I am having some trouble with this problem and don't know if I am doing it right:

$$\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$$

so the steps I have taken so far are, I split it into

$$\int_0^\infty \dfrac{x^6+6}{x^8+8} + \int_{-\infty}^0 \dfrac{x^6+6}{x^8+8}$$

for the second integral I made a change of variable $x = -u$

so it would look the same as the first integral and then use $$ \lim_{x\to \infty} \dfrac{f(x)}{g(x)}$$ where $g(x) = \dfrac1{x^2}$.

Just wondering if my approach is correct.

  • 1
    $\begingroup$ It is basically fine. Note that $\frac{1}{x^2}$ misbehaves at $0$, so if you are going to use a limit comparison, you need to split the integral into $0$ to $1$ (fine) and $1$ to $\infty$ (limit comparison). Or else you can do a limit comparison with $\frac{1}{1+x^2}$, exploiting fact $\int_0^\infty \frac{dx}{1+x^2}=\frac{\pi}{2}$. $\endgroup$ May 16 '13 at 5:16
  • 1
    $\begingroup$ thank you to the people who help me edit my post. This is my first time using mathexchange $\endgroup$
    – Henry
    May 16 '13 at 5:16
  • $\begingroup$ @Siddhant Trivedi : Why change from $g(x) = \frac{1}{x^2}$ to $g(x) = 1/x^2$ ? $\endgroup$
    – jimjim
    May 16 '13 at 5:25
  • $\begingroup$ oh...i forgot it...sorry... $\endgroup$ May 17 '13 at 4:34

Your approach is fine, as long as you compare with $1/x^2$ when $|x|\gt1$ and then use the continuity of $\frac{x^6+6}{x^8+8}$ for $|x|\le1$. We can illustrate this with a slightly different comparison function.

First, note that $$ \lim_{|x|\to\infty}\frac{(x^6+6)(x^2+1)}{x^8+8}=1\tag{1} $$ Thus, $(1)$ says that there is an $m$ so that if $|x|\ge m$, then $\frac{(x^6+6)(x^2+1)}{x^8+8}\le2$. Since $\frac{(x^6+6)(x^2+1)}{x^8+8}$ is continuous on the compact set $[-m,m]$, it is bounded there. Therefore, there is an $M$ so that $$ \frac{(x^6+6)(x^2+1)}{x^8+8}\le M\tag{2} $$ Therefore, $$ \frac{x^6+6}{x^8+8}\le\frac{M}{1+x^2}\tag{3} $$ Now, simply use $$ \int_{-\infty}^\infty\frac{\mathrm{d}x}{1+x^2}=\pi\tag{4} $$ to show that $$ \int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x\tag{5} $$ converges by comparison.

Although this answer uses contour integration, it does give the value for $$ \int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x=\frac{\pi}{8}\csc\left(\frac{\pi}{8}\right)\left(2^{5/8}+3\cdot2^{-5/8}\right)\tag{6} $$


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