bound. Consequently, in this setting, a nonzero coefficient \(A\) on the linear part of the numerator creates a logarithmic term that determines divergence.
Why the Formula Works
Write \(D(x)=x^2+px+q=(x+p/2)^2+c\). Since \(D'(x)=2x+p\), the numerator decomposition separates the integrand into a logarithmic derivative and a constant multiple of a shifted quadratic reciprocal. For \(c>0\), the derivative of the arctangent term follows from the chain rule.
\(Ax+B=\frac{A}{2}D'(x)+(B-Ap/2)\). Dividing by \(D(x)\) gives the sum of \(\frac{A}{2}\frac{D'(x)}{D(x)}\) and \(\frac{B-Ap/2}{(x+p/2)^2+c}\).
By the chain rule, \(\frac{d}{dx}\bigl(\frac{A}{2}\ln D(x)\bigr)=\frac{A}{2}\frac{2x+p}{D(x)}\).
\(\frac{d}{dx}\arctan\bigl(\frac{x+p/2}{\sqrt c}\bigr)=\frac{\sqrt c}{(x+p/2)^2+c}\). Multiplying by \(\frac{B-Ap/2}{\sqrt c}\) gives the required constant numerator over \(D(x)\).
Adding the derivatives in the last two steps recovers \((Ax+B)/D(x)\), proving the stated antiderivative formula. If \(A\ne0\), the logarithmic term behaves like \(A\ln R\) as \(R\to\infty\), up to a bounded difference, while the arctangent term approaches a finite limit. The integral therefore diverges. If \(A=0\), only the arctangent term remains, and its limit is finite; the integral converges. These conclusions assume \(c>0\), so the denominator has no zero on the interval.
Worked Examples
Worked Example: A Shifted Quadratic Tail
Evaluate \(\displaystyle \int_1^\infty \frac{2}{(x+1)^2+4}\,dx\).
First, integrate on a finite interval \([1,R]\). The chain rule gives \[ \frac{d}{dx}\arctan\!\left(\frac{x+1}{2}\right) =\frac{1/2}{1+((x+1)/2)^2} =\frac{2}{(x+1)^2+4}. \] Thus \(F(x)=\arctan((x+1)/2)\) is an antiderivative of the integrand, and \[ \int_1^R \frac{2}{(x+1)^2+4}\,dx =\arctan\!\left(\frac{R+1}{2}\right)-\arctan(1). \] As \(R\to\infty\), \((R+1)/2\to\infty\), so the first arctangent tends to \(\pi/2\). Also, \(\arctan(1)=\pi/4\). Therefore, \[ \int_1^\infty \frac{2}{(x+1)^2+4}\,dx =\frac{\pi}{2}-\frac{\pi}{4} =\frac{\pi}{4}. \] Check: differentiating the stated antiderivative gives \(2/((x+1)^2+4)\), and substituting the endpoints into it gives the value \(\pi/4\).
Worked Example: A Linear Numerator That Gives Divergence
Determine whether \(\displaystyle \int_0^\infty \frac{x+3}{x^2+2x+5}\,dx\) converges.
Complete the square in the denominator: \[ x^2+2x+5=(x+1)^2+4. \] Its derivative is \(2x+2\), and the numerator decomposes as \[ x+3=\frac12(2x+2)+2. \] An antiderivative is therefore \[ F(x)=\frac12\ln(x^2+2x+5)+\arctan\!\left(\frac{x+1}{2}\right). \] To check this, differentiate both terms: \[ F'(x)=\frac{x+1}{x^2+2x+5} +\frac{2}{(x+1)^2+4} =\frac{x+1+2}{x^2+2x+5} =\frac{x+3}{x^2+2x+5}. \] The finite-interval integral is \[ \int_0^R \frac{x+3}{x^2+2x+5}\,dx =\frac12\ln\!\left(\frac{R^2+2R+5}{5}\right) +\arctan\!\left(\frac{R+1}{2}\right) -\arctan\!\left(\frac12\right). \] The lower-end arctangent is \(\arctan(1/2)\), since \(x=0\) makes \((x+1)/2=1/2\). As \(R\to\infty\), the logarithmic term tends to \(+\infty\), and the arctangent tends to \(\pi/2\). The remaining lower-end terms are constants. Hence the finite-interval integrals tend to \(+\infty\), so the improper integral diverges. This also agrees with the positive integrand and its positive logarithmic growth.
