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Test the improper integral ∫ 1 / (x3 + 1)1/2 for convergence

Test the following improper integral for convergence:

    \[ \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \]

The integral converges.

Proof. We know the integral \int_0^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx converges (example #1 on page 417 of Apostol). Applying the limit comparison test (by the note to Theorem 10.25 on page 418, which says that if \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 then the convergence of \int g(x) implies the convergence of \int f(x).) we have

    \begin{align*}  \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{ \frac{1}{\sqrt{x^3+1}}}{\frac{1}{x}} \\[9pt]  &= \lim_{x \to +\infty} \frac{x}{\sqrt{x^3+1}} \\[9pt]  &= \lim_{x \to +\infty} \frac{1}{x^{\frac{1}{2}} \sqrt{1 + \frac{1}{x}}} \\[9pt]  &= 0. \end{align*}

Since we know \int_0^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx converges the theorem establishes the convergence of

    \[ \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \qquad \blacksquare \]


  1. Sorry, but integral of q/x^{3/2} do not converge from 0 to infinity, but from 1 to infinity. It seems you need to split the integral in two: one from zero to 1 (converges because its a proper integral), and 1 to infinity (converges because of the comparison limit test).
    Is that ok?

    • Anonymous says:

      However, when you split the integral, the integral of x^(-3/2) between 0 and 1 diverges because this integral is not well defined on 0 and it’s limit diverges.

      • Bruno Tavares says:

        you can split the question’s integral in two integrals from 0 to 1 and from 1 to inf.The first exists and it is finite because the function inside is continuous from 0 to 1.The second can be comparated with the integral of x^(-3/2) from 1 to inf without the problem of inexistence of this integral at 0.

      • Anonymous says:

        I take it back, you can’t compare with 1/x^(-3/2) because 10.25 needs it to exist on 0,infty but it doesn’t.

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