Arctangent Tails

Why an Arctangent Appears in an Infinite Tail

A function can be positive at every point and still have a finite total area over an unbounded interval. The integral of $1/(x^2+a^2)$, where $a$ is a nonzero constant, is a useful example: its antiderivative involves the inverse tangent, and the improper integral can be evaluated by taking a limit of ordinary definite integrals.

The calculation combines two ideas. First, a suitable rescaling turns the denominator into the standard form associated with the arctangent. Second, the value of the improper integral is not obtained by substituting infinity into an antiderivative. Instead, we evaluate at a finite cutoff and take the limit as that cutoff increases without bound.

Key result: For $a>0$, the right-hand tail from zero is convergent, and $$ \int_0^\infty \frac{dx}{x^2+a^2}=\frac{\pi}{2a}. $$

Definition and Antiderivative

For an integrand defined on $[c,\infty)$, the right-infinite integral is defined as the limit of integrals over finite intervals, provided that limit exists and is finite. For this tutorial, take $a>0$. The integrand $1/(x^2+a^2)$ is continuous on the real line, since $x^2+a^2$ is always positive.

Definition: For a finite real number $c$, the improper integral $$ \int_c^\infty \frac{dx}{x^2+a^2} $$ converges to a finite value if the limit $$ \lim_{R\to\infty}\int_c^R \frac{dx}{x^2+a^2} $$ exists and is finite. Here $R$ is a finite cutoff with $R>c$.

The standard derivative formula for the inverse tangent is $\frac{d}{du}\arctan(u)=1/(1+u^2)$. Applying the chain rule to $\arctan(x/a)$ gives an antiderivative for the integrand.

\frac{d}{dx}\left(\frac{1}{a}\arctan\left(\frac{x}{a}\right)\right) = \frac{1}{a}\cdot\frac{1}{1+(x/a)^2}\cdot\frac{1}{a} = \frac{1}{a^2+x^2}.

The condition $a>0$ matters in writing this particular form: it ensures the scale $a$ is positive and the antiderivative has coefficient $1/a$. If the original parameter is allowed to be any nonzero real number, the denominator depends only on its square, and the corresponding half-line formula uses $|a|$ in place of $a$.

Deriving the Right-Hand Tail Formula

We first calculate from zero. For each finite $R>0$, the Fundamental Theorem of Calculus applies because the integrand is continuous on $[0,R]$, and the antiderivative above is differentiable there. Thus the finite-cutoff integral has a direct expression.

Use a finite cutoff.
For $R>0$, integrate from $0$ to $R$, not directly to infinity.

Apply the antiderivative.
Evaluate $\frac{1}{a}\arctan(x/a)$ at both finite endpoints.

Take the limit.
Use $\lim_{u\to\infty}\arctan(u)=\pi/2$ to evaluate the cutoff limit.

In detail, because $a>0$, $R/a$ tends to positive infinity when $R$ tends to positive infinity. The inverse tangent approaches $\pi/2$ from below. The value at the lower endpoint is $\arctan(0)=0$, so the limit is finite.

\begin{aligned} \int_0^\infty \frac{dx}{x^2+a^2} &=\lim_{R\to\infty}\int_0^R \frac{dx}{x^2+a^2}\\ &=\lim_{R\to\infty} \left[\frac{1}{a}\arctan\left(\frac{x}{a}\right)\right]_0^R\\ &=\lim_{R\to\infty}\frac{1}{a} \left(\arctan\left(\frac{R}{a}\right)-\arctan(0)\right)\\ &=\frac{1}{a}\left(\frac{\pi}{2}-0\right) =\frac{\pi}{2a}. \end{aligned}

Why this proves convergence: The finite-cutoff integrals approach the finite number $\pi/(2a)$. By the definition of a right-infinite integral, this establishes that the improper integral converges and has that value.

Worked Examples

Worked Example: The Unit-Scale Tail

Evaluate $\int_0^\infty \frac{dx}{x^2+1}$. Here $a=1$, so the antiderivative is $\arctan(x)$. Start with a finite cutoff $R>0$:

\begin{aligned} \int_0^R \frac{dx}{x^2+1} &=\left[\arctan(x)\right]_0^R\\ &=\arctan(R)-\arctan(0)\\ &=\arctan(R). \end{aligned}

As $R\to\infty$, $\arctan(R)\to\pi/2$. Therefore,

\int_0^\infty \frac{dx}{x^2+1} =\lim_{R\to\infty}\arctan(R) =\frac{\pi}{2}.

Check the antiderivative: $\frac{d}{dx}\arctan(x)=1/(1+x^2)$, which is the given integrand. The cutoff calculation and its limit therefore give the claimed value.

Worked Example: A Larger Positive Parameter

Evaluate $\int_0^\infty \frac{dx}{x^2+9}$. Since $9=3^2$, the parameter is $a=3$. An antiderivative is $\frac{1}{3}\arctan(x/3)$. With $R>0$:

\begin{aligned} \int_0^R \frac{dx}{x^2+9} &=\left[\frac{1}{3}\arctan\left(\frac{x}{3}\right)\right]_0^R\\ &=\frac{1}{3}\arctan\left(\frac{R}{3}\right) -\frac{1}{3}\arctan(0). \end{aligned}

When $R\to\infty$, $R/3\to\infty$, so the first arctangent tends to $\pi/2$. Thus

\int_0^\infty \frac{dx}{x^2+9} =\frac{1}{3}\left(\frac{\pi}{2}-0\right) =\frac{\pi}{6}.

Check by differentiation:

\frac{d}{dx}\left(\frac{1}{3}\arctan\left(\frac{x}{3}\right)\right) = \frac{1}{3}\cdot\frac{1}{1+x^2/9}\cdot\frac{1}{3} = \frac{1}{9+x^2}.

