# Does the convergence of $\int_{0}^\infty |f(x)|\,dx$ imply the convergence of $\int_{0}^\infty f(x)\,dx$?

I am facing a question where I am asked to prove the convergence of the following improper integral:

$$\int_{0}^{+\infty} \frac{\cos x}{\sqrt{x}(1+x)}dx$$

To do this, I separate the integral into:

$$\int_{0}^{1} \frac{\cos x}{\sqrt{x}(1+x)}dx+ \lim_{t \to +\infty} \int_{1}^{t} \frac{\cos x}{\sqrt{x}(1+x)}dx$$

Now, my concern is in evaluating the right hand side. My professor told me that to do this, we can compare the convergence of $$\frac{|\cos x|}{\sqrt{x}(1+x)}$$ as such:

$$0 \leq \frac{|\cos x|}{\sqrt{x}(1+x)} \leq \frac{1}{x^{3/2}}$$

And since $$\int_{0}^{+\infty} \frac{1}{x^{3/2}}dx$$ converges ($$p >\frac{3}{2}$$) then $$\int_{0}^{+\infty} \frac{|\cos x|}{\sqrt{x}(1+x)}dx$$ converges.

My question is, how does this imply that $$\int_{0}^{+\infty} \frac{\cos x}{\sqrt{x}(1+x)}dx$$ converges? Is there a way to prove this?

And also does the convergence of $$\int_{0}^\infty |f(x)|dx$$ imply the convergence of $$\int_{0}^\infty f(x)dx$$ generally?

• Lebesgue and Riemann integral are monoton functions, then by $-|f|\leq f \leq |f|$ the desired implication follows. Commented Apr 21, 2023 at 8:11

## 2 Answers

Sure!

First, if you know that $$\int_0^\infty |f|$$ converges, you know that there exists a bound $$M > 0$$ such that, for every $$t \in \mathbb{R}_+$$, $$\int_0^t |f| \leq M$$.

Now, let us write $$f = f_+ + f_-$$, where $$f_+(x) := \max \{ f(x), 0 \}$$ and $$f_-(x) = \min \{ f(x), 0 \}$$. For each $$x \in \mathbb{R}_+$$, you have $$- |f(x)| \leq f_-(x) \leq 0 \leq f_+(x) \leq |f(x)|$$.

We will prove that $$\int_0^\infty f_+$$ converges. You can prove the same for $$f_-$$. And so $$\int_0^\infty f$$ converges since $$f = f_+ + f_-$$.

Let us define $$g(t) := \int_0^t f_+(x) \mathrm{d}x$$. Since $$f_+ \geq 0$$, $$g$$ is a non-decreasing function of $$t$$. Moreover, for every $$t \in \mathbb{R}_+$$, $$g(t) \leq \int_0^t |f(x)| \mathrm{d}x \leq M.$$ Thus $$g$$ is a non-decreasing bounded function, so it converges to some limit as $$t \to + \infty$$.

We recall The Cauchy Criterion for functions when $$x\to+\infty$$:

$$\lim\limits_{x\to+\infty} f(x)$$ exists $$\iff$$ $$\forall \epsilon>0$$, $$\exists M>0$$ s.t. $$x_1>M$$, $$x_2>M$$ $$\implies$$ $$|f(x_1)-f(x_2)|<\epsilon$$

Apply this to $$F(t)=\int_a^t |f(x)|dx$$: If $$\lim\limits_{t\to+\infty}F(t)=\int_a^{+\infty} |f(x)|dx$$ exists, then $$\forall \epsilon>0$$, $$\exists M>0$$ s.t. $$\forall\ t_1,\ t_2>M$$, $$\epsilon>|F(t_1)-F(t_2)|=\mid\int_{t_1}^{t_2}\mid f(x)\mid dx\mid\ge |\int_{t_1}^{t_2} f(x)dx|=|G(t_1)-G(t_2)|$$ where $$G(t)=\int_a^t f(x)dx$$ $$\implies$$ $$\int_a^{+\infty}f(x)dx=\lim\limits_{x\to\infty}G(x)$$ exists.

Also $$\int_{0}^{+\infty} \frac{1}{x^{3/2}}dx$$ does not converge since $$0$$ is the boundary point of the interval on which you do the intergration. But $$\int_{1}^{+\infty} \frac{1}{x^{3/2}}dx$$ converges, and $$\int_{0}^{1} \frac{\cos x}{\sqrt{x}(1+x)}dx$$ converges since $$|\frac{\cos x}{\sqrt{x}(1+x)}dx|\le \frac{1}{\sqrt{x}(1+x)}$$ and $$\frac{1}{\sqrt{x}(1+x)}$$~$$\frac{1}{\sqrt{x}}$$ as $$x\to 0$$. Combine these two estimation will give a proof of your problem.