Dgdrxrt

The question is asking to evaluate the following integral:

$$int_0^infty fracx^1/3x^2+7x+6dx$$

I am required to use complex analysis methods to solve this integral but I cannot seem to find a method that would work. I know I need to use Branch Cuts/points and the Estimation and Jordan's Lemma, but that is all I know. I tried using u-substitution to see if I would get the right answer and that did not work either. Any help is appreciated.

Let $x=t^3$ to make
$$I=int fracx^1/3x^2+7x+6,dx=3intfrac t^3t^6+7 t^3+6,dt$$ Using partial fraction decomposition
$$frac t^3t^6+7 t^3+6=frac185 left(t^3+6right)+fract-25 left(t^2-t+1right)-frac15(t+1)$$ and the fist term can again be decomposed writing
$$t^3+6=(t-a)(t-b)(t-c)$$ making
$$frac1t^3+6=-frac1(a-b) (b-c) (t-b)-frac1(a-c) (c-b) (t-c)+frac1(a-b) (a-c)
(t-a)$$ making that you face quite simple antiderivatives.

We can save some effort here by using the result:

$$I(p) = int_0^inftyfracx^-p1+xdx = fracpisin(pi p)$$

This is valid for $0<p<1$. It's easy to derive this result using contour integration involving the same formalism needed to tackle the desired integral directly, so you could try to derive this one first.

To use this result to evaluate the desired integral, let's write the denominator of the integrand as:

$$frac1x^2 + 7 x + 6 = frac15left(frac1x+1 - frac1x+6right)$$

We can then substitute this partial fraction expansion in the integral, and write it as the sum of two integrals of the same form as $I(p)$. However, the two integrals we obtain this way don't converge for $p = -frac13$, so it seems that the path to using this shortcut is blocked.

To get around this problem, we can use that $I(p)$ has a unique continuation as a meromorphic function of $p$ on the complex plane. We can then start with evaluating the two integrals for values of $p$ for which these integrals do converge. We then add up the two expressions to get to the expression of the desired integral with a general value of the exponent $p$. This expression then as been derived for $0<p<1$, but it has a unique analytic continuation to the complex plane, allowing us to simply substitute $p = -frac13$ in the result.

We then need to evaluate:

$$int_0^inftyfracx^-p6+xdx = fracpi6^psin(pi p) $$

This yields the result:

$$J(p) = int_0^inftyfracx^-px^2 + 7 x + 6dx = fracpileft(1-6^-pright)5sin(pi p)$$

For $p = -frac13$ this yields:

$$int_0^inftyfracx^frac13x^2 + 7 x + 6dx = frac2pisqrt315left(sqrt[3]6-1right)$$