Worked Example: A Convergent Tail with a Different Endpoint
Evaluate \(\displaystyle \int_2^\infty \frac{3}{x^2+9}\,dx\).
For finite \(R>2\), the chain rule gives \[ \frac{d}{dx}\arctan\!\left(\frac{x}{3}\right) =\frac{1/3}{1+(x/3)^2} =\frac{3}{x^2+9}. \] Therefore, \[ \int_2^R \frac{3}{x^2+9}\,dx =\arctan\!\left(\frac{R}{3}\right)-\arctan\!\left(\frac{2}{3}\right). \] As \(R\to\infty\), \(\arctan(R/3)\to\pi/2\), so \[ \int_2^\infty \frac{3}{x^2+9}\,dx =\frac{\pi}{2}-\arctan\!\left(\frac{2}{3}\right). \] Check: the derivative calculation reproduces the integrand, and the upper endpoint contributes \(\pi/2\); the lower endpoint contributes exactly \(\arctan(2/3)\).
A Common Coefficient Error
When differentiating an arctangent with a scaled input, it is easy to lose a factor from the chain rule. For example, for \(G(x)=\arctan((x+1)/2)\), a common incorrect calculation is \[ G'(x)=\frac{1}{(x+1)^2+4}. \] That is missing a factor of \(2\). The correct calculation is \[ G'(x)=\frac{1/2}{1+((x+1)/2)^2} =\frac{2}{(x+1)^2+4}. \] So \(G\) is an antiderivative of \(2/((x+1)^2+4)\), not of \(1/((x+1)^2+4)\). For the latter integrand, the correct antiderivative is \(\frac12G(x)\). Differentiating the proposed antiderivative before evaluating the improper limit is a reliable way to catch this error.
A second useful check is to substitute the actual finite lower endpoint into the antiderivative. In the second example, \(x=0\) gives \((x+1)/2=1/2\), not \(1\). Thus the lower-end arctangent is \(\arctan(1/2)\), not \(\pi/4\). Keeping the lower-end value explicit prevents a correct antiderivative from producing an incorrect definite integral.
How to Read the Tail from the Antiderivative
For a denominator \(x^2+px+q\) with \(c=q-p^2/4>0\), the antiderivative separates the two possible tail behaviors. A constant numerator has \(A=0\), so there is no logarithmic term: the arctangent approaches a finite number, and the half-line integral converges. A numerator with a nonzero coefficient of \(x\) has \(A\ne0\), so the antiderivative contains \(\frac A2\ln(x^2+px+q)\). Since \(x^2+px+q\) grows like \(x^2\), this term grows like \(A\ln x\), and the bounded arctangent cannot cancel it.
This is a direct-antiderivative test for this family of rational functions, not a claim that every rational tail can be decided by the same formula. It also relies on the denominator staying nonzero on the integration interval. When the denominator has a zero there, the integral raises a different issue: the integrand is singular at a finite point, and the analysis of such endpoints is separate.
Check Your Understanding
Use finite-interval antiderivatives first, then take the upper-endpoint limit.
- Find an antiderivative of \(\frac{4}{(x-2)^2+9}\), and use it to evaluate its integral from \(2\) to infinity.
- For \(\frac{2x+1}{x^2+4x+8}\), identify the logarithmic and arctangent parts of an antiderivative. Does the integral from \(0\) to infinity converge?
- Why does a bounded arctangent term not prevent divergence when the antiderivative also contains a nonzero multiple of \(\ln(x^2+px+q)\)?
- Differentiate \(\arctan((x+1)/2)\). Which numerator over \((x+1)^2+4\) does it produce?
- In evaluating an improper integral, why must the finite lower endpoint be substituted into the antiderivative rather than replaced by a familiar angle without checking its input?
