Improper Integral of a periodic function converges

Question

Given that $f(x)$ is a $p$-periodic function and $\int_0^p{f(x)}dx=0$. Show $\int_1^\infty\frac{f(x)}{x}dx$ converges.

1. I know this integral can be broken into $\int_1^\infty\frac{f(x)}{x}dx=\int_0^\infty\frac{f(x)}{x}dx-\int_0^1\frac{f(x)}{x}dx$ for easier dealing if needed

2. My first thought it so say that with $f(x)$ periodic, we can break the integral into parts such that: $\int_0^\infty\frac{f(x)}{x}dx=\int_0^p\frac{f(x)}{x}dx+\int_p^{2p}\frac{f(x)}{x}+...+\int_{(n-1)p}^{np}\frac{f(x)}{x}dx+...$ from here we can somehow use the given fact that $\int_0^pf(x)dx=0$ and that $\int_a^{a+np}f(x)dx=n\int_0^pf(x)dx$

We also can use that $\lim_{x->\infty}\frac{1}{x}=0$

But can somehow help me put these thoughts together in such a way that my mathematical proof is rigorous?

Answer 1

Define $$ F(x) = \int_{0}^x f(t) \, dt. $$ The $\int_0^p f = 0$ condition gives us that $F$ is also periodic, and we also have $$ \lvert F(x)\rvert \leqslant \int_0^x \lvert f(x)\rvert \, dx \leqslant \int_0^p \lvert f(x)\rvert \, dx = A, $$ say, for all $x$.

Now consider the limit used in the definition of the improper integral, and integrate by parts: $$ \int_1^R \frac{f(x)}{x} \, dx = \left[ \frac{F(x)}{x} \right]_1^R + \int_1^R \frac{F(x)}{x^2} \, dx = \frac{F(R)}{R}-F(1) + \int_1^R \frac{F(x)}{x^2} \, dx $$ The first term tends to zero since $\lvert F(R)\rvert $ is bounded by $A$, and for the last term, $$ \left\lvert\int_R^{\infty} \frac{F(x)}{x^2} \, dx\right\rvert \leqslant \int_R^{\infty} \frac{\lvert F(x)\rvert}{x^2} \, dx \leqslant A\int_R^{\infty} \frac{dx}{x^2} = \frac{A}{R} \to 0, $$ so this improper integral exists as a well-defined limit. Hence the original integral exists as an improper integral.

Answer 2

I will assume that $f$ is continuous. Let $F(x)=\int_1^xf(t)\,dt$. Then $F(1)=0$, $F'(x)=f(x)$ and, since $\int_{1+np}^{1+(n+1)p}f(t)\,dt=0$ for all $n\in\mathbb{N}$, $F$ is bounded. Then for any $R>1$, integrating by parts we have $$ \int_1^R\frac{f(x)}{x}\,dx=\frac{F(x)}{x}\Bigr|_1^R+\int_1^R\frac{F(x)}{x^2}\,dx=\frac{F(R)}{R}+\int_1^R\frac{F(x)}{x^2}\,dx. $$ Since $F$ is bounded $F(R)/R\to0$ as $R\to\infty$, and $$ \int_1^\infty\frac{F(x)}{x^2}\,dx. $$ is absolutely convergent.

Answer 3

Let us define $$F(x):=\int_1^x{\frac{f(y)}{y}dy}$$ In order to prove convergence of $F$ as $x\rightarrow \infty$ we will show that $F$ is bounded with derivative that tends to zero.

Since $f$ periodic with $\int_0^pf(x)dx=0$ there is some constant $c>0$ such that $|\int_0^xf(y)dy|\leq c$ for all $x\in[0,\infty)$ (uniformly bounded). Then $F$ can be written as $$F(x)=\int_1^x{\frac{1}{y}\frac{d}{dy}\bigg[\int_0^y{f(u)du}\bigg]dy}=\frac{1}{x}\int_0^x{f(u)du}-\int_0^1{f(u)du}+\int_1^x{\frac{1}{y^2}\bigg[\int_0^y{f(u)du}\bigg]dy}$$ Note that for $x>1$ $$\Bigg|\int_1^x{\frac{1}{y^2}\bigg[\int_0^y{f(u)du}\bigg]}dy\Bigg|\leq c\int_1^x{\frac{1}{y^2}dy}\leq c$$ and $$\Bigg|\frac{1}{x}\int_0^x{f(u)du}\Bigg|\leq c $$ Combining the previous inequalities with the above identity boundedness of $F$ can be deduced. Also $\lim_{x\rightarrow\infty}\frac{dF(x)}{dx}=\lim_{x\rightarrow \infty}\frac{f(x)}{x}=0$. Since $F$ is bounded with $\lim_{x\rightarrow\infty} F'(x)=0$ it converges.

Answer 4

Te method describe by other solutions can be used to solve a slightly more general solution to the OP.

This is problem in T. Apostol's analysis book (2nd ed)

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Problem: Suppose $f$ is $p$ periodic and integrable (either as in the sense or Riemann or Lebesgue) over $[0,p]$ Then $\lim_{A\rightarrow \infty}\int^A_p x^{-s}f(x)\,dx$ exists for all $s>0$.

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Let $F(x)=\int^x_p f(t)\,dt$, for $x\geq p$. It is easy to check that $F$ can be extended as a continuous periodic function (period $p$) over the real line.

Integration by parts gives

\begin{aligned} \int^A_p x^{-s}f(x)\,dx&=\int^A_p x^{-s} dF(x)= x^{-s}f(x)|^A_p +s \int^A_p \frac{F(x)}{x^{s+1}}\,dx\\ &=A^{-a}F(A)+s\int^A_p\frac{F(x)}{x^{s+1}}\,dx \end{aligned}

Since $F$ is bounded and $\int^\infty_0\frac{1}{x^{s+1}}\,dx <\infty$ for all $s>0$, we have that $\int^\infty_A\frac{F(x)}{x^{s+1}}\,dx$ converges (both as an improper Riemann integral and also in the sense of Lebesgue) and so

$$\lim_{A\rightarrow\infty}\int^A_p x^{-s}\,f(x)\,dx=s\int^\infty_p\frac{F(x)}{x^{s+1}}\,dx $$