This is a problem from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming exam. I'm struggling with this combination of a fractional exponent and logarithm. Below is my attempt, but I need help with how to finish it or if there's a better method for this type of problem.
Problem:
Evaluate the improper integral:
$$
I = \int_0^{+\infty} \frac{x^{1/3}\ln x}{x^2+4}\; dx
$$
Explain all steps carefully (show contours, introduce branches, etc.).
Attempted Solution:
Let $f(z) = \frac{z^{1/3}\ln z}{z^2+4}$ and consider the following contour with a branch cut along the positive real axis, where $\Gamma$ is the entire closed curve, $C_R$ is the outside circle of radius $R$, $C_r$ is the inside circle of radius $r$, $A$ is the line from $r$ to $R$, $B$ is the line from $R$ back to $r$, and $0\leq \arg z \leq 2\pi$.
Using the Residue Theorem, we have
$$
\begin{split}
\oint_\Gamma f(z) dz
&= 2\pi i \Big(\text{Res}(f,2i)+\text{Res}(f,-2i)\Big) \\
&= \int_{C_R} f(z)dz + \int_{C_r} f(z)dz
+ \int_r^R f(z)dz + \int_R^r f(z)dz,
\end{split}
$$
where $\int_r^R f(z)dz$ uses $\arg z = 0$ and $\int_R^r f(z)dz$ uses $\arg z = 2\pi$.
Note that $z^{1/3} = e^{\frac{1}{3}(\ln z)} = e^{\frac{1}{3}(\ln |z|+i\arg z)}$, thus we can write
$$
f(z) = \frac{e^{\frac{1}{3}(\ln |z|+i\arg z)}\Big(\ln|z|+i\arg z\Big)}{z^2+4}.
$$
Using the ML estimation lemma, we find that $\int_{C_r} f(z)dz$ goes to zero as $r\to 0$ and $\int_{C_R}$ goes to zero as $R\to\infty$. (Work omitted here but can be shown if needed.) We're left with
$$
\begin{align*}
2\pi i \Big(&\text{Res}(f,2i)+\text{Res}(f,-2i)\Big) \\
&= \int_r^R f(z)dz + \int_R^r f(z)dz\\
&= \int_r^R \frac{e^{\frac{1}{3}(\ln |z|)}\ln|z|}{z^2+4}dz + \int_R^r \frac{e^{\frac{1}{3}(\ln |z|+2\pi i)}\Big(\ln|z|+2\pi i\Big)}{z^2+4}dz\\
&= \int_r^R \frac{x^{1/3}\ln x}{x^2+4}dx – e^{\frac{2\pi}{3}i} \int_r^R \frac{x^{1/3}(\ln x+2\pi i)}{x^2+4} dx
\end{align*}
$$
Now from here, I know how to evaluate the residues and I see that we can isolate the term we're looking for as
$$
\lim_{r\to 0}\lim_{R\to\infty} \int_r^R \frac{x^{1/3}\ln x}{x^2+4}dx,
$$
but I don't know what to do with the remaining integral term on the righthand side. How do I complete this? Or should I have used a different contour or branch cut?
Best Answer
$$I=\int_0^\infty \frac{x^a}{x^2+4}dx$$ Let $x=2\sqrt{t}$
$$I=2^{a-2}\int_0^\infty \frac{t^{\frac{a-1}{2}}}{t+1}dt=2^{a-2}\text{B}(\frac{a+1}{2},\frac{1-a}{2})=2^{a-2}\frac{\pi}{\sin(\pi\frac{a+1}{2})}=2^{a-2}\frac{\pi}{\cos(\pi\frac{a}{2})}$$
Your integral is the following. $$\left(\frac{d}{da}I\right)_{a=\frac{1}3}$$