I noticed that $\displaystyle \int_{-a}^{b} \frac{x^3}{(x^2+4)(x^2+1)}$ will converge to $0$ whenever $a=b$ and will converge to some value whenever $a,b$ are in the reals (excluding infinity). How would you show that this integral does not converge over the interval $(-\infty, \infty)$ to someone who is not convinced?

When I evaluated this integral using complex residues for the interval $(-\infty, \infty)$ $(Res[z=2i] + Res[z=i])$,I got a complex value as an answer.

  • $\begingroup$ the highest power in the nominator the nominator is three, in the denominator it's four. therefore as $x->\infty$, the integral behaves as $\int (1/x)dx=\ log(x)$ which is clearly divergent. $\endgroup$ – tired May 17 '15 at 13:43
  • $\begingroup$ last comment: the integral is also zero on the infinite real line in a slightly more general sense: adding $1/x-1/x$ and split the integral in an intelligent way $\endgroup$ – tired May 17 '15 at 14:08

$\displaystyle \frac{x^3}{(x^2+4)(x^2+1)} \geq \frac{x^3}{4x^4}=\frac{1}{4x}$

for sufficiently large $x$.

OBS: And what do you mean by "will converge to $0$ whenever $a=b$ and will converge to some value whenever a,b are in the reals (excluding infinity)"? The integral on those values is a bona fide, well-defined, riemann-integral. There is no "convergence".

  • $\begingroup$ What about the case when $x<0$? $\endgroup$ – Samrat Mukhopadhyay May 17 '15 at 13:43
  • $\begingroup$ you can apply the same reasoning, switching x to -x $\endgroup$ – tired May 17 '15 at 13:45
  • $\begingroup$ btw, it should be 1/x instead of 1/4x because the highest order polynomial in the denominator has coefficent 1 $\endgroup$ – tired May 17 '15 at 13:47
  • $\begingroup$ @tired my reasoning is: $\frac{x^2+4}{x^2} \rightarrow 1$, so $x^2 +4\leq 2x^2$ for sufficiently large $x$. Analogously, $x^2+1 \leq 2x^2$ $\endgroup$ – Aloizio Macedo May 17 '15 at 14:08
  • $\begingroup$ If $a=b$ you integrate an odd function on $[-a,a]$ so you get $0$. $\endgroup$ – Beni Bogosel May 17 '15 at 15:28

Beside the simple point from Aloizio Macedo which would apply to many cases, may be, going further, you could use the fact that $$\frac{x^3}{(x^2+4)(x^2+1)}=\frac{4 x}{3 \left(x^2+4\right)}-\frac{x}{3 \left(x^2+1\right)}=\frac 23 \frac{2 x}{x^2+4}-\frac 16 \frac{2 x}{x^2+1}$$ So $$\int\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\frac{2}{3} \log \left(x^2+4\right)-\frac{1}{6} \log \left(x^2+1\right)$$ Already, the doefficients show what would be the problem $\frac 23 -\frac 16=+\frac 12$.

Since you already noticed that, if $a=b$, the result is $0$ then $$\int_{-a}^b\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\int_{a}^b\frac{x^3}{(x^2+4)(x^2+1)}\,dx=\frac{1}{6} \log \Big(\frac{(b^2+4)^4\,(a^2+1)}{(a^2+4)^4\,(b^2+1)}\Big)$$ which, forgetting a constant which depend on $a$, is basically $\log(b)$ for large values of $b$.


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