3
$\begingroup$

Is the following function known to be integrable? It is supposed to be a probability density function, i.e., integrates to one. However, it leaves the online Mathematica integrator stumped:

\begin{equation*} f(x) = \frac{k-1}{2(1+|x|^k)},\; x \in (-\infty, \infty),\; k \geq 2 \end{equation*}

  • 0
    Yes, it is integrable, but the Abs[] makes the problem trickier for a computer algebra system. Break it up into positive and negative for a piecewise solution.2011-10-02
  • 1
    Sure it's integrable: pulling out the constant $\frac{k-1}{2}$, the function is continuous and for all $x \geq 1$ it is at most $\frac{1}{1+x^2}$, which you can see is integrable directly by finding the antiderivative. I have no experience with Mathematica, but perhaps you could try removing the absolute value, integrating from $0$ to $\infty$ instead, and multiplying by $2$?2011-10-02
  • 0
    Note though that for $k = 2$ we have $\int_{-\infty}^{\infty} \frac{dx}{1+x^2} = \arctan(\infty) - \arctan(-\infty) = \pi/2 - (-\pi/2) = \pi$. Multiplying this by $\frac{1}{2}$ does not give $1$. In other words, your function doesn't seem to be a probability density function...2011-10-02
  • 0
    Thanks Pete and Ed. But clearly, even for higher values of $k$, the function is not straightforwardly integrable. For example, what is $\int_0^{\infty} 1/(1+x^3)\, dx$ or $\int_{-\infty}^0 1/(1-x^3)\, dx$?2011-10-02
  • 2
    According to Maple, $\displaystyle \int_0^\infty 1/(1+x^3)\ dx = 2 \pi \sqrt{3}/9$. More generally, $\displaystyle \int_0^\infty 1/(1+x^n)\ dx = (\pi/n) \csc(\pi/n)$.2011-10-03
  • 0
    @Praetoria: you know you can factor $1+x^3$, yes? Split into partial fractions, evaluate the integral from $0$ to $u$, and take the limit as $u\to \infty$...2011-10-03
  • 0
    @Praetoria: I took your question "is this function integrable?" to mean "Does the improper integral converge?" and my comment was intended to show that it does. If you mean to find the explicit value -- well, this can be done in principle for every rational function and in practice when one knows explicit formulas for the roots of the denominator, as is the case here: they are complex roots of unity. Robert Israel's comment gives an explicit formula (which I haven't checked but have no reason whatsoever to doubt...).2011-10-03
  • 0
    @Robert: perhaps your comment should be made into an answer (perhaps with a little more explanation). In the long run it is tidiest if questions are answered in the answers, rather than in the comments...2011-10-03
  • 0
    Thanks everyone for the answers! If Robert puts in an answer I will mark it.2011-10-03

0 Answers 0