# Dirichlet's test for convergence of improper integrals

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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Your version of Dirichlet's test seem to be wrong: take $f\equiv 1$ and $g(x) = 1/x$. – Dirk May 4 '12 at 18:36
@Ory: Your condition should read there exists $M$ such that $\int_a^x f(t) dt \leq M$ for all $x > a$ – user17762 May 4 '12 at 18:40
@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. – user17762 May 4 '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 5 '12 at 9:41
Check this reference books.google.it/… – Siminore May 5 '12 at 10:09

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.