Integration of $F(z)=\int_1^2 \frac{1}{(x-z)^2}dx$

Question

Consider the function $$F(z)=\int_1^2 \frac{1}{(x-z)^2}dx$$, where $\Im(z)>0$

Question is further given but i want to calculate the integration first .

Solution i tried - I am confused here because i am not getting how to solve integration,i am using two methods here

1 method-$$F(z)=\int_1^2 \frac{1}{(x-z)^2}dx$$ $$F(z)=\int_1^2 \frac{1}{(x-(x+iy))^2}dx$$

$$F(z)=\int_1^2 \frac{1}{(-iy)^2}dx$$

$$F(z)=\int_1^2 \frac{1}{-y^2}dx$$

which will give

$$F(z)=\frac{-1}{y^2}$$

Second method- $$F(z)=\int_1^2 \frac{1}{(x-z)^2}dx$$ $$F(z)=\int_1^2 (x-z)^{-2}dx$$ $$F(z)=\left |\frac{ (x-z)^{-1}}{-1} \right |_1^{2}$$

which will give $$F(z)=\frac{1}{z-2}+\frac{1}{1-z}$$

From above which method is right ,i am confused here

Please Help

Thank you

Answer 1

In the first method you are treating $z$ as a variable, namely z=x+iy, which is not the case, if the question states that $z\in\mathbb{C}$. The second one it is.