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Does the following integral converge? I will post my solution, but I am unsure if it is true.

$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$


My solution

Let $$ g(x) = \frac{2x}{\sqrt {x^3}}$$ $$ f(x) = \frac{2x +3}{\sqrt {x^3 + 2x + 5}} $$ Then $$\lim_{k \to \infty} \frac{f(x)}{g(x)} = 1$$

Therefore whatever one does, so does the other.

$$ \int_{0}^{\infty} g(x) = \int_{0}^{\infty} \frac{2}{\sqrt {x}} = +\infty $$ Therefore g(x) diverges, thus $$\int_{0}^{\infty} f(x) = \int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx = + \infty$$ diverges too

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    $\begingroup$ This works, but be careful about using something like '$f\sim g$ therefore whatever one does, so does the other' - it works for the way you're using it, but there are many instances where this can get you into trouble, especially when the functions can oscillate between positive and negative. $\endgroup$ – Thomas Bloom Mar 4 '20 at 12:24
  • $\begingroup$ Thanks man, that was a concern I had! $\endgroup$ – Dimitris Mar 4 '20 at 12:28
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    $\begingroup$ A more rigorous way to do it might be something like 'there exists $X$ such that if $x>X$ then $f(x) \geq x^{-1/2}$. Therefore $\int_X^\infty f(x) \geq \int_X^\infty x^{-1/2}=\infty$.' $\endgroup$ – Thomas Bloom Mar 4 '20 at 12:33
  • $\begingroup$ @ThomasBloom The limit form of the comparison test is completely rigorous provided its hypotheses are satisfied. But I agree that someone could use it in a careless way and end up with something wrong. In fact, OP's use is a bit questionable since the second integral introduces a singularity at $x=0$. $\endgroup$ – bjorn93 Mar 5 '20 at 0:59
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Notice that: $$\frac{2x+3}{\sqrt{x^3+2x+5}}\approx x^{-0.5}$$ for large $x$ and we know that: $$\int_0^\infty\frac{1}{x^n+c}dx$$ only converges for $c>0,n>1$

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Your argument is correct. It's an example of a comparison test (described here, albeit with explicit squeezing, which $\sim$ implies can be done).

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$$\int_{3}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx > \int_{3}^{\infty} \frac{x}{\sqrt {2x^3 }} \,dx= \bigg| \sqrt {2x} \bigg|_{3}^{\infty}=\infty$$ You're correct. The integral diverges.

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Perhaps redundant:

Let $x \ge 2$.

$\dfrac{x}{\sqrt{3x^3}}\lt \dfrac{2x+3}{\sqrt{x^3+2x+5}}$, or

$0<g(x):=(1/√3)x^{-1/2} <$

$\dfrac{2x+3}{\sqrt{x^3+2x+5}}=:f(x);$

$\int_{2}^{\infty} g(x)dx$ diverges, so does $\int_{2}^{\infty}f(x)dx$ (Monotonicity of integral).

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