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I would like to know the value of the following improper integral: $$\int \limits_{-\infty}^{\infty}\frac{2x}{1+x^2}dx$$

as the function $f(x)=\displaystyle\frac{2x}{1+x^2}$ satisfies $f(-x)=-f(x)$ can I immediately conclude that the integral is equal to zero?

Thank you in advance for any suggestion.

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This looks as if it might be an attempt to calculate the mean of a Cauchy distribution. If so, then it cannot be calculated –  Henry Mar 8 '11 at 23:56
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3 Answers

This improper integral does not exist. It falls off like $1/x$ and that is not integrable. What you note works if you can show that it converges.

Now I have taken the definition as computing the integral from $a$ to $b$ and send $a$ to negative infinity and then $b$.

If you send them to infinity at the same time, then the integral is zero by what you have noted. You can also note that the indefinite integral of your integrand is equal to $\ln(1 + x^2)$.

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You can't conclude that the integral is zero. What you can conclude though is that the Cauchy principal value is zero. The integral is defined as below. $$\int_{-\infty}^{\infty} \frac{2x}{1+x^2} dx = \lim_{R_1 \rightarrow - \infty, R_2 \rightarrow \infty} \int_{R_1}^{R_2} \frac{2x}{1+x^2} dx = \lim_{R_1 \rightarrow - \infty} \int_{R_1}^c \frac{2x}{1+x^2} dx + \lim_{R_2 \rightarrow \infty} \int_{c}^{R_2} \frac{2x}{1+x^2} dx $$ where $c \in \mathbb{R}$.

Unless both the integrals are finite, you cannot make sense of the integral since you will get something "like $(-\infty) + (+ \infty$)"

However note that if $R_1 \rightarrow -\infty$ and $R_2 \rightarrow \infty$ at the same pace i.e. $R_2 = -R_1 = R$, then it is called the Cauchy principal value and is zero.

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The Cauchy principal value of your integral is defined to be $$ \lim_{a \to \infty} \int_{-a}^a {2x \over 1+x^2} \: dx $$ and this is zero by symmetry. So there is a sense in which the answer is $0$. However, if we let the limits go to infinity at different rates then we can make this integral "equal" whatever we want; for example, for any positive real constant $k$, $$ \lim_{a \to \infty} \int_{-a}^{ka} {2x \over 1+x^2} \: dx = \lim_{a \to \infty} \log \left( {1+k^2 a^2 \over 1+a^2} \right) = 2 \log k. $$ Basically, since there is infinitely much area between this curve and the $x$-axis both in the first and third quadrants, we have to be careful about how we arrange things in order to make them "cancel out"; this is why we just throw up our hands and say that the integral doesn't converge.

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