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

Question

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?

Answer 1

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 $t \to + \infty$.

Answer 2

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.