Question (Schaum Advanced Calculus). Test for convergence:

$\displaystyle \int^{-1}_{-{infty}} \frac{\exp{x}}{x}\;{dx} $

My thinking. Substitute y=-x and then deal with a simpler construction $\displaystyle \frac{\exp{(-y)}}{y}$. This is what the model answer is saying, too: -$\displaystyle \int^{infty}_{1}} \frac{\exp{(-y)}}{y}\;{dy} $ converges by comparison to the similar integral of $\displaystyle \exp{(-y)}$

My problem is to understand why the new integral is negative. Here is my list of changes of signs during the substitution y=-x:

1) x in the integrand changes to -y, so there is the first new '-' sign

2) dy/dx is -1 (differentiating y=-x gives a '-')

3) since the limits of integration are changing from -$\displaystyle \infty$ to $\displaystyle \infty$ and from -1 to 1, we need to swap them (put the infinity to the top, and the 1 to the bottom), resulting in yet another '-' sign - number 3.

Therefore, total minuses: 3; overall, the integral becomes negative.Is that right?