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

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

• 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. – Thomas Bloom Mar 4 '20 at 12:24
• Thanks man, that was a concern I had! – Dimitris Mar 4 '20 at 12:28
• 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 $\int_X^\infty f(x) \geq \int_X^\infty x^{-1/2}=\infty$.' – Thomas Bloom Mar 4 '20 at 12:33
• @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 $x=0$. – bjorn93 Mar 5 '20 at 0:59

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 $$c>0,n>1$$

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

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

Perhaps redundant:

Let $$x \ge 2$$.

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

$$0

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

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