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Prove or disprove: If lim f(x) = 0 and lim In = A then ∫ f(x) converges to A

by RoRi

The following function f is defined for all x \geq 1, and n is a positive integer. Prove or provide a counterexample to the following statement.

Assume

    \[ \lim_{x \to \infty} f(x) = 0 \qquad \text{and} \qquad \lim_{n \to \infty} I_n = A. \]

Then

    \[ \int_1^{\infty} f(x) \, dx = A. \]

Incomplete.

4 comments

  1. Evangelos says:

    Proof. For some positive real x such that n \leq x \leq n+1, we have

        \begin{align*} \int_{1}^{x} f(x)\, dx &= \int_{1}^{n} f(x)\, dx + \int_{n}^{x} f(x)\, dx \\ &= I_{n} + (F(x) - I_{n}) \end{align*}

    where F(x) = \int_{1}^{x} f(x)\, dx. But as n \rightarrow \infty and hence x \rightarrow \infty, f(x) \rightarrow 0, and thus F(x) - I_{n} = \int_{n}^{x} f(x)\, dx \rightarrow 0. But we know \lim_{n \to \infty} I_{n} = A, which implies \lim_{x \to \infty} F(x) = A. Thus, we have shown that

        \begin{align*} \lim_{x \to \infty} \int_{1}^{x} f(x)\, dx = A \end{align*}

    This completes the proof.

    • William C says:

      A clearer way to show that the integral from n to x of f(x) approaches zero could be to show that f(x) is bounded by some M and -M for large enough x (which follows from the limit of f(x), so that the integral from n to x of f(x) <= integral from n to x of M <= M (by the same logic the integral must also be greater than -M). Since M goes to zero (which also follows from the limit of f(x), the integral must also go to zero (follows from the squeeze theorem). I'm not sure if this is necessary but thought I'd put this here in case anyone finds it helpful.

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