Contour Integration: What is the function?

Question

I know that the integral around a closed path = 0 since there are no poles.

Why is the integral along the slanted path $\int_0^R e^{-x^2w^2} w dx$? If $w$ is defined to be along the slope, shouldn't it simply be $\int_0^R e^{-w^2} dw$?

Answer 1

If $w$ is the intersection between of the slant ray and the unit circle, then the segment can be parametrized with negative direction as $z=xw$ with $0 \leq x \leq R$.

Then $z^2=x^2w^2$ and $dz=w dx$.

$$\int_{slant}e^{-z^2} dz = -\int_{0}^R e^{-x^2w^2} wdx$$