if f(x)

Let f,g,h:R→Rf,g,h:\mathbb{R}\rightarrow\mathbb{R}, and a∈Ra\in\mathbb{R} such that ∀x∈[a,∞] f(x)0 ∃M∈R: ∀m>M |m∫af(x)dx−L|<ϵ2⟹∀m1,m2>M|m2∫m1f(x)dx|<ϵ\int\limits_a^\infty f(x)dx = L \Longrightarrow \forall \epsilon > 0\ \exists M \in \mathbb{R}:\ \forall m > M\ \left\vert\int\limits_a^m f(x)dx – L\right\vert < \frac{\epsilon}{2} \Longrightarrow \forall m_1, m_2 > M \left\vert\int\limits_{m_1}^{m_2} f(x)dx\right\vert < \epsilon Thus limk→∞∞∫kf(x)dx=0\lim\limits_{k\rightarrow\infty} \int\limits_k^\infty f(x)dx = 0. Same applies for h(x)h(x). We get \forall \epsilon > 0 \exists M > a:\ \forall M 0 \exists M > a:\ \forall M m_1m_2 > m_1, hence \forall \epsilon > 0 \exists M>a:\ \forall m_1,m_2 >M\ \left\vert\int\limits_{m_1}^{m_2}g(x)dx\right\vert< \epsilon\forall \epsilon > 0 \exists M>a:\ \forall m_1,m_2 >M\ \left\vert\int\limits_{m_1}^{m_2}g(x)dx\right\vert< \epsilon If \int\limits_{a}^{\infty}g(x)dx \leq \epsilon\int\limits_{a}^{\infty}g(x)dx \leq \epsilon diverges, then \exists \epsilon>0:\ \forall M>a \exists m_1,m_2>M:\ \left\vert\int\limits_{a}^{m_1}g(x)dx – \int\limits_{a}^{m_2}g(x)dx\right\vert > \epsilon \Rightarrow \left\vert\int\limits_{m_1}^{m_2}g(x)dx\right\vert > \epsilon\exists \epsilon>0:\ \forall M>a \exists m_1,m_2>M:\ \left\vert\int\limits_{a}^{m_1}g(x)dx – \int\limits_{a}^{m_2}g(x)dx\right\vert > \epsilon \Rightarrow \left\vert\int\limits_{m_1}^{m_2}g(x)dx\right\vert > \epsilon. But we already showed that there exists such an MM such that this isn’t true for. Hence, the integral converges.

I considered this after studying about the comparison test for positive functions, and thought one could argue a a more general one. Thanks!

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1 Answer
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I assume you are using Riemann integration.

The answer is no. Consider the following counter-example.

Denote by \alpha\alpha the function \alpha:\mathbb{R}\rightarrow\mathbb{R}\alpha:\mathbb{R}\rightarrow\mathbb{R} defined for every x \in \mathbb{R}x \in \mathbb{R} as follows.

\alpha(x) := \frac{1}{x^2}\mathbb{1}_{\mathbb{R}\setminus[-1,1]}.

\alpha(x) := \frac{1}{x^2}\mathbb{1}_{\mathbb{R}\setminus[-1,1]}.

Now consider the following functions.

\begin{align}
f &:= -\mathbb{1}_{[-1,1]} + -\alpha, \\
g &:= \mathbb{1}_{[-1,1]\cap\mathbb{Q}}, \\
h &:= 2\mathbb{1}_{[-1,1]} + \alpha.
\end{align}

\begin{align}
f &:= -\mathbb{1}_{[-1,1]} + -\alpha, \\
g &:= \mathbb{1}_{[-1,1]\cap\mathbb{Q}}, \\
h &:= 2\mathbb{1}_{[-1,1]} + \alpha.
\end{align}

Then

f < g < hf < g < h. ff and hh are Riemann integrable on all of \mathbb{R}\mathbb{R}. gg is not Riemann integrable on any interval that contains [0,1][0,1], as gg coincides there with the Dirichlet function. Addendum If gg is Riemann integrable on [a,m][a,m] for every m \in [a,\infty)m \in [a,\infty), then the improper integral \int_a^\infty g(x)\ dx\int_a^\infty g(x)\ dx does converge. This can be derived from the Cauchy criterion for convergence of an improper integral in a manner similar to the argument you detailed in your post.      Oh right! I dropped over one of the comparison-test's demands, all three functions have to be Riemann integrable on [a,m][a,m] for any m>am>a. Apologize, I’ve updated the post.
– galra
2 days ago

  

 

@gaira: I’ve rolled back your changes. I have already answered your original question. Please accept my answer. If you want to post another question, you may do this.
– Evan Aad
2 days ago

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@galra: I have added an addendum to my answer to cover the additional condition of integrability.
– Evan Aad
2 days ago