Evaluate $\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$

Question

Evaluate $$\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$$ by any method. In short I am interested in any method that overcomes the lack of convergence of this integral and gives an "number" to it.

EDIT

As I'm getting answers regarding convergence test, this should clear the question. This is integral is divergent. Do other integration theories (Lebesgue, Henstock-Kurweil,..) overcome this problem and actually help assign a non-infinite value to this. (P.S. I'm engineering guy, need incentive to go beyond Riemann and this is a start)

Answer 1

Using the trivial measure, let $\int_{\mathcal{D}} f d\mu = 0$ for any function $f$ and any subset $\mathcal{D}$ of $\mathbb{R}$. Under this definition, your integral is convergent and its value is equal to $0$. But what are the applications of this stupid integral? None. To have a sensible integral, you want, for example, "the area under the line $y=3$ between $0\leq x\leq 2$" to be equal to $2\times 3 = 6$. When we construct an integral that has such natural and useful properties (which is another story), it turns out that your integral does not converge, as shown in the other answers.

Answer 2

By definition

$$\int\limits_1^\infty\frac x{1+x^2}dx=\frac12\int\limits_1^\infty\frac{2x}{1+x^2}dx=\lim_{x\to\infty}\frac12\log(1+x^2)=\infty$$

Thus, the integral doesn't converge.

Answer 3

$$\int{\frac{x}{1+x^2}}dx=\frac{1}{2}\ln(1+x^2)+C$$

$$\int_0^\infty{\frac{x}{1+x^2}}dx=\lim_{b\to\infty}\int_0^b\frac{x}{1+x^2}dx=\lim_{b\to\infty}\frac{1}{2}\ln(1+b^2)=\infty$$

Answer 4

$$\begin{array} \\ \text{1. Given expression} & \int_1^\infty \dfrac{x}{1 + x^2} dx\\ \ & \\ \text{2. Let } u = 1 + x^2 \text{ and work out the} &= \dfrac{1}{2}\int_{x = 1}^\infty \dfrac{du}{u}\\ \text{and substitution.}& \\ & \\ \text{3. Work out the integration.} &= \dfrac{1}{2} \ln(u) \quad \vert_{x = 1}^\infty \\ & \\ \text{4. You can either work out this}&= \dfrac{1}{2} \ln(1 + x^2) \vert_{x = 1}^\infty \\ \text{out by substitution and applying} &= \lim_{t \rightarrow \infty} \frac{1}{2}\ln(1 + x^2) \vert_{x = 1}^t\\ \text{Fundamental Theorem of Calculus...} &= \lim_{t \rightarrow \infty} \frac{1}{2}(\ln(1 + t^2) - \ln(2)) \\ &= \mathrm{dne} \end{array}$$

...or work out the substitution shorthand, rewriting the given limits as the new ones (That is: Substitute $x = 1$ and $x = \infty$ for $u = 1 + x^2$) . . .

$$\begin{array}\frac{1}{2} \int_{u = 2}^{\infty} \frac{du}{u} \\ = \frac{1}{2} \ln(u) \vert_2^\infty \\ = \frac{1}{2} (\ln(\infty) - \ln(2)) \\ = \mathrm{dne}\\ \end{array}$$

Done with the computation. Lord Soth's response is also good to follow if you want to work out the problem with some sort of real analysis.

Answer 5

Here is a heuristic reasoning why one should not expect to assign a number to this integral, even if we speak about a different type of "convergence/divergence".

Lets assume we can get, in some sense, a number $I$ so that

$$\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx =I \,.$$

After the substitution $x=\frac{u}{a}$ with $a>1$ we get

$$I= \int \limits_{0}^{\infty} \frac{\frac{u}{a}}{1+\frac{u^2}{a^2}} \frac{1}{a} du = \int \limits_{0}^{\infty} \frac{u}{a^2+u^2} du< \int \limits_{0}^{\infty} \frac{u}{1+u^2} du =I$$

Thus $I <I$.

Of course, the "convergence" could be weak enough so that standard properties of integrals are not true anymore, but then it is unprobable that that type of convergence would be helpful.

Note that for all $R >0$

$$ \int \limits_{0}^{R} \frac{x}{1+x^2} dx =\int \limits_{0}^{aR} \frac{x}{a^2+x^2} dx \,.$$

So any type of convergence, must either make those two limits different, or fail the following property:

• $f <g , f,g$ continuous implies $\int_a^ \infty f < \int_a^ \infty g$.

[and in our example $f(x) < g(x)$ for all $x$, which means that $\int_a^\infty f =\int_a^b f +\int_b^\infty f$ should probably also fail..]