Confused about Cauchy Integral Theorem

Question

Cauchy integral theorem states that $\oint\limits_{\gamma}f(x)\,\mathrm{dx}= 0$ if $f(x)$ is analytical(differentiable) on a region containing the contour. But I am confused. I read that if you draw a semi-circle contour with the diameter on the real axis, then the integral from $ \int_{-R}^{R} f(x) \,dx = \oint\limits_{\gamma}f(x)\,\mathrm{dx}$. But if $f(x) = x^{2}$ which is always analytical how can it be equal to $0$? Because that would require it to be negative for some parts of the $x$-axis, but it is not. Could someone please explain this? Thank you

Answer 1

Assuming you mean $f(z)=z^2,$ from scratch: on the semicicircle, take $\gamma_1(t)=Re^{it}$ and on the $x$-axis, $\gamma_2(t)=t.$ Then, $\int_{\gamma_1+\gamma_2}f=iR^3\int_0^\pi e^{3it}dt+\int_{-R}^Rt^2dt=\frac{1}{3}\left(-2R^3+2R^3\right)=0.$