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Does $$I=\int\limits_{-\infty}^{+\infty}\frac{2xdx}{x^2+1}$$ converge ? And should I use definition when calculating improper integrals ?

  • $I=\ln|x^2+1|_{-\infty}^{+\infty}=\infty-\infty$ ??

  • Let $x=-t$, we obtain $\displaystyle I=\int\limits_{+\infty}^{-\infty}\frac{2tdt}{t^2+1}=-I\implies I=0$

Could anybody explain the different of these solution, please ?

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  • $\begingroup$ Tried $\lim_{s\to\infty} \int_{-s}^s \frac{2x}{x^2+1}dx$? $\endgroup$ – Zelos Malum Oct 17 '15 at 10:41
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    $\begingroup$ @DDK: You forgot that if $x = -t$ then $\mathrm{d} x = - \mathrm{d} t$. $\endgroup$ – Éric Guirbal Oct 17 '15 at 10:44
  • $\begingroup$ @ÉricGuirbal you forgot that minus times minus is plus, and the limits are not reversed back. $\endgroup$ – uniquesolution Oct 17 '15 at 11:01
  • $\begingroup$ @uniquesolution: you're right :-) $\endgroup$ – Éric Guirbal Oct 17 '15 at 11:07
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No, the integral diverges because if it did converge it would have also converged on subsets of the line, say -- $[0,\infty)$, but

$$\int_0^{\infty}\frac{2x}{x^2+1}\,dx=\lim_{M\to\infty}\int_0^{M}\frac{2x}{x^2+1}\,dx=\lim_{M\to\infty}\log(M^2+1)=\infty$$

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  • $\begingroup$ Thank you. But how about my answer: $\displaystyle I=\int\limits_{+\infty}^{-\infty}\frac{2tdt}{t^2+1}=-I\implies I=0$. Can I conclude that if a improper intergral has result $0$, it will diverge ? $\endgroup$ – mja Oct 17 '15 at 10:58
  • $\begingroup$ it is true that for every real $M$, the integral $I(M)=\int_{-M}^M\frac{2t\,dt}{t^2+1}$ satisfies $I(M)=-I(M)$ and therefore equal to zero. However, by the usual definition of convergence of integrals, you cannot deduce that the improper integral (when $M\to\infty$) converges. You can however say that the principal value is zero. It's a matter of definition. $\endgroup$ – uniquesolution Oct 17 '15 at 11:05
  • $\begingroup$ @DDK: for improper integrals, the integrals $\int_a^b f(x)\,\mathrm{d}x$ and $\int_\alpha^\beta f(\phi(t))\phi'(t)\,\mathrm{d}t$ converge or diverge together. If they converge, they are equals. $\endgroup$ – Éric Guirbal Oct 17 '15 at 11:21
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The integral is undefined as written ($\infty - \infty$ is an undefined form). However, it is possible to define its Cauchy principal value (https://en.wikipedia.org/wiki/Cauchy_principal_value) as:

$$\lim_{a \to \infty} \int_{-a}^{a} \frac{2x}{x^2+1}dx$$

and the value of this is zero. Note that this is not an unqualified evaluation of the improper definite integral (which, as mentioned, is undefined), but merely a way to assign a value to it in a particular context.

Your manipulation by changing the variable (and limits) is invalid because that can only be done if the definite integral is defined (in general), which it isn't in this case. It would, however, be valid for proving that the Cauchy principal value is zero.

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