Finding integral of a function

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:

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?

Solution 1:

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

Solution 2:

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 $2\pi \sqrt{3}/9$.