This is the original integrand, and the cutoff limit confirms the value $\pi/6$.

Worked Example: A Tail Starting at a Positive Number

Evaluate $\int_2^\infty \frac{dx}{x^2+4}$. Here $a=2$. For a finite cutoff $R>2$, use the antiderivative $\frac{1}{2}\arctan(x/2)$:

\begin{aligned} \int_2^R \frac{dx}{x^2+4} &=\left[\frac{1}{2}\arctan\left(\frac{x}{2}\right)\right]_2^R\\ &=\frac{1}{2}\arctan\left(\frac{R}{2}\right) -\frac{1}{2}\arctan(1). \end{aligned}

As $R\to\infty$, the first arctangent tends to $\pi/2$. Also, $\arctan(1)=\pi/4$. Therefore,

\begin{aligned} \int_2^\infty \frac{dx}{x^2+4} &=\frac{1}{2}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\\ &=\frac{1}{2}\cdot\frac{\pi}{4} =\frac{\pi}{8}. \end{aligned}

The derivative of the proposed antiderivative is $\frac{1}{2}\cdot\frac{1}{1+x^2/4}\cdot\frac{1}{2}=1/(x^2+4)$, as required. The lower-endpoint term is subtracted, so this is only the part of the full tail lying to the right of $x=2$.

Any Finite Lower Endpoint

The same cutoff argument works if the lower endpoint is a finite real number $c$, not just zero or a positive number. The integrand remains continuous, and the arctangent antiderivative applies on the whole real line. The only limit needed is still $\arctan(R/a)\to\pi/2$ as $R\to\infty$.

Theorem: If $a>0$ and $c\in\mathbb{R}$, then $$ \int_c^\infty \frac{dx}{x^2+a^2} = \frac{1}{a}\left(\frac{\pi}{2}-\arctan\left(\frac{c}{a}\right)\right). $$

To justify the formula, evaluate first on $[c,R]$, where $R>c$. The Fundamental Theorem of Calculus gives the difference between the antiderivative at $R$ and at $c$. Then take $R\to\infty$. This proves the stated value and its convergence, rather than treating infinity as an ordinary endpoint.

\begin{aligned} \lim_{R\to\infty}\int_c^R \frac{dx}{x^2+a^2} &=\lim_{R\to\infty}\frac{1}{a} \left(\arctan\left(\frac{R}{a}\right)-\arctan\left(\frac{c}{a}\right)\right)\\ &=\frac{1}{a}\left(\frac{\pi}{2}-\arctan\left(\frac{c}{a}\right)\right). \end{aligned}

Taking $c=0$ recovers the half-line formula because $\arctan(0)=0$. Taking $c>0$ gives a shorter tail, as the third worked example illustrates. For $c<0$, the same expression remains valid; the integrand has no singularity at zero because $a>0$.

Common Pitfalls

A frequent error is to omit one of the chain-rule factors when differentiating the proposed antiderivative. Another is to treat the upper endpoint infinity as if it were a number that can be substituted before taking a limit.

Common Mistake: Losing a Scale Factor

For $a=3$, a wrong calculation is to use $\arctan(x/3)$ alone as an antiderivative. Differentiating shows the problem:

\frac{d}{dx}\arctan\left(\frac{x}{3}\right) = \frac{1}{1+x^2/9}\cdot\frac{1}{3} = \frac{3}{x^2+9},

which is three times the required integrand. The correction is to include the coefficient $1/3$:

\frac{d}{dx}\left(\frac{1}{3}\arctan\left(\frac{x}{3}\right)\right) = \frac{1}{3}\cdot\frac{3}{x^2+9} = \frac{1}{x^2+9}.

Consequently the integral is $\pi/6$, not $\pi/2$. Always differentiate a proposed antiderivative to check the scale.

A second limit error is to say that $\arctan(R/a)$ becomes exactly $\pi/2$ at a finite $R$. It does not: for every finite positive $R/a$, the arctangent is less than $\pi/2$. The value $\pi/2$ appears only as the limit. Also, replacing $a>0$ by a negative parameter without adjusting the formula can introduce a sign error. Since the denominator contains $a^2$, a general nonzero parameter gives $\pi/(2|a|)$ for the integral from zero to infinity.

Why the Result Is Useful

The arctangent tail is a concrete model of how an improper integral can converge: the interval extends without bound, but the accumulated area approaches a finite limit. Its value also records how the parameter changes the scale. Increasing $a$ makes the denominator larger at each fixed $x$, and the formula $\pi/(2a)$ correspondingly decreases.

The key habits are broadly useful in evaluating infinite-interval integrals: retain a finite cutoff, apply an antiderivative only on a finite interval, and take the limit explicitly. The resulting answer is justified only when that limit exists. Here the arctangent limit proves both convergence and the exact value.

Check Your Understanding

Answer each question by using a finite cutoff and the arctangent limit where appropriate.

What condition on $a$ was used for the formula $\pi/(2a)$ as stated, and what is the result for a nonzero real parameter of either sign?
Find an antiderivative of $1/(x^2+16)$ and verify it by differentiation.
Evaluate $\int_0^\infty \frac{dx}{x^2+16}$ by writing the cutoff integral and taking its limit.
State the formula for $\int_c^\infty \frac{dx}{x^2+a^2}$ when $a>0$ and $c$ is finite.
Why is it incorrect to substitute infinity directly into the antiderivative as though infinity were a finite endpoint?

What You'll Learn

Why an Arctangent Appears in an Infinite Tail

Definition and Antiderivative

Deriving the Right-Hand Tail Formula