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Dirichlet's test claims that for two continuous functions $f,g\in[a,\infty]$ where $f,g\geq 0$, if a certain $M$ exists such that $\left|\int_a^bf(x)dx\right|\leq M$ for every $a\leq b$, and $g(x)$ is monotonically decreasing, and $\lim_{X\to\infty}g(x)=0$, then $\int_a^\infty fg$ is convergent.

Does this also apply for a non-continuous $f(x)$? $g(x)$ is still continuous.

This question relates to another question of mine, regarding a specific integral problem. If this is true, then my other problem will be solved.

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Ory Band
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    Your version of Dirichlet's test seem to be wrong: take $f\equiv 1$ and $g(x) = 1/x$. – Dirk May 04 '12 at 18:36
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    @Ory: Your condition should read there exists $M$ such that $\int_a^x f(t) dt \leq M$ for all $x > a$ –  May 04 '12 at 18:40
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    @Ory: Also the restriction that $f$ is continuous might be a bit too strong. I think it should still work if $f$ is continuous except on a measure zero set. Hence, it will work for the problem you have posed in the other thread. –  May 04 '12 at 18:41
  • @Marvis I've stated that such $M$ should exists by stating that $f(x)$ is upper-bounded. Will fix the question text, thanks. – Ory Band May 05 '12 at 09:41
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    Check this reference http://books.google.it/books?id=UkQ8ANdjJUcC&pg=PA441&lpg=PA441&dq=dirichlet+test+improper+integrals&source=bl&ots=IC4VQmn272&sig=zyhp9cfkSB6_bMW1vDz85WUtldQ&hl=it&sa=X&ei=AvykT7WgFfHP4QSLr-jACQ&ved=0CGsQ6AEwAw#v=onepage&q=dirichlet%20test%20improper%20integrals&f=false – Siminore May 05 '12 at 10:09
  • @Dirk I found what's missing. It's $\int_a^b f(x)dx$ that needs to be bounded, not the original $f(x)$ (just like Marvis said). Your example where $f\equiv1$ makes $\int_a^\infty 1=\infty$. – Ory Band May 05 '12 at 11:13
  • Anyway, I've still haven't found the answer I was looking for (just fixes for my typos) - Is this test also true for a non-continuous $f(x)$? – Ory Band May 05 '12 at 11:17
  • @Marvis regarding the problem on my other question - This doesn't help solve the problem with the second integral, since there're an infinite number of non-continuous points at $[a,\infty]$ for $(-1)^{\lfloor x \rfloor}$. Either that or that I didn't understand what you've explained. I haven't studied measure theory yet. – Ory Band May 05 '12 at 11:32
  • @OryBand The set of points at which the function $(-1)^{\lfloor x \rfloor}$ is ${1,2,\ldots}$ which is a measure zero set. –  May 05 '12 at 15:10

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I paste here a theorem from Mathematical Analysis by S. C. Malik,Savita Aror (see also Google Books here)

Theorem. If $\phi$ is bounded and monotonic in $[a,+\infty)$ and tends to zero at $+\infty$, and $\int_a^X f$ is bounded for $X \geq a$, then $\int_a^\infty f \phi$ is convergent.

As you can see, no continuity is really necessary. Honestly, the continuity assumption appears often for the sake of simplicity. Improper integrals can be defined as limits of Riemann integrals: all you need is local integrability. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught.

Siminore
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