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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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    A change of variables should be for a map from an interval to an interval.2012-10-17

1 Answers 1

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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.

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    Okay, the problem arose when calculating $\int e^{\frac{m}{x}} x^{m} dx$. I have opened up a separate topic for that: http://math.stackexchange.com/questions/215706/how-to-calculate-int-e-fracmx-xm-dx-for-positive-m Thanks for your help!2012-10-17