# Thread: change of variable question (changing of signs)

1. ## change of variable question (changing of signs)

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?

2. Originally Posted by Volga
Is that right?
Yes:

$\displaystyle \displaystyle\int_{-\infty}^{-1}\dfrac{e^x}{x}dx=\displaystyle\int_{+\infty}^{1} \dfrac{e^{-y}}{-y}(-dy)=\displaystyle\int_{+\infty}^{1}\dfrac{e^{-y}}{y}dy=-\displaystyle\int_{1}^{+\infty}\dfrac{e^{-y}}{y}dy$

Fernando Revilla

3. Thank you! I am happy now )))