Evaluate the integral $$\int_{-\infty}^{+\infty}\dfrac{x}{1+x^2}dx .$$

enter image description here

Intuitive approach

As you see, it's an odd function,and we can say that value of the integral is $0$, because its symetric point is $0$ in opposite sign each other, and I think $0$ is mid point of $(-\infty,\infty)$.

Solution (wrong)

It's a improper integral and let's change its form:

$$\displaystyle\int_{-\infty}^{+\infty}\dfrac{x}{1+x^2}dx=\displaystyle\int_{-a}^{+a}\dfrac{x}{1+x^2}dx=\lim_\limits{a\rightarrow \infty}\left[\dfrac{1}{2}\ln|x^2+1|\right]^{^{a}}_{_{-a}}=\lim\limits_{a\rightarrow \infty}\dfrac{1}{2}[0]=0$$

I've checked in Wolfram, but it says that this integral is not defined. Why can't we just calculate simply? There are no improper points, and it is a very simple function. What is the big deal? What I miss?

  • $\begingroup$ "positive" and 'negative" areas can certainly cancel each other out, so that the answer becomes zero, provided that the individual areas are finite. With the answer of Travis, your integral does not qualify for that, hence it is divergent. Now if you replace that $x^2$ in the denom with $x^4$, you anti derivative would look like an arctan. Graph is "similar" but both areas are now finite (do you see why?) and that integral would be convergent (and thus zero) $\endgroup$
    – imranfat
    Apr 12, 2017 at 2:34

3 Answers 3


The point is that, for this improper integral to exist (in the Riemann sense), you would need to prove that the limits $$ \lim_{M,N\to+\infty}\int_{-N}^{M}\frac{x}{1+x^2}dx $$ exist separately in $M$ and $N$. However, this is false because $$ \int_{-N}^M \frac{x}{1+x^2}dx=\frac{1}{2}\log\frac{1+M^2}{1+N^2}\xrightarrow[M\to+\infty]{N=\text{ fixed}}+\infty. $$ For the integral to exist in the Lebesgue sense, you need to have first control on the positive and negative parts separately, i.e. $$ \int_{\mathbb R}\frac{x}{1+x^2}dx=\left[\,\int_{\mathbb R_-}\frac{x}{1+x^2}dx\right]+\left[\,\int_{\mathbb R_+}\frac{x}{1+x^2}dx\right] $$ if both square brackets exist separately. Again this is not true.

What is correct, and is the content of your statement, is the fact that $$ \lim_{M\to+\infty}\int_{-M}^{+M}\frac{x}{1+x^2}dx=0 $$ as you proved. This is commonly defined as the Cauchy Principal Value $\mathrm{PV}$ of the otherwise singular integral: summing up $$ \mathrm{PV}\int_{-\infty}^{+\infty}\frac{x}{1+x^2}dx=\lim_{M\to+\infty}\int_{-M}^{+M}\frac{x}{1+x^2}dx=0. $$


We could have also declared, albeit less symmetrically, for any $\lambda > 0$ that $$\int_{-\infty}^{\infty} \frac{x \,dx}{x^2 + 1} = \lim_{a \to \infty} \int_{-a}^{\lambda a} \frac{x \,dx}{x^2 + 1},$$ and evaluating this limit gives $$\lim_{a \to \infty} \left.\frac{1}{2} \log (x^2 + 1)\right\vert_{-a}^{\lambda a} = \lim_{a \to \infty} \frac{1}{2} \log\left(\frac{\lambda^2 a^2 + 1}{a^2 + 1}\right) = \log \lambda .$$ In particular, by choosing $\lambda$ appropriately, we can make the limit take any real value we want!

To avoid this situation categorically, we usually restrict our attention to improper integrals for which the value does not depend on the relative rate at which we "exhaust" the integrand in the leftward and rightward directions. We formalize this situation by defining an integral of the form $$\int_{-\infty}^{\infty} g(x) \,dx$$ to be the sum $$\int_{-\infty}^a g(x) \,dx + \int_a^{\infty} g(x) \,dx ,$$ provided that both of the constituent improper integrals exist, so that one only deals with one limit at a time. (It is straightforward to verify that if this value exists, it is independent of $a$.)

Now, in some circumstances, one wants to assign a value to an integral $\int_{-\infty}^{\infty} g(x) \,dx$ for which the above definition does not give a value, because one or both of the summands does not converge. In this case, one can define the Cauchy Principal Value as the limit of the integrals over intervals symmetric about the origin, that is, as $$PV \int_{-\infty}^{\infty} g(x) \,dx := \lim_{a \to \infty} \int_{-a}^a g(x) \,dx ,$$ and for odd $g$ this limit is always zero. But as your example shows, this notion does not in general coincide with the usual notion of improper integral.

  • 3
    $\begingroup$ The effect of $\lambda$ on the limit is very enlightening. $\endgroup$ Apr 11, 2017 at 23:38
  • $\begingroup$ I agree, Austin $\endgroup$
    – imranfat
    Apr 12, 2017 at 2:30

Because $$\int \frac{x}{1 + x^2}dx = \frac{1}{2} \ln(1+x^2) + c$$ we know that:

$$\int\limits_{- \infty}^{+ \infty} {\frac{xdx}{1 + x^2}} = \int\limits_{- \infty}^{0} {\frac{xdx}{1 + x^2}} + \int\limits_{0}^{+ \infty} {\frac{xdx}{1 + x^2}} = \lim\limits_{a \to -\infty}\int\limits_{a}^{0} {\frac{xdx}{1 + x^2}} + \lim\limits_{b \to +\infty}\int\limits_{0}^{b} {\frac{xdx}{1 + x^2}}$$

$$ = \lim\limits_{a \to -\infty}\frac{1}{2} \ln(1+x^2)\vert_{a}^{0} + \lim\limits_{b \to +\infty} \frac{1}{2} \ln(1+x^2)\vert_{0}^{b}$$

$$=\lim\limits_{a \to -\infty} -\frac{1}{2}\ln(1+a^2) + \lim\limits_{b \to +\infty} \frac{1}{2}\ln(1+b^2) $$

from which we certainly cannot conclude it is $0$ (indeterminate form)

Your mistake was to think that because the function is odd, the integral is zero, because if $f$ is odd, then $\forall a \in \mathbb{R}$ (if there are no problems with continuity etc.), $$\int\limits_{-a}^{a} f(x) dx= 0$$ However, this does not work if the areas are infinite! For the same reason $$\int\limits_{-\infty}^{+\infty} xdx$$ diverges, while the function is in fact odd.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.