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I was searching a proof of the following Abel-Dirichlet test :

Theorem : If $\Phi$ is bounded and monotonic in $[,+∞)$ and tends to zero at $+\infty$, and $\int_{a}^{x} f dt$ is bounded for $x \geq a$, then $\int_{a}^{+\infty} f \cdot \Phi dt$ is convergent.

The most similar question asked i found is the following : Dirichlet's test for convergence of improper integrals

But no proofs were given,

I was obviously interested in a proof that requires minimal hypothesis, in particular with $f$ that has not to be necessarly continuous.

Since i can't consult the pdf of the book quoted in the link above, i was wondering wether we could discuss about the proof of the Theorem, in the easiest way, since my knowledge involves the first year of analysis\calculus at university and some notions about Null set and its consequences as the Vitali-Lebesgue Theorem.

jacopoburelli
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2 Answers2

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The theorem is proved by applying the second mean value theorem for integrals (given that $\Phi$ is monotone and $f$ is simply integrable).

For $c_2 > c_1 > a$, there exists $\xi \in (c_1,c_2)$ such that

$$\tag{*}\left|\int_{c_1}^{c_2}\Phi(x) f(x) \, dx\right| = \left|\Phi(c_1)\int_{c_1}^{\xi}f(x) \, dx + \Phi(c_2)\int_{\xi}^{c_2}f(x) \, dx \right| \\ \leqslant |\Phi(c_1)|\left|\int_{c_1}^{\xi}f(x) \, dx\right| + |\Phi(c_2)|\left|\int_{\xi}^{c_2}f(x) \, dx\right|.$$

Under the hypotheses of the Dirichlet test, the integrals on the RHS are uniformly bounded for all $a < c_1 < \xi < c_2$ and $\Phi(c_1),\Phi(c_2) \to 0$. Choosing $c_2 > c_1$ sufficiently large, (*) can be made smaller than any $\epsilon >0$. Thus, the improper integral is convergent by virtue of the Cauchy Criterion.

RRL
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The assumptions can be relaxed if one works in the setting of Lebebgue integration.

Propoisiton A: Suppose that

  • (a) $f\in L^{loc}_s[a,\infty)$ (i.e. $f$ is Lebesgue integrable in any closed and bounded subinterval of $[a,\infty)$).
  • (b) $C=\sup_{x\geq a}\Big|\int^x_a f(t)\,dt\Big|<\infty$
  • (c) $g$ is nonincreasing, bounded, nonnegative on $[a,\infty)$ with $g(x)\xrightarrow{x\rightarrow\infty}0$.

Then $I=\lim_\limits{b\rightarrow\infty}\int^b_a f(t)g(t)\,dt$ exists (as a real number).

Here is another result with slightly different conditions.

Propositon B: Suppose that

  • (a) $f\in L^{loc}_s[a,\infty)$ (i.e. $f$ is Lebesgue integrable in any closed and bounded subinterval of $[a,\infty)$
  • (b) $\lim_\limits{b\rightarrow\infty}\int^b_af(t)\,dt$ exits (as a real number)
  • (c) $g$ is monotone and bounded

Then $I=\lim_\limits{b\rightarrow\infty}\int^b_a f(t)g(t)\,dt$ exists (as a real number).

Proof of Propositions: By the second mean value theorem for integrals, for any $a<M<N$, there is $x_0=x_0(M,N)\in[M,N]$ such that \begin{align} \int^N_M f(t)g(t)\,dt=g(M)\int^{x_0}_Mf(t)\,dt+g(N)\int^N_{x_0}f(t)\,dt\tag{0}\label{zero} \end{align}

Under the assumptions of Proposition A, for any $\varepsilon>0$, there is $K>a$ such that $|g(t)|<\frac{\varepsilon}{2(C+1)}$ whenever $t\geq K$. From \eqref{zero}, if $N>M>K$, $$\Big|\int^N_M f(t)g(t)\,dt\Big|<\varepsilon$$

Under the conditions of Proposition B, for any $\varepsilon>0$, there is $K>a$ such that $\big|\int^p_qf(t)\,dt\big|<\frac{\varepsilon}{2(\|g\|+1)}$ whenever $p\geq q\geq K$. From \eqref{zero}, if $K<M<N$, $$\Big|\int^N_M f(t)g(t)\,dt\Big|<\varepsilon$$

In both cases, it follows from Cauchy's criteria that $\lim_\limits{b\rightarrow\infty}\int^b_a f(t)g(t)\,dt$ exists.

Mittens
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