Improper integral arctg(x²/(x+1))/(x²+4) from 0 to ∞

This problem asks to investigate the convergence of an improper integral of mixed type: with a potential singularity at zero and an infinite upper limit.

Problem

$\int_{0}^{+ \infty} \frac{arctg \frac{x^{2}}{x + 1}}{x^{2} + 4} d x$

Solution

Step 1

The integrand f(x) = arctg(x²/(x+1))/(x²+4) may have a singularity at x = 0 and has an infinite upper limit. Split the integral at the point x = 1 into two:

Why was the point x = 1 chosen for the split?

The point x = 1 is a convenient boundary: on [0, 1] the function is bounded (we will verify this separately), and on [1, +∞) an upper-bound estimate can be applied to investigate convergence. This split allows us to handle each part independently.

\int_{0}^{+ \infty} \frac{arctg \frac{x^{2}}{x + 1}}{x^{2} + 4} d x = \int_{0}^{1} \frac{arctg \frac{x^{2}}{x + 1}}{x^{2} + 4} d x + \int_{1}^{+ \infty} \frac{arctg \frac{x^{2}}{x + 1}}{x^{2} + 4} d x

Step 2

Consider the behaviour of f(x) as x → 0+. The argument of arctangent x²/(x+1) → 0, so arctg(x²/(x+1)) ~ x²/(x+1) → 0. The denominator x²+4 → 4. Hence f(x) → 0/4 = 0. The singularity at x = 0 is removable, and the integral on [0, 1] is a proper Riemann integral, not improper.

Why is the singularity at zero removable?

The limit of the function as x → 0+ is finite (equal to 0). Hence the function can be extended at x = 0 with the value 0 and becomes continuous on [0, 1]. Therefore the integral on [0, 1] converges as an ordinary Riemann integral.

Step 3

Now investigate the behaviour of f(x) as x → +∞. The argument of arctangent x²/(x+1) → +∞, so arctg(x²/(x+1)) → π/2. The arctangent function is bounded: for all t > 0, 0 < arctg t < π/2. Hence the numerator is bounded as well.

Step 4

We obtain the upper bound for x ≥ 1:

0 \leq f (x) \leq \frac{π / 2}{x^{2}}, x \geq 1

Step 5

Consider the reference integral:

\int_{1}^{+ \infty} \frac{d x}{x^{2}}

Why does the reference integral converge?

The integral ∫₁⁺∞ dx/xᵅ converges for α > 1 and diverges for α ≤ 1. Here α = 2 > 1, so the integral of g(x) = 1/x² converges.

Step 6

By the comparison test in upper-bound form: the integral on [1, +∞) converges. The integral on [0, 1] is proper. By additivity, the original integral converges.

Answer

Converges.

Alternative path: via limit comparison

The same problem can be solved using the comparison test in limit form. Consider the limit of the ratio f(x)/g(x) with g(x) = 1/x² at infinity:

lim_{x \to + \infty} \frac{f (x)}{1 / x^{2}} = lim_{x \to + \infty} \frac{x^{2} \cdot arctg \frac{x^{2}}{x + 1}}{x^{2} + 4} = \frac{π}{2}

The limit is finite and non-zero (equal to π/2). Hence the integrals ∫₁⁺∞ f(x) dx and ∫₁⁺∞ dx/x² converge or diverge simultaneously. The second integral converges (α = 2 > 1), so the original integral converges as well.

Frequently asked questions

How do I investigate an improper integral of mixed type?

Such an integral is split into parts: at the singularity (if it exists inside or at the endpoint) and at infinity — separately. Each part is investigated by its own method: near a singularity — via equivalence and the reference integral of the second kind, at infinity — via an upper-bound estimate or limit comparison with the reference integral of the first kind.

Why is arctangent bounded and how is this used?

The function arctg t for all t > 0 takes values in (0, π/2). This allows us to replace the numerator by π/2 in the upper bound: 0 ≤ arctg(x²/(x+1)) ≤ π/2. Such a replacement does not change convergence, since π/2 is a finite constant.

What is a removable singularity and why is it important?

A singularity is removable if the limit of the function at that point exists and is finite, but the function itself is either undefined or has a different value at that point. For the integral this means that on a segment near this point the integral is proper (Riemann), not improper. In our case f(x) → 0 as x → 0+, so the singularity is removable.

Which method is preferable — upper-bound estimate or limit comparison?

Both methods give the same result. The upper-bound estimate is simpler when the function is bounded and easily dominated by the reference. The limit comparison is more convenient when the behaviour of the function at infinity needs to be determined precisely (equivalence). In this problem both methods work equally well.