Changing signs of integration limits
By Matthew Cannon
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?
$\endgroup$2 Answers
$\begingroup$In fact, there is another sign change involved. Formally, it is best to perform the substitution $u = -x$. Then $du = -dx$ and so
$$ \int_{-R}^{-\varepsilon} \frac{e^{ix}}{x} \, dx = \int_{R}^{\varepsilon} \frac{e^{-iu}}{-u} \, (-du) = \int_{R}^{\varepsilon} \frac{e^{-iu}}{u} \, du = -\int_{\varepsilon}^{R} \frac{e^{-iu}}{u} \, du.$$
$\endgroup$ 1 $\begingroup$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 $
Thanks a lot to Stackexchange !.
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