Real VS Complex for integrals: $\int_0^\infty \frac{dx}{1 + x^3}$ The integral
$$\int_0^\infty \frac{dx}{1 + x^4} = \frac{\pi}{2\sqrt2}$$
can be evaluated both by a complex method (residues) and by a
real method (partial fraction decomposition).
The complex method works also for the integral 
$$\int_0^\infty \frac{dx}{1 + x^3} = \frac{2\pi}{3\sqrt3}$$
but partial fraction decomposition does not give convergent integrals. I would like to know if there is some real method for evaluating this last integral.
 A: Note that for $a > 0$, $$\int_0^N \frac{1}{x+a}\ dx = \ln(N+a) - \ln(a) = \ln(N) - \ln(a) + o(1)\  \text{as} \ N \to \infty$$ while
$$\eqalign{\int_0^N \frac{x+a}{(x+a)^2 + b^2}\ dx &= \frac{1}{2} \left(\ln((N+a)^2+b^2) - \ln(a^2+b^2)\right)\cr &= \ln(N) - \ln(a^2+b^2) + o(1) \  \text{as} \ N \to \infty\cr}$$
and (if $b > 0$)
$$ \eqalign{\int_0^N \frac{1}{(x+a)^2+b^2}\ dx = \frac{\arctan\left(\frac{N+a}{b}\right) - \arctan\left(\frac{a}{b}\right)}{b} = \frac{\pi}{2b} - \frac{\arctan\left(\frac{a}{b}\right)}{b} + o(1)
 \  \text{as} \ N \to \infty\cr}$$
In particular, from the partial fraction decomposition
$$ \frac{1}{1+x^3} = \frac{1/3}{x+1} + \frac{(2-x)/3}{x^2 - x + 1} = \frac{1/3}{x+1} +
\frac{1/2}{(x-1/2)^2+3/4} - \frac{(x-1/2)/3}{(x-1/2)^2 + 3/4}$$
you get
$$ \int_0^N \frac{1}{1+x^3} \ dx = \frac{\ln(N) - \ln(1))}{3} + \frac{\pi/2 + \arctan(1/\sqrt{3})}{\sqrt{3}} - \frac{\ln(N) - \ln((1/2)^2 + 3/4)}{3}  + o(1)$$
i.e.
$$\int_0^\infty \frac{1}{1+x^3} \ dx = \frac{\pi}{\sqrt{3}} + \frac{\arctan(1/\sqrt{3})}{\sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$
A: For what is worth:
Your integral evaluates in terms of the sine function:
$$\int\limits_0^\infty \frac{1}{1+x^a}=\frac{\pi}{a}\sec\frac{\pi}{a}$$
refer to this question and the link in it.
A: 
I would like to know if there is some real method for evaluating this last integral.

Actually, all integrals of the form $\displaystyle\int_0^\infty\frac{x^n}{1+x^m}dx$ can be solved by substituting $t=\dfrac1{1+x^m}$ , and then recognizing the expression of the beta function in the new integral, which can be written as a product of gamma functions. Then we use the reflection formula in order to finally arrive at the desired result, $I=\dfrac\pi m\cdot\csc\left[(n+1)\dfrac\pi m\right]$ — See my answer here for more information.
A: Make the substitution $x = \frac{1}{t}$ and you get
$$ \int_{0}^{\infty} \frac{t}{1+t^3} \text{d}t$$
Write the one you want as 
$$ \int_{0}^{\infty} \frac{1}{1+t^3} \text{d}t$$
Now you can add both and cancel that pesky $1+t$ factor.
btw, a straightforward approach using partial fractions also works.
You consider
$$F(x) = \int_{0}^{x} \frac{1}{1+t^3} \text{d}t$$
Using partial fractions you can find that (I used Wolfram Alpha, I admit)
$$F(x) = \frac{1}{6}\left(2\log(x+1) - \log(x^2 - x -1) + 2\sqrt{3} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right) + \frac{\pi}{6\sqrt{3}}$$
Now as $x \to \infty$, we have that $2\log(x+1) - \log(x^2 - x + 1) \to 0$ .
