# Is the function $\frac{x \cos x}{1+x^2}$ improperly integrable function?

I was wondering if the function $$\frac{x \cos x}{1+x^2}$$ is improperly integrable on $$[0,\infty)$$, i.e., $$\int_0^\infty \frac{x \cos x}{1+x^2}dx$$ exists in the sense of improper integration. More precisely,

$$\displaystyle \lim_{R \to \infty} \int_0^R \frac{x \cos x}{1+x^2}dx$$ converges in $$\mathbb{R}.$$

Clearly, the Cauchy principal value $$\int_{-\infty}^{\infty} \frac{x \cos x}{1+x^2}dx$$ is $$0,$$ since $$\frac{x \cos x}{1+x^2}$$ is an odd function.

By similar argument as in the answer of Per Manne [Convergence/absolute convergence of $\int_0^\infty \frac{\cos x}{1+x}dx$ (Baby Rudin P6.9), I can conclude $$\displaystyle \lim_{n \to \infty} \int_{\frac{\pi}{2}}^{(n+\frac{1}{2})\pi}\frac{x \cos x}{1+x^2}dx$$ converges. Here $$n \in \mathbb{N}$$ is crucial for the proof, since the alternating series test was used in the proof. But, I was wondering if it implies the convergence of $$\displaystyle \lim_{R \to \infty} \int_0^R \frac{x \cos x}{1+x^2}dx$$.

Please give me any comment for my question. Thanks in advance!

• It is simpler to reduce the integration to $\int_0^R\frac{(1-x^2)\sin x}{(1+x^2)^2}dx$, which converges absolutely. – user Feb 14 at 14:28

You are making a good observation.

It happens in this case that the improper integral is truly convergent and

$$\lim_{R \to \infty}\int_{\frac{\pi}{2}}^R \frac{x \cos x}{1 + x^2} \, dx = \lim_{n \to \infty}\int_{\frac{\pi}{2}}^{(n + \frac{1}{2})\pi} \frac{x \cos x}{1 + x^2} \, dx,$$

In general, convergence of $$\int_a^R f(x) \, dx$$ as the (real) upper limit $$R \to \infty$$ implies convergence of $$\int_a^n f(x) \, dx$$ as the (integer) upper limit $$n \to \infty$$.

However, the converse is not true unless $$f$$ is eventually nonnegative or nonpositive so that $$F(x) = \int_a^x f(t) \, dt$$ is monotonic.

For a counterexample, take $$f(x) = \cos \pi x$$ where

$$\lim_{n \to \infty} \int_0^n \cos \pi x \, dx = \lim_{n \to \infty} \frac{1}{\pi}\sin \pi n = 0,$$

but for $$R \in \mathbb{R}^+ \setminus \mathbb{N}$$,

$$\lim_{R \to \infty} \int_0^R \cos \pi x \, dx = \lim_{R \to \infty} \frac{1}{\pi}\sin \pi R\quad \text{does not exist}$$.

We can prove the improper integral in this question converges in a variety of ways -- for example, by showing that the Cauchy criterion is satisfied. For all $$x > 1$$ the function $$x \mapsto x/(1+x^2)$$ is decreasing. For any $$b > a > 1$$, there exists by the second mean value for integrals $$\xi \in (a,b)$$ such that

$$\left|\int_a^b \frac{x \cos x}{1 + x^2} \, dx \right| = \left|\frac{a}{1 + a^2}\int_a^\xi \cos x \, dx \right| = \frac{a}{1+a^2} |\sin \xi - \sin a| \leqslant \frac{2a}{1+a^2}$$

Since the RHS converges to $$0$$ as $$a \to \infty$$, the Cauchy convergence criterion is satisfied. That is, for any $$\epsilon > 0$$ we have for all sufficiently large $$a$$,

$$\left|\int_a^b \frac{x \cos x}{1 + x^2} \, dx \right| < \epsilon$$

Yes it does imply the desired convergence, and to the same value.

More general result: Suppose $$f$$ is continuous on $$[0,\infty).$$ Set $$I(R) = \int_0^R f.$$ Assume $$R_1 < R_2 < \cdots \to \infty.$$ If i) $$I(R_n)\to L$$ as $$n\to \infty,$$ and ii) $$\int_{R_n}^{R_{n+1}}|f| \to 0,$$ then $$\int_0^\infty f$$ converges to $$L.$$

This is pretty easy to prove; I'll leave it to you for now. In your problem we would apply this with $$f(x)=x\cos x/(1+x^2)$$ and $$R_n = (n+1/2)\pi.$$