$$f(z)=\frac{e^{iz}-1-iz}{z^3}$$
What is the value of $$\int_{C} f(z) dz$$ if C is the arc of the semicircle with radius $R\to \infty$ ,going counterclockwise from point (R,0) to (-R,0)
Attempt: I am making D = C + C' where C' is a line going from (-R,0) to (R,0). From there, I tried to use the residue theorem $\int_{C} f(z) dz + \int_{C'} f(z) dz = \pi i$ buth I cannot find the value of $\int_{C'} f(z) dz$ as R approach infinity
any help is appreciated. Thx
Note that $$\int_Cf(z)dz=\int_0^\pi f(Re^{it})iRe^{it}dt$$ for any $z=Re^{it}$ on $C$, we have $$\begin{aligned}&|f(Re^{it})iRe^{it}|=\frac{|e^{iRe^{it}}-1-iRe^{it}|}{R^2}=\frac{|e^{iR\cos t}e^{-R\sin t}-1-iRe^{it}|}{R^2}\\ \leq &\frac{|e^{iR\cos t}e^{-R\sin t}|+1+|iRe^{it}|}{R^2}=\frac{e^{-R\sin t}+1+R}{R^2}\\ \leq &\frac{2+R}{R^2} \end{aligned}$$
Then it follows that $$\left|\int_C f(z)dz\right|\leq \int_0^\pi |f(Re^{it})iRe^{it}|dt\leq \pi\frac{2+R}{R^2}\to 0$$