Changing signs of integration limits

When we reverse the limits and then change the signs of the limits of the definite integral $$\int\limits_{-R}^{-\epsilon}\frac{e^{ix}}{x}dx $$ the integral becomes $$-\int\limits_{\epsilon}^{R}\frac{e^{-ix}}{x}dx $$

This is how the process appears to go:

(1) Changing the order of the limits of integration adds the minus sign before the integral. This is clear.

(2) Changing the signs of the limits changes the signs of the $x$'s, but also the sign of $dx$ appears to have changed as well, for otherwise there wouldn't be the minus sign before the integral.

Part (2) is not completely clear to me, because $dx$ arises from the Riemann sums, in which $\Delta x$ is always a positive quantity. So what exactly happens when we change the limits' signs?

Solution 1:

In fact, there is another sign change involved. Formally, it is best to perform the substitution $u = -x$. Then $du = -dx$ and so

$$ \int_{-R}^{-\varepsilon} \frac{e^{ix}}{x} \, dx = \int_{R}^{\varepsilon} \frac{e^{-iu}}{-u} \, (-du) = \int_{R}^{\varepsilon} \frac{e^{-iu}}{u} \, du = -\int_{\varepsilon}^{R} \frac{e^{-iu}}{u} \, du.$$

Solution 2:

You will use

$$* \int_a^b f = -\int_b^af $$

and the substitution formula

$$** \int_a^bf (\phi (t))\phi'(t)dt=\int_{\phi (a)}^{\phi (b)}f (x)dx $$

with $\phi (t)=-t $

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