Evaluate $\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$ 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)
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 $a&gt;1$ we get

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

Thus $I &lt;I$.

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 $R &gt;0$

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

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


$f &lt;g , f,g$ continuous implies $\int_a^ \infty f &lt; \int_a^ \infty g$.


[and in our example $f(x) &lt; g(x)$ for all $x$, which means that $\int_a^\infty f =\int_a^b f +\int_b^\infty f$ should probably also fail..]
This is not difficult:

$$\operatorname{st}\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx=\frac{\pi ^2}{2}-4+\psi\left(\frac{3}{2}\right)=0.971292...$$