16
$\begingroup$

I have stumbled upon an exercise that reads thus:

$$\int\limits_{-\infty}^0\frac{x^x}{x^3-1}\mathrm{d}x=\frac{2\sqrt3}{9}\pi,$$

and I am guessing it is asking to prove the above equality. Calculating a primitive (indefinite integral, antiderivative) of that function seems quite impossible. My only other idea was to find the power series for $x^x$, which is $x^x=\sum\limits_{i=0}^\infty\frac{1}{i!}x^i\log^i(x)$ according to Wolfram Alpha, somehow prove I can swap series and integral after plugging that series into the integral, and compute each term, but the calculations are really "crazy" for the general term $\int\limits_{-\infty}^0\frac{x^i\log^i(x)}{i!(x^3-1)}\mathrm{d}x$. Note that $i$ is an index, not the imaginary unit. I have a hunch this may require some complex analysis, especially since $x^x$ is not really defined on those negative numbers, but is complexly. Since I can't go anywhere on my own, I thought I'd come here and ask: is there some smart and not overly calculation-y way to compute this?

Edit: It does seem that the integral has a complex value. Here is the screenshot of the exercise:

enter image description here

A a first glance I thought it was a Calculus exercise. Then $x^x$ is not defined for negative numbers, or at least, not for all of them. The integral seems to have a complex value as written. So this has become a really misterious exercise. Possible interpretations:

  • The real part of the integral is required;
  • The integral of the real part is required;
  • The imaginary part of the integral is required.

Does any of these get close to the given value?

$\endgroup$
18
  • $\begingroup$ The curve would have to be the $[-\infty,0]$ segment of the real axis, which is not closed! So I can't apply that theorem, can I? $\endgroup$ – MickG Sep 24 '14 at 9:56
  • 1
    $\begingroup$ The question doesn't make sense, for $x \in (-\infty,0)$, what is the sign of numerator $x^x$? If it is positive, then the integrand is negative! $\endgroup$ – achille hui Sep 24 '14 at 10:10
  • 2
    $\begingroup$ The main issue here is how $x^x$ is defined for $x<0$. From $x^x=e^{x\log x}$, assuming that we take the branch of the logarithm for which $\log(-x)=\pi i+\log x$, we are left with $$-\int_{0}^{+\infty}\frac{x^x e^{-\pi i x}}{x^3+1}\,dx$$ whose value is neither purely imaginary nor purely real. $\endgroup$ – Jack D'Aurizio Sep 24 '14 at 10:14
  • 1
    $\begingroup$ @MickG: yes, you are right, there is a $x^{-x}$ in place of $x^x$. $\endgroup$ – Jack D'Aurizio Sep 24 '14 at 10:32
  • 1
    $\begingroup$ @MickG That is the power series for $e^z$ for $z=x\ln(x)$. It is valid as a complex power series centered at $z=0$ $\endgroup$ – David P Sep 24 '14 at 10:51
4
$\begingroup$

Interestingly enough, since this exercise seems off, I have found (using "proof by wolfram alpha") that

$$\int_{-\infty}^0 \dfrac{x}{x^3-1} dx = \dfrac{2\pi\sqrt{3}}{9}$$

So unless there's a notation disparity, it was probably a typo.

The above integral although it does not answer the question, is not hard

$\endgroup$
5
  • $\begingroup$ Yep, that is "easy". Just involves splitting the integrand into the sum of two fractions one for each factor of the denominator, standard way of dealing with rational functions, then integrating each one. One of the sub-integrals requires a not-too nice substitutions, two infinities cancel each other out, and voilà, $\frac{2\sqrt{3}}{9}\frac{\pi}{2}$ is served. Oh… I have a half too much. But the arctangent does tend to $\frac{\pi}{2}$, right? Here's the splitting, and… $\endgroup$ – MickG Sep 24 '14 at 17:18
  • $\begingroup$ … evaluating the primitive in 0 and $-\infty$ has the logs cancel each other out, so what we are left with is $\frac{2\sqrt{3}}{6}\arctan(\frac{2\cdot0+1}{\sqrt{3}})-\frac{2\sqrt{3}}{6}\lim \limits_{x\to-\infty}\arctan\frac{2x+1}{\sqrt{3}}= \frac{\sqrt{3}\pi}{18}+\frac{\sqrt{3}\pi}{6}=\frac{2\pi\sqrt{3}}{9}$. Umm… I'll just check my calculations :). $\endgroup$ – MickG Sep 24 '14 at 17:25
  • $\begingroup$ There is surely some kind of malfunction in MathJax there :). $\endgroup$ – MickG Sep 24 '14 at 17:26
  • $\begingroup$ Yep, the integral amounts to $\dfrac{\tan^{-1}(1/\sqrt{3})}{\sqrt{3}} - \dfrac{-\pi/2}{\sqrt{3}} $ $\endgroup$ – David P Sep 24 '14 at 17:29
  • $\begingroup$ OK I found my mistake: hadn't changed the integration limits when making a substitution :). I have accepted your answer because it seems to solve the exercise, instead of accepting the one below which seems to address the question better, though it uses a Wolfram approximation and doesn't give a way to compute the integral. $\endgroup$ – MickG Sep 24 '14 at 17:29
3
$\begingroup$

I'm sorry I must be missing something here because numerically I get a complex number$$\int_{-\infty}^0\frac{x^x}{x^3-1} = \int_0^{\infty}\frac{-(-x)^{-x}}{x^3+1}\approx -0.08459+0.64991 i$$

This is nowhere near $2\pi \sqrt{3}/9$.

$\endgroup$
2
  • $\begingroup$ Please see the edit. This exercise is quite strange indeed. $\endgroup$ – MickG Sep 24 '14 at 16:16
  • $\begingroup$ @GiulyB I saw your answer before you deleted it. I am posting this comment to tell you \infty is the infinity sign, in case you don't know, and also to ask whether or not you are the «friend of mine» who «emailed me a screenshot of» this exercise. One curiosity: does deleting a voted answer undo the rep changes the votes caused? So did you gain your 4 rep points back after deleting the answer? $\endgroup$ – MickG Sep 24 '14 at 17:33

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.