Consider an (improper) integral like $$ \int_{-\infty}^{+\infty} \frac{x^4 \sin x}{x^2+1}dx $$ If we treat this as the sum of two separate one-sided improper integrals $$ \int_{-\infty}^{a} \frac{x^4 \sin x}{x^2+1}dx + \int_{a}^{+\infty} \frac{x^4 \sin x}{x^2+1}dx = \lim_{t\to-\infty}\int_{t}^{a} \frac{x^4 \sin x}{x^2+1}dx + \lim_{u\to+\infty}\int_{a}^{u} \frac{x^4 \sin x}{x^2+1}dx $$ then we should say the original integral does not converge, since neither of these two limits exists.

But if we treat this as the limit of a two-sided integral $$ = \lim_{s\to\infty}\int_{-s}^{s} \frac{x^4 \sin x}{x^2+1}dx $$ then this limit exists and in fact is zero.

Do we say that such a 2-sided integral converges or diverges? And either way, is there terminology for an improper integral such that for some joint approaches to $\pm \infty$ of its endpoints the integral converges, yet for others it does not?

For example, if $$f(x) = \left\{ \begin{array}{ll} \frac{x^4 \sin x}{x^2+1} & x\geq 0 \\\frac{16x^4 \sin 2x}{4x^2+1} & x<0 \end{array}\right. $$ we would have, for $d(s) = -\frac{s}{2}$ and $u(s) = s$, $$ \lim_{s->\infty}\left[ \int_{d(s)}^0f(x)\,dx + \int_0^{u(s)}f(x)\,dx \right] = 0 $$ while for $d(s) = -s$ and $u(s) = s$, $$ \lim_{s->\infty}\left[ \int_{d(s)}^0f(x)\,dx + \int_0^{u(s)}f(x)\,dx \right] $$ diverges. So for this example, if we insist on taking the two endpoints to $\pm\infty$ together, the integral converges or diverges depending on how we take those endpoints to infinity. Is there a name for this sort of integral?

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    $\begingroup$ Cauchy-Principle Value $\endgroup$ – Mark Viola Mar 13 '17 at 17:11
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    $\begingroup$ this integral doesn't converge $\endgroup$ – Dr. Sonnhard Graubner Mar 13 '17 at 17:19
  • $\begingroup$ As Graubner says it does not converge but the function is odd so any sub-integral from -R to R will be 0. If it would converge it would therefore have to be 0. $\endgroup$ – mathreadler Mar 13 '17 at 18:39

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