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How to correctly calculate the integral: $$\int_0^\infty \frac{1}{x^2 +x + \sqrt x}dx$$

Edit: I tried to figure out if the limit exists:

Step 1: break the integral to two parts: from 0 to 1, from 1 to infinity.

Step 2: use limit comparison test for both of the integral: the first integral compared at 1 to 1/sqrt(x) and the second is compared at infinity to 1/x^2.

Step 3: conclude that both converge, hence the original integral also converges.

Step 4: (this is the one im trying to figure out, how to actually calculate it, because the limit exists).

Blurred_Vision
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    What have you tried? I could recommend the substitution $y=\sqrt{x}$, but that only helps if you can solve cubic equations. – J.G. Sep 05 '22 at 11:09
  • Here's how to ask a good question. Follow these guidelines to get help in this forum. – jjagmath Sep 05 '22 at 11:10
  • @J.G. I tried to break the integral to 0-1 and 1-infinity and use comparison tests, so i found that the 0-1 converges. – Blurred_Vision Sep 05 '22 at 11:12
  • Edit the question to include your efforts. You'll get better feedback if you show that you really tried to solve it on your own before asking for help, and you're not just trying to get others solve it for you. – jjagmath Sep 05 '22 at 11:15
  • @jjagmath edited, thats the farthest i got to. – Blurred_Vision Sep 05 '22 at 11:21
  • If you do what @J.G. suggested, there is no more problem at $y=0$ – Claude Leibovici Sep 05 '22 at 11:28
  • Perhaps tell us where it came from. Are you to calculate the integral, or does it merely ask whether it converges? Is it a course where they expect you to use residues in the complex plane? – GEdgar Sep 05 '22 at 11:28
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    Hint expanding on my last comment: write the only real root of $y^3+y+1$ as $-a$ (if you want a formula for it use Cardano's method), so the integral is$$\int_0^\infty\frac{2dy}{(y+a)(y^2-ay+1/a)}.$$Now use partial fractions. – J.G. Sep 05 '22 at 11:30
  • Solutions in the complex plain allowed. The complexity is too high to calculate it via wolfram alpha, so any prediction also allowed. – Blurred_Vision Sep 05 '22 at 11:31
  • @RobertLee not really. Taking denominator as a whole there should not be dependence on constants. Otherwise, i wouldnt post the question, and use the complex-valued logarithm to calculate the limit. But its very close to my question using Cardano`s method as J.G. mentioned. Thanks for sharing. – Blurred_Vision Sep 05 '22 at 13:03

2 Answers2

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We have \begin{align*} I &= \int_0^\infty \frac{1}{x^2 + x + \sqrt{x}} \ \mathrm{d}x\\ &= \int_0^\infty \frac{2}{y^3 + y + 1} \ \mathrm{d}y && \text{using $y = \sqrt{x}$}\\ \end{align*} Clearly, this only has one real root, denoted $r$, where $r < 0$. Since $r^3 + r + 1 = 0$, we have $r^{-1} = -r^2 - 1$, so $y^3 + y + 1 = (y - r)\left (y^2 + ry + (r^2 + 1) \right)$. Then we have \begin{align*} \frac{2}{y^3 + y + 1} &= \frac{A}{y - r} + \frac{By + C}{y^2 + ry + (r^2 + 1)}\\ 2 &= A\left (y^2 + ry + (r^2 + 1) \right ) + \left (By + C \right )(y - r)\\ y=r \implies A &= -\frac{2r}{2r + 3}\\ B &= \frac{2r}{2r + 3}\\ C &= \frac{4r^2}{2r+3}\\ \implies \frac{2}{y^3 + y + 1} &= \frac{2r}{2r + 3}\left (\frac{y + 2r}{y^2 + ry + (r^2 + 1)} - \frac{1}{y - r} \right ) \end{align*} Thus we have \begin{align*} I &= \frac{2r}{2r + 3} \int_0^\infty \frac{y + 2r}{y^2 + ry + (r^2 + 1)} - \frac{1}{y - r} \ \mathrm{d}y\\ &= \frac{2r}{2r + 3} \int_0^\infty \frac{y + \frac{r}{2}}{y^2 + ry + (r^2 + 1)} + \frac{\frac{3r}{2}}{\left ( y + \frac{r}{2}\right )^2 + \left (\frac{3r^2}{4} + 1\right )} - \frac{1}{y - r} \ \mathrm{d}y\\ &= \frac{6r}{2r + 3} \left ( \frac{1}{2} \ln(-r) - \frac{(-r)^{\frac{3}{2}}}{\sqrt{3 - r}} \left (\frac{\pi}{2} + \tan^{-1}\left ( \frac{(-r)^{\frac{3}{2}}}{\sqrt{3 - r}}\right )\right )\right ) \end{align*} Taking $r = -\frac{2}{\sqrt{3}}\sinh\left ( \frac{1}{3} \sinh^{-1} \left (\frac{3\sqrt{3}}{2}\right ) \right )$ gives a numerical evaluation for the above term of about $1.8435267\dots$.

Sharky Kesa
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$$\int \frac{dx}{x^2 +x + \sqrt x}=2\int \frac{dy}{y^3 +y + 1}$$

Write $$\frac{1}{y^3 +y + 1}=\frac 1{(y-a)(y-b)(y-c)}$$ Use partial fraction decomposition $$\frac{1}{y^3 +y + 1}=\frac{1}{(a-b) (a-c) (y-a)}+\frac{1}{(b-a) (b-c) (y-b)}+\frac{1}{(c-a) (c-b) (y-c)}$$ Then three logarithms to be recombined before using the bounds.

Look how nice is the real root of the cubic $$a=-\frac{2}{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 \sqrt{3}}{2}\right)\right)$$ Then from Vieta $$b=-\frac{a}{2}-i\frac{\sqrt{a^3+4}}{2 \sqrt{|a|}}\quad \text{and} \quad c=-\frac{a}{2}+i\frac{ \sqrt{a^3+4}}{2 \sqrt{|a|}}$$

Claude Leibovici
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