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Tell if the following improper integral converges or not. $$\int_0^{+\infty} \dfrac{1}{x^2+7}\ \text{d}x$$

I know that a necessary but not sufficient condition is that $\lim_{x\to +\infty} f(x) = 0$. Here this holds, so the integral might converge.

Now, I thought of the comparison:

$$\int_0^{+\infty} \dfrac{1}{x^2+7}\ \text{d}x < \int_0^{+\infty} \dfrac{1}{x^2}\ \text{d}x$$

Since the second integral diverges, and the first one is $<$, it converges.

Is this method valid?

Átila Correia
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Heidegger
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    No. To prove convergence this way you need to find a larger integral that converges, Hinr: why/where does the second integral diverge? Can you fix that? – Ethan Bolker Dec 15 '22 at 21:23
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    @EthanBolker Can I say that $$\dfrac{1}{x^2+7} < \dfrac{1}{x^2+1}$$ and then conclude? The integral of $\dfrac{1}{x^2+1}$ is the arctangent so it's convergent – Heidegger Dec 15 '22 at 21:25
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    Yes, your solution is the simplest. I suggest you edit it as an answer to your own question. – Anne Bauval Dec 15 '22 at 21:34
  • @AnneBauval Thank you!! – Heidegger Dec 15 '22 at 21:36
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    Incidentally, it is not a necessary condition for convergence of an improper integral that $\lim_{x\to \infty} f(x) = 0$, in general. (Though it is necessary for a nonnegative and decreasing function $f$.) – Daniel Schepler Dec 15 '22 at 21:57
  • Indeed, look. – Anne Bauval Dec 15 '22 at 22:06
  • @DanielSchepler Oh, I did not know. On my notes it was underlined as necessary... Thank you for having specified when it really is. Also, say I have $$\int_1^{+\infty} \dfrac{e^x}{(x+4)^2}\ \text{d}x$$

    In this case, does noticing that $\lim_{x\to +\infty} f(x) = +\infty$ straightly tell me that the integral diverges instead?

     – Heidegger Dec 16 '22 at 13:18
  • Yes, it's true that if $\lim_{x\to \infty} f(x) = \infty$, then $\int_a^\infty f(x),dx$ converges. And also, if $\lim_{x\to \infty} f(x) = L \ne 0$, then $\int_a^\infty f(x),dx$ diverges. It's when $\lim_{x\to \infty} f(x)$ does not exist that it's still possible for $\int_a^\infty f(x),dx$ to converge. – Daniel Schepler Dec 16 '22 at 17:05
  • @DanielSchepler I'm confused: when as $x\to +\infty$, $f(x) \to \infty$, the integral converges or diverges? – Heidegger Dec 16 '22 at 21:20
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    Oh, sorry, I meant diverges. – Daniel Schepler Dec 16 '22 at 21:51

3 Answers3

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According to your efforts, I would recommend you to split the integral into two parts as follows: \begin{align*} \int_{0}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7} & = \int_{0}^{1}\frac{\mathrm{d}x}{x^{2} + 7} + \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7} \end{align*}

The first part is integrable because its argument is a continuous function on the compact interval $[0,1]$, hence uniformly continuous and integrable. The second part is also integrable. That is because we can apply the monotonicity of the integral to conclude that: \begin{align*} \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{2} + 7} < \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{2}} = -\frac{1}{x}\bigg\rvert_{1}^{\infty} = 1 < + \infty \end{align*}

Hopefully this helps!

Átila Correia
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In a simpler way, we can say

$$\dfrac{1}{x^2+7} < \dfrac{1}{x^2+1}$$

Thence

$$\int_0^{+\infty} \dfrac{1}{x^2+7} \text{d}x < \int_0^{+\infty} \dfrac{1}{x^2+1}\ \text{d}x$$

The second integral gives the arctangent function which evaluates to $\pi/2$. Hence

$$\int_0^{+\infty} \dfrac{1}{x^2+7} \text{d}x < \dfrac{\pi}{2}$$

Whence it converges.

To prove that the integral of $\dfrac{1}{x^2+1}$ converges indeed, see @Atila Correia's answer!

Heidegger
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    Your last sentence is kind but inappropriate. You already proved "that the integral of $\frac1{x^2+1}$ converges indeed"! – Anne Bauval Dec 15 '22 at 22:09
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For $a>0$ we have $$\int\limits_0^\infty {1\over x^2+a^2}\,dx= {1\over a}\arctan\left ({x\over a}\right )\bigg\rvert_{0}^{\infty}={\pi\over 2a}$$ In OP case we have $a=\sqrt{7}.$

Ryszard Szwarc
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