Improper integral arccos(1/x)/(x√x) from 1 to ∞

This problem asks to investigate the convergence of an improper integral of the first kind with an infinite upper limit.

Problem

$\int_{1}^{+ \infty} \frac{arccos \frac{1}{x}}{x \sqrt{x}} d x$

Solution

Step 1

The integrand f(x) = arccos(1/x)/(x√x) is defined and continuous on [1, +∞): for x ≥ 1 the argument 1/x lies in (0, 1], so arccos(1/x) takes values in [0, π/2]; the denominator x√x = x^3/2 ≥ 1 is positive. Hence the integral is improper of the first kind. To investigate convergence, we apply the comparison test in upper-bound form.

Why is the function continuous on the whole interval?

The function arccos u is defined and continuous for u ∈ [−1, 1]. For x ≥ 1, 1/x ∈ (0, 1], which lies in this interval. The denominator x√x = x^3/2 does not vanish for x ≥ 1. Thus, on the whole [1, +∞) the integrand is continuous, and the only singularity of the integral is the infinite upper limit.

Step 2

Let us bound the integrand above by a simpler function whose integral is easy to investigate.

Step 3

For any x ≥ 1 the value arccos(1/x) is bounded:

0 \leq arccos \frac{1}{x} \leq \frac{π}{2}

Step 4

Dividing numerator and denominator, we obtain the bound for the integrand:

0 \leq f (x) \leq \frac{π / 2}{x^{3 / 2}}

Why is bounding the numerator sufficient?

We replace the numerator arccos(1/x) by its maximum value π/2. Since the denominator is positive, the fraction only increases. The resulting function g(x) = 1/x^3/2 is a power function whose integral on [1, +∞) is known and converges. This allows us to apply the comparison test in upper-bound form.

Step 5

Consider the reference integral of the comparison function:

\int_{1}^{+ \infty} \frac{d x}{x^{3 / 2}}

Why does the reference integral converge?

The integral ∫₁⁺∞ dx/x^α converges for α > 1 and diverges for α ≤ 1. This is the classical reference improper integral of the first kind. In our case α = 3/2 > 1, so the integral of g(x) converges.

Step 6

By the comparison test in upper-bound form: if 0 ≤ f(x) ≤ C·g(x) for x ≥ a and ∫ g(x) dx converges, then ∫ f(x) dx also converges. In our case f(x) ≤ (π/2)·(1/x^3/2), and the integral of 1/x^3/2 converges. Hence the original integral converges.

Answer

Converges.

Frequently asked questions

How do I investigate an improper integral for convergence?

The main method is the comparison test. Choose a simpler function g(x) equivalent to f(x) as x → +∞ (or as x → a at a singular point), and compare the integrals. If 0 ≤ f(x) ≤ C·g(x) and ∫ g converges, then ∫ f converges. If f(x) ≥ C·g(x) and ∫ g diverges, then ∫ f diverges.

How do we choose the comparison function g(x)?

Look at the behaviour of the numerator and denominator as x → +∞. In our case the numerator arccos(1/x) is bounded (does not grow), while the denominator x√x = x^3/2 behaves like a power function. Hence g(x) = 1/x^3/2.

What is the reference integral ∫₁⁺∞ dx/xᵅ?

This is an improper integral of the first kind whose convergence depends on α. For α > 1 it converges, for α ≤ 1 it diverges. It is used as a reference: if the integral under study is dominated by the integral of C/x^α with α > 1, then it converges.

Why can the integral not be evaluated directly?

The integrand contains arccos(1/x) and x√x. The antiderivative of such a combination is not expressed in elementary functions. So the only way is to investigate convergence using comparison tests.