Basically, if we integrate $\oint \ \frac{ \zeta'(z)} {\zeta(z)}dz$ over a rectangle of vertices, say (1+15i), (0+15i), (0+10i) and (1+10i) (counter-clockwise), then, according to Cauchy's Residue Theorem, it should give us number of zeroes of the Riemann zeta function in that rectangle times $2\pi i$ (as there aren't any poles in that region) which is, in fact, just $2\pi i$. However, if the integral $\oint \ \frac{ \zeta'(z)} {\zeta(z)}dz$ is integrated individually on each line of the rectancle, i.e $\int_{1}^0 \frac{ \zeta'(x+15i)} {\zeta(x+15i)}dx+i\int_{15}^{10} \frac{ \zeta'(iy)} {\zeta(iy)}dy+\int_{0}^{1} \frac{ \zeta'(x+i10)} {\zeta(x+i10)}dx+i\int_{10}^{15} \frac{ \zeta'(1+ix)} {\zeta(1+ix)}dx$ it doesn't, in fact, give the desired value.
Can someone please tell me where I made the mistake? Any kind of help will be appreciated.