Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I tried my luck with Wolfram Alpha, with $p \in \mathbb{R}$

$$\int_{-\infty}^{\infty} \frac{x^p}{1+x^2} dx = \frac{1}{2} \pi ((-1)^p+1) \sec(\frac{\pi p}{2})$$ for $-1<p<1$, and doesn't exist for other $p$.

I wonder how to integrate it myself? Especially given that $(-1)^p$ may be a non-real complex number. Thanks in advance!

PS: Does Mathematica or some other (free) CAS give the process of deriving the result?

share|cite|improve this question
Check here. – Sangchul Lee Feb 21 '12 at 13:18
@sos440: Thanks! Why does the integral exists if and only if $-1<p<1$ ? – steveO Feb 21 '12 at 14:21
The integrand is $\approx x^{p}$ near $x = 0$. Thus if $p \leq -1$, the integral diverges. Similarly, the integrand is $\approx x^{p-2}$ near $x = \infty$. Thus if $p \geq 1$, the integral diverges. – Sangchul Lee Feb 21 '12 at 14:46
@sos440:Thanks! Could that be formulated formally? I don't quite get why the approximations to the integrand at $x=0$ and $x=\infty$ explain? – steveO Feb 21 '12 at 14:58
@steveO What you should think is that $\dfrac{1}{1+x^2} \sim 1$ at $x=0$, and $\dfrac{1}{1+x^2} \sim \dfrac{1}{x^2}$ at $x=\infty$ so the results are immediate. – Pedro Tamaroff Feb 21 '12 at 23:10
up vote 4 down vote accepted

Split integration over $\mathbb{R}$ into integration over $\mathbb{R}_{\geqslant 0}$ and $\mathbb{R}_{<0}$ and perform a change of variables $x \mapsto -x$ in the latter one: $$ \int_{-\infty}^\infty \frac{x^p}{1+x^2} \mathrm{d} x = \left(1 + (-1)^p \right) \int_0^\infty \frac{x^p}{1+x^2} \mathrm{d} x $$ Now the idea is to reduce the integral to the Euler's beta integral. To this end perform substitution $x^2 = \frac{t}{1-t}$ so that $t$ ranges from 0 to 1. $$ x \mathrm{d} x = \frac{1}{2} \cdot \frac{\mathrm{d} t}{(1-t)^2} $$ Thus $$ \begin{eqnarray} 2 \int_0^\infty \frac{x^p}{1+x^2} \mathrm{d} x &=& \int_0^1 \left( \frac{t}{1-t} \right)^{(p-1)/2} \frac{1}{1+ \frac{t}{1-t}} \frac{\mathrm{d} t}{(1-t)^2} \\ &=& \int_0^1 t^{(p+1)/2 - 1} \left( 1-t \right)^{-1-(p-1)/2} \mathrm{d} t \\ &=& \operatorname{Beta}\left( \frac{p+1}{2}, \frac{1-p}{2} \right) = \frac{ \Gamma\left( \frac{1-p}{2} \right) \Gamma\left( \frac{p+1}{2}\right)}{\Gamma\left( \frac{p+1}{2} + \frac{1-p}{2} \right)} \\ &=& \frac{\pi}{ \sin\left( \pi \frac{p+1}{2} \right)} = \pi \sec\left( \frac{\pi p}{2} \right) \end{eqnarray} $$

share|cite|improve this answer
Thanks, Sasha! Why does the integral exists if and only if $-1 < p < 1$? – steveO Feb 21 '12 at 14:08
Consider $\int_0^\infty x^p/(1+x^2) \mathrm{d} x$. Near the origin the integrand behaves as $x^p + \mathcal{o}(x^{p})$. This is integrable is $p > -1$. For large $x$, $x^p/(1+x^2) = \frac{1}{x^{2-p}} + \mathcal{o}\left( \frac{1}{x^{2-p}} \right)$, which is integrable if $2-p > 1$. Combining these gives $-1<p<1$. – Sasha Feb 21 '12 at 15:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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