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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous and 1 - periodic. If $ \int_0^1f(x)dx = 0$. Then $$\int_1^{\infty}\frac{f(x)}{x}dx $$ converges.

My question is that does a bound on the $\int_1^{\infty}\frac{f(x)}{x}dx$ will imply convergence of the integral? Please justify as well.

And if not then how does one show convergence.

My Attempt

Let $ G(x) = \int_0^xf(x)dx$ then $G'(x)=f(x)$ we have to show that $\int_1^{\infty}\frac{G'(x)}{x}dx$ converges. We note that $f(x)$ is bounded on $[0,1]$ say by $M$ and hence on $[0, \infty)$. Then so is $G(x) =\int_0^xf(x)dx= \int_{[x]}^xf(x)dx <M(x- [x])\leq M$.

Using integration by parts we have $$ \left.\frac{G(x)}{x}\right|_1^{\infty} + \int_1^{\infty}\frac{G(x)}{x^2}dx < M(2)$$

na1201
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  • " does a bound on the [integral] will imply convergence of the integral?" I'm not sure if this is exactly what you mean, but note that $\int_0^x \cos t,dt = \sin x$ is bounded between $-1$ and $1$, but $\int_0^\infty \cos t,dt$ doesn't converge. However, in general, bounding $\int_{a}^{\infty}|f(x)|,dx$ does imply that $\int_a^\infty f(x),dx$ converges. (Of course, that might not help here). – JimmyK4542 Dec 30 '16 at 20:02
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    For a bound to imply convergence, we need to have that $\int_1^\infty \frac{f(t)}{t}\mathrm{d}t$ to be monotonically increasing (which happens if the integrand is non-negative, which isn't true here unless $f$ is identically zero). – Mark Schultz-Wu Dec 30 '16 at 20:05
  • Perhaps prove that $\int_1^N\frac{f(x)}x\ dx$ is bounded and monotone? – Simply Beautiful Art Dec 30 '16 at 20:07
  • You might want to take a look at Abel-Dirichlet criterion for convergence of improper integrals, if you want something more general : If $f,g : [a, \infty) \to \mathbb R$ are two functions such that $g$ is monotonic, differentiable and $\lim \limits_{x \rightarrow \infty} g(x) = 0$, and $f$ is continuous such that $F(x) = \int_a^x f(u)du$ is bounded on $[a, \infty)$, then $\int_a^{\infty} f(x) g(x) dx$ converges. – Desura Dec 30 '16 at 20:15
  • @Desura. Thanks. Can you give me a source where I can find the proof of the theorem you have quoted. – na1201 Dec 30 '16 at 21:42
  • Related: Improper Integral of a periodic function converges – Winther Dec 30 '16 at 21:55
  • I like your approach. Another way to do it is to rewrite the integral as follows: $$\int_1^\infty \frac{f(x)}{x}{\rm d}x = \sum_{n=1}^\infty \int_0^1\frac{f(x)}{n+x}{\rm d}x = \sum_{n=1}^\infty\int_0^1 \left[\frac{f(x)}{n+x} - \frac{f(x)}{n}\right]{\rm d}x = \sum_{n=1}^\infty\int_0^1 \frac{xf(x)}{n(n+x)}{\rm d}x$$ In the first equality I used the period nature of $f$ to say $f(n+x) = f(x)$ and I subtracted $0 = \int_0^1 f(x)/n {\rm d}x$ in the second equality. This last integral(s) are bounded by $\frac{\sup_{x\in[0,1]}|xf(x)|}{n^2}$ whose sum converges. – Winther Dec 30 '16 at 22:05
  • @manhattan If you are okay with comparison test for improper integral and absolute convergence implies convergence, I added a proof of the criterion below. – Desura Dec 30 '16 at 22:43

3 Answers3

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Dirichlet's test is a consequence of summation/integration by parts.
In our case, it is useful in its integral formulation:

If $f(x)$ has a bounded primitive and $g(x)$ is decreasing to zero on $\mathbb{R}^+$,
the integral $\int_{0}^{+\infty}\frac{f(x)}{x}\,dx$ exists.

If $f(x)$ is continuous, $1$-periodic and with mean zero, surely $$\left|\int_{0}^{M}f(x)\,dx\right| = \left|\int_{\left\lfloor M\right\rfloor}^{M}f(x)\,dx\right|\leq \max_{x\in[0,1]}|f(x)| = C $$ for any $M\in\mathbb{R}^+$, so the claim trivially follows from Dirichlet's test.

Jack D'Aurizio
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You are on the right track. Integrate by parts to see that $$ \int_1^z{f(x)\over x}\,dx = {G(z)\over z}-{G(1)\over 1}+\int_1^z {G(x)\over x^2}\,dx,\quad z>1. $$ As you observe, $G$ is bounded. This ensures that the first term on the right of the display above converges to $0$, and that the integral on the right converges.

John Dawkins
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Your problem is a special case that can be solved by Abel-Dirichlet Criterion. You only need to prove that the function $F(x) = \int_1^x f(u)du$ is bounded on $[1, \infty)$ to use it.

Proof of Abel-Dirichlet Criterion :

Let $f, g : [a, \infty) \mapsto \Bbb R$ be two functions such that $f$ is continuous, $F : [a, \infty) \mapsto \Bbb R$ defined by $F(x) = \int_a^x f(u)du$ is bounded (say by a positive constant $M$), and such that $g$ is monotonic, continuously differentiable, with $\lim \limits_{x \rightarrow \infty} g(x) = 0$. Then $\int_a^{\infty} f(u)g(u)du$ converges.

Indeed, consider $\int_a^x f(u)g(u)du$.

We integrate it by parts : \begin{equation}\tag{1} \int_a^x f(u)g(u)du = \left.F(u)g(u)\right|_a^{x} - \int_a^x F(u)g'(u)du = F(x)g(x) - F(a)g(a) - \int_a^x F(u)g'(u)du = F(x)g(x) - \int_a^x F(u)g'(u)du \end{equation} because $F(a) = 0$.

Since $F$ is bounded, $\lim \limits_{x \rightarrow \infty} g(x) = 0$, we have $\lim \limits_{x \rightarrow \infty} F(x)g(x) = 0$.

Finally, notice that the improper integral $\int_a^{\infty} F(u)g'(u)du$ converges absolutely (so it converges) :

Indeed, $\lvert F(u)g'(u) \rvert \leq \lvert M g'(u) \rvert = M \lvert g'(u) \rvert $. Since $g$ is monotonic, either $g'$ is positive, or $g'$ is negative. Without loss of generality, it is positive. So $\lvert F(u)g'(u) \rvert \leq M g'(u)$ (just add a minus sign if $g'$ is negative)

Now, $\int_a^{\infty} Mg'(u)du = \lim \limits_{x \rightarrow \infty} \int_a^{x} Mg'(u)du = \lim \limits_{x \rightarrow \infty} (Mg(x) - Mg(a)) = -Mg(a)$. So $\int_a^{\infty} Mg'(u)du$ converges. We conclude by comparison test that $\int_a^{\infty} F(u)g'(u)du$ converges absolutely.

Thus, by taking the limit $x \rightarrow \infty$ in $(1)$, we have that $\int_a^{\infty} f(u)g(u)du$ converges.

Desura
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  • Thanks. That solves my problem. :) – na1201 Dec 30 '16 at 23:12
  • No problem. By the way, I made a mistake in my hypotheses. You need $g$ continuously differentiable (otherwise, you cannot always integrate $g'$). I corrected it. It still applies to your problem. – Desura Dec 31 '16 at 00:01