# Change of variable $x \rightarrow \frac{1}{x}$ and integration limits

Suppose I have an integral

$$\int_{0}^{m}\dots dx$$

and I introduce a new variable $y = \frac{1}{x}$. Supposed that I can correctly treat the integrand, this yields

$$\int_{\infty}^{1/m}\dots dy$$

Now compare this with

$$\int_{-0}^{m}\dots dx$$

which leads to $$\int_{-\infty}^{1/m}\dots dy$$

Which one is valid and why?

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Depends on whether $m$ is positive or negative. –  André Nicolas Oct 17 '12 at 14:39
A change of variables should be for a map from an interval to an interval. –  GEdgar Oct 17 '12 at 14:58

The crux is that you need to treat the integral as $\displaystyle \int_{+0}^m dx$ as
$$\displaystyle \lim_{\epsilon \to 0^+} \int_\epsilon^m dx$$
where $\epsilon \to 0^+$ indicates a limit "from above".
For the other integral, you would have $\displaystyle \lim_{\epsilon \to 0^-} \int_\epsilon^m dx$; since each interval $(\epsilon, m)$ then contains $0$ as an interior point, the substitution $y = \frac 1x$ is invalid.
Okay, the problem arose when calculating $\int e^{\frac{m}{x}} x^{m} dx$. I have opened up a separate topic for that: math.stackexchange.com/questions/215706/… Thanks for your help! –  Aton Oct 17 '12 at 15:40