Real integral getting numerically evalued to complex number I tried feeding th following integral into WolframAlpha:
$$\int_0^2\frac{1}{3\sqrt{x}(\log(x))^{\frac13}},$$
to get an idea of its value. Result:

http://bit.ly/strangeint
$$\int_0^2\frac{1}{3\sqrt{x}(\log(x))^{\frac13}}=0.810563-0.620515.$$

And I'm like "How exactly could a real integral have a complex value?". No seriously, the integrand is real for all real values, so how did that complex part come about?
 A: WolframAlpha distinguishes between x^(1/3) and cbrt(x) - the first returns the principal root while the second returns the real root.  For example, cbrt(-8) returns $-2$ while (-8)^(1/3) returns $1 + 1.73205i$.  Thus, you can use cbrt in your WolframAlpha query.
integrate (1/(3*sqrt(x)*cbrt(log(x))), x=0..2)


Note that the output states that the computation assumes that cbrt is the real root and offers you a button to switch to the principal.  Your computation with log(x)^(1/3) should have assumed the principal and offered a button to get to the real.
A: The problem is that the logarithm has a branch point in $0$, as well as $\sqrt[3]{z}$, so we must be careful when defining the branches we are taking. If we simply set $x=e^t$ we have:
$$ I = \int_{-\infty}^{\log 2}\frac{e^{t/2}}{3 t^{1/3}}\,dt = \int_{0}^{\log 2}3t^{-1/3}e^{t/2}\,dt+\int_{-\infty}^{0}3t^{-1/3}e^{t/2}\,dt$$
where the first integral can be computed through a Taylor series and the second one depends on a value for the $\Gamma$ function, once we define $t^{-1/3}$ for $t<0$ as $-(-t)^{-1/3}$. Wolfram Alpha is unaware of such assumption, so it probably takes another branch, resulting in a complex-valued integral.
