Consider \[ I = \frac{1}{2\pi i} \int_C \frac{\sin z}{1 - \cos(z^3)} \, dz, \] where \( C = \{ z \in \mathbb{C} : z = x + iy, |x| + |y| = 1, x, y \in \mathbb{R} \} \) is oriented positively as a simple closed curve. Then, the value of \( 120I \) is equal to _________ (in integer).
Step 1: Use local behavior instead of solving for all singularities
Rather than explicitly solving \( \cos(z^3)=1 \), we note that for small \(z\), the only possible singularity of \[ f(z)=\frac{\sin z}{1-\cos(z^3)} \] inside the contour \(C:\ |x|+|y|=1\) is at \(z=0\). All other solutions of \(z^3=2n\pi\) satisfy \(|z|>(2\pi)^{1/3}>1\), hence lie outside \(C\).
Step 2: Determine the nature of the singularity at \(z=0\)
We analyze the order of the zero in the denominator using Taylor expansions:
\[ \sin z = z - \frac{z^3}{6} + O(z^5), \]
\[ 1-\cos(z^3)=\frac{(z^3)^2}{2}+O(z^9)=\frac{z^6}{2}+O(z^9). \]
Hence, near \(z=0\),
\[ f(z)=\frac{z-\frac{z^3}{6}+O(z^5)}{\frac{z^6}{2}+O(z^9)} = \frac{2}{z^5} - \frac{1}{3z^3} + O\!\left(\frac{1}{z}\right). \]
So \(z=0\) is a pole of order 5, and the residue is the coefficient of \(\tfrac{1}{z}\) in the Laurent expansion.
Step 3: Extract the residue using series division
Carrying the expansion to the \(z^5\)-term carefully, the coefficient of \(\tfrac{1}{z}\) turns out to be:
\[ \operatorname{Res}(f,0)=\frac{1}{60}. \]
Step 4: Apply the Residue Theorem
By the Residue Theorem,
\[ \int_C \frac{\sin z}{1-\cos(z^3)}\,dz =2\pi i\cdot \frac{1}{60}. \]
Thus,
\[ I=\frac{1}{2\pi i}\int_C \frac{\sin z}{1-\cos(z^3)}\,dz =\frac{1}{60}. \]
Step 5: Final computation
\[ 120I = 120\times\frac{1}{60}= \boxed{2}. \]
Final Answer: \(\boxed{2}\)