Improper integral (1+arcsin(1/x))/(1+x√x) from 1 to ∞

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

Find the leading terms of the numerator and denominator as x → +∞ (or as we approach a singular point). In our case the leading term in the numerator is the constant 1 + π/2 (arcsin(1/x) is bounded), and in the denominator it is x^(3/2) (since x√x = x^(3/2) grows faster than 1). So g(x) = 1/x^(3/2).

Q: 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 (value 1/(α−1)), for α ≤ 1 it diverges. This integral is used as a reference: if the integral under study is equivalent to it, the conclusion on convergence carries over automatically.

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

Problem

$\int_{1}^{+ \infty} \frac{1 + arcsin (1 / x)}{1 + x \sqrt{x}} d x$

Solution

Step 1

The integrand f(x) = (1+arcsin(1/x))/(1+x√x) is continuous on [1, +∞) since arcsin(1/x) is defined for x ≥ 1, and the denominator is positive. Hence, the integral is improper of the first kind. To investigate convergence, we apply the comparison test in limit form.

Why is the comparison test chosen?

The integrand is a complex combination of the arcsine and a root, whose antiderivative is not expressed in elementary functions. Direct evaluation of the integral is impossible. However, as x → +∞, the function behaves like a power function, which allows us to choose a simpler comparison function g(x) and apply the limit comparison test. This test works when lim f(x)/g(x) is finite and nonzero.

Step 2

Bound the numerator from above. Since arcsin(1/x) for x ≥ 1 takes values in [0, π/2], we have 1 + arcsin(1/x) ≤ 1 + π/2. Then:

| f (x) | = \frac{| 1 + arcsin (1 / x) |}{1 + x \sqrt{x}} \leq \frac{1 + \frac{π}{2}}{1 + x \sqrt{x}} \leq \frac{1 + \frac{π}{2}}{x \sqrt{x}}

Step 3

This shows that f(x) = O(1/x^(3/2)) as x → +∞. Take g(x) = 1/x^(3/2) as the comparison function.

g (x) = \frac{1}{x^{3 / 2}}

Step 4

The integral of the comparison function:

Why does the integral of g(x) converge?

The integral ∫₁⁺∞ dx/xᵅ converges when α > 1 and diverges when α ≤ 1. This is the reference improper integral of the first kind. In our case α = 3/2 > 1, so the integral converges.

\int_{1}^{+ \infty} \frac{d x}{x^{α}}

Here α = 3/2 > 1, so the integral converges.

Step 5

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

α = \frac{3}{2} > 1

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 inside the interval). If lim f(x)/g(x) = c with 0 < c < ∞, then the integrals ∫ f and ∫ g converge or diverge simultaneously.

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