# Do improper integrals like $\int_{-\infty}^{+\infty} f$ converge if $xf(x)\rightarrow 0$?

My teacher assumes without proof in his notes that, given a rational function $R(x)$, the improper integral $\int_{-\infty}^{+\infty} R(x)dx$ converges if $\lim_{|x|\rightarrow\infty} xR(x) = 0$. He then proceeds to explain the Estimation lemma, that is a very similar result.

However it is unclear for me whether this fact is true, and if it is bidirectional (i.e. if $\int_{-\infty}^{+\infty} R(x)dx$ converges then $\lim_{|x|\rightarrow\infty} xR(x) = 0$) and holds true even for non-rational functions

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Presumably your rational function has no poles. Then the result is true, and bidirectional. One cannot expect anything similar for general functions. The standard example is $\frac{1}{x\log x}$. This needs to be fixed up if you want the interval to be $(-\infty,\infty)$, easy to do. Just let $f(x)=1/(|x|\log(|x|))$ for $|x|\ge 2$, and anything nice between $-2$ and $2$. – André Nicolas Jun 12 '12 at 15:23
In general the function you're integrating can have no limit at infinity. The same holds for $x f$. Take for example comb function d.pr/i/ElNd+ or if you want to deal with formulas, take sth like $f(x) = x^2 \exp (-x^8 \sin^2 (40x))$. – qoqosz Jun 12 '12 at 15:58

One cannot extend the result to general functions. For example, let $f(x)=\frac{1}{|x|\log(|x|)}$ if $|x|\ge 2$, and let $f(x)=f(2)$ for $-2 \lt x\lt 2$.

Then $\lim_{x\to\infty} xf(x)=\lim_{x\to -\infty} xf(x)=0$, but the improper integral does not converge.

The result does hold, bidirectionally, for rational functions whose denominator vanishes nowhere. It is essentially obvious, though writing out the proof is typographically unpleasant. For a rational function $R(x)=\frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials with $Q(x)$ nowhere $0$ (and $P$ not identically $0$), there is convergence iff $\deg(Q)\ge 2+\deg(P)$. This is precisely the condition for $\lim_{|x|\to \infty} xR(x)$ to be $0$. The reason is basically that degrees can only take on integer values.

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Consider $\displaystyle \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} dx$ where $P,Q$ are polynomials and $x$ is a real variable. This converges absolutely if $\deg(P) \leq \deg(Q) -2.$ To evaluate this family of integrals, we consider $\displaystyle \int_C \frac{P(z)}{Q(z)} dz$ where $C$ is the standard semi-circular contour in the upper half plane with radius $R.$ The Estimation Lemma yields that the contribution from the arc is $\mathcal{O}(1/R)$ as $R\to\infty.$

Hence $$\int^{\infty}_{-\infty} \frac{P(x)}{Q(x)} dx = 2\pi i \sum \text{ Residues of P/Q in the upper half plane }.$$

When $\deg(P)=\deg(Q)-1$ this same calculation holds true, but now the calculated value does not represent $\displaystyle \int^{\infty}_{-\infty} \frac{P(x)}{Q(x)} dx$ ; that integral does not exist because we get different values if the upper and lower limits tend to infinity in different ways. What does exist (and we calculated) is the Cauchy Principle Value: $$\text{ C.P.V } \int^{\infty}_{-\infty} \frac{P(x)}{Q(x)} dx = \lim_{R\to \infty} \int^R_R \frac{P(x)}{Q(x)} dx.$$

If $Q$ has a simple zero at $x_0 \in \mathbb{R}$ our calculation still returns a value we can make sense of, another Cauchy principle value: $$\text{ C.P.V} \int^b_a \frac{P(x)}{Q(x)} dx = \lim_{\epsilon\to 0} \int^{x_0+\epsilon}_a\frac{P(x)}{Q(x)} dx + \int^b_{x_0-\epsilon}\frac{P(x)}{Q(x)} dx.$$

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Thanks for your contribution, I chose the other answer because it answers well just what I have asked, so it is good for a reference, but your answer helps me to understand better what I'm studying – Cauchy Jun 12 '12 at 16:17
@Cauchy Good to hear, I had hoped that it would. – Ragib Zaman Jun 12 '12 at 16:19