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$$\int_{0}^{\infty} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx$$

What's the method for determining if this integral converges or diverges? The integral seems to converge if I put it into Wolfram Alpha.

But do we assume it's similar to $$\int_{0}^{\infty} \frac{1}{x^4} dx$$

Because if so I can't get that one to converge since it's $-\frac{1}{3x^3}$ where $x=\infty$ and $x=0$ (I don't know how to write the "right bar" notation for integrals) which goes to $-0 + \infty$ which is divergent. So I am not sure what is right.

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    $\begingroup$ Quick observation: The numerator "behaves" like a linear term and the denominator is fourth degree. Therefore the difference is of degree 3 in favor of the denominator. If the denominator does not become zero on given interval, the integral is convergent. For comparison you may consider interval $(1,\infty)$ $\endgroup$ – imranfat Feb 2 '18 at 16:54
  • $\begingroup$ with a numerical method i get $$-3.13184$$ $\endgroup$ – Dr. Sonnhard Graubner Feb 2 '18 at 16:55
  • $\begingroup$ @Dr.SonnhardGraubner: which numerical method? Otherwise, it is a pretty irrelevant comment. $\endgroup$ – Jack D'Aurizio Feb 2 '18 at 17:31
  • $\begingroup$ i used the Trapezoidal rule $\endgroup$ – Dr. Sonnhard Graubner Feb 2 '18 at 17:35
  • $\begingroup$ @Dr.SonnhardGraubner Isn't the integrand positive? $\endgroup$ – Teepeemm Feb 2 '18 at 21:20
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Note that

  • $1-x^2+x^4>0$
  • $\frac{\sqrt{1-x+x^2}}{1-x^2+x^4}\sim \frac1{x^3}$

thus

$$\int_{0}^{\infty} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx=\int_{0}^{1} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx+\int_{1}^{\infty} \frac{\sqrt{1-x+x^2}}{1-x^2+x^4} dx$$

Since the first integral is a proper integral and the second converges by comparison with

$$\int_{1}^{\infty}\frac1{x^3} dx$$

the given integral converges.

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The functions in the numerator and denominator are positive for all $x$, so the simplest is to use equivalents, as the integrand is positive: we have $\; 1-x+x^2\sim_\infty x^2$, $\;1-x^2+x^4\sim_\infty x^4$, so $$\frac{\sqrt{1-x+x^2}}{1-x^2+x^4}\sim_\infty \frac{\sqrt{x^2}}{x^4}=\frac 1{x^3}.$$ As $\displaystyle\int_1^\infty\!\frac{\mathrm d x}{x^3}$ converges, so does $\;\displaystyle\int_{0}^{\infty}\!\frac{\sqrt{1-x+x^2}}{1-x^2+x^4}\,\mathrm d x $.

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You can say that it's similar to $1/x^3$ for large $x$, but at values closer to $0$ your function to integrate goes to $1$, not to $\infty$. So in principle write your integral as a sum of integrals from $0$ to $1$, and from $1$ to $\infty$.

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To prove convergence of $\int_0^{\infty}f(x)dx$ for a continuous $f(x)$ it often helps to find a crude but effective upper bound for $|f(x)|$ for large $x$.

When $x>1$ we have $\sqrt {1-x+x^2}\; <\sqrt {1+2x+x^2}=x+1<2x .$

And when $x>2$ we have $x^4-x^2+1>x^4-x^2>x^4/2.$

So when $x>2$ we have $$0<\frac {\sqrt {1-x+x^2}}{1-x^2+x^4}<\frac {2x}{x^4/2}=\frac {4}{x^3}$$ so the integral converges.

We can also observe that when $x>0$ we have $\sqrt {1-x+x^2}=x(1+g(x))$ and $1-x^2+x^4=x^4(1+h(x))$ where $g(x)$ and $h(x)$ converge to $0$ as $x\to \infty.$

So for all sufficiently large $x$ we have $|1+g(x)|<3/2$ and $|1+h(x)|>1/2.$ Thus $$0<\left|\frac {x(1+g(x)}{x^4(1+h(x)}\right|<\frac {x(3/2)}{x^4(1/2)}=\frac {3}{x^3}$$ for all sufficiently large $x.$

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For large $x>M$ your integral behaves like $\int_{M}^\infty\frac{1}{x^3}\mathrm dx$ which converges. To dress this up with more rigor, note that $$ x\geq2\implies \sqrt{1-x+x^2}\leq x $$ and so $$ \frac{\sqrt{1-x+x^2}}{1-x^2+x^4}\leq \frac{x}{x^4-x^2}\leq\frac{1}{x^3-x}\\ \leq\frac{1}{x^2-1} $$ Now integrate this final guy by partial fractions to find $$ \int_2^\infty\frac{1}{x^2-1}\mathrm dx=\frac{\log(3)}{2} $$ Now examine any potential poles: I.e. when $1-x^2+x^4=0$ or when $$ x=\sqrt{\frac12(1\pm 3i}) $$ i.e. never on the real line. So your integral converges by comparison to the $p=2$ integral above.

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