# 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.

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As I'm getting answers regarding convergence test, this should clear the question. This is integral is divergent. Do other integration theories (Lebegue, 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)

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No method can show this integral is convergent, or yield a number, since it is not convergent... – N. S. Jul 9 '13 at 19:24
If you could overcome lack of convergence, no one would bother with convergence tests. – Javier Jul 9 '13 at 19:25
@N.S. no method can show this integral is convergent,...,since it is not convergent: how do you know it is not convergent if there is no method to show that? I'm a tad confused here. :-( – nt.bas Jul 9 '13 at 19:29
My point is that it is easy to prove that it is divergent, or if you preffer there are methods to prove it is NOT convergent... Simplest is the substitution $u=1+x^2$. Alternately comparison test. – N. S. Jul 9 '13 at 19:33
I think what OP is looking for is some sort of method for divergent integrals. That is: is there something that does for some divergent integrals what Abel and Cezaro summation do for divergent sums? – Omnomnomnom Jul 9 '13 at 19:39

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.

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This is as a close to an answer as I can get, but before I accept it, one thing to clarify: this measure theory stuff is related to Lebesgue integration and the given integral can be shown, using Lebegue theory, to be convergent and its value is zero. Is that what are you are saying? – nt.bas Jul 9 '13 at 19:59
Well, yes and no, depending on hopefully I do not confuse you. What I am saying is that you may assign every integral you can possible think of the value $0$, and that integral would still be valid and you may say "my integral is convergent." But such a definition would be too restrictive, and your definition of the integral will not be useful at all. – Lord Soth Jul 9 '13 at 20:05
Yes, you may formally construct such an integral using "measure theory," but I wouldn't say "Lebesgue measure theory," as usually in this case what is understood is the Lebesgue integral, and the Lebesgue integral of your integral does not exist. – Lord Soth Jul 9 '13 at 20:06
Thanks! Everything clear now. – nt.bas Jul 9 '13 at 20:11
@nt.bas You are welcome. – Lord Soth Jul 9 '13 at 20:14

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.

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I think the OP already knows that the integral does not converge. – Lord Soth Jul 9 '13 at 19:33
Yes, but DonAntonio provides this for clarification. – NasuSama Jul 9 '13 at 19:33
After reading the OP's comment below his question I realized he might not be sure the integral doesn't converge or how to show this... – DonAntonio Jul 9 '13 at 19:34
@nt.bas Ah... This seems like the different type of question you are asking! – NasuSama Jul 9 '13 at 19:44
@nt.bas , I can't understand how a comment written after you posted your question can confuse you about your question... – DonAntonio Jul 9 '13 at 19:45

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..]

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If you assigned the integral a value of zero, you wouldn't have the specific problem you mention, but you would have a non-zero non-negative function whose integral is zero. – Omnomnomnom Jul 9 '13 at 20:31
:..a non-zero non-negative continuous functions..." – DonAntonio Jul 10 '13 at 9:02

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

...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.

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I would never write "$=\text{dne}$". One may say that something does not exist; one should not say that something is equal to "does not exist". – Michael Hardy Jul 9 '13 at 20:33
You wrote $u=1+x$. That must have been a typo, so I changed it to $u=1+x^2$. – Michael Hardy Jul 9 '13 at 20:35

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

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

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