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I think $$\lim_{R\rightarrow\infty} \int_0^R e^{x^2 - R^2}dx = 0$$ but I cannot show it. It is easy to see that for any $\epsilon > 0$ we have $$\lim_{R\rightarrow\infty} \int_0^{R-\epsilon} e^{x^2 - R^2}dx = 0$$ but then I cannot justify interchanging limits in $$\lim_{R\rightarrow\infty} \int_0^R e^{x^2 - R^2}dx = \lim_{R\rightarrow\infty}\lim_{\epsilon \rightarrow 0} \int_0^{R-\epsilon} e^{x^2 - R^2}dx = \lim_{\epsilon \rightarrow 0}\lim_{R\rightarrow\infty} \int_0^{R-\epsilon} e^{x^2 - R^2}dx = 0$$ Any idea?

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2 Answers 2

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There is another way: L'Hospital's Rule Can be applied here. Note that

$$\lim_{R\rightarrow \infty} \int_0^R e^{x^2-R^2}dx = \lim_{R\rightarrow \infty} \frac{\int_0^R e^{x^2}dx}{e^{R^2}}$$

Note that both numerator and denominator are differentiable and both tend to infinity. By L'Hospital's rule, you get $$\lim_{R\rightarrow \infty} \int_0^R e^{x^2-R^2}dx = \lim_{R\rightarrow \infty} \frac{e^{R^2}}{2Re^{R^2}} = \lim_{R\rightarrow \infty} \frac{1}{2R} = 0$$

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  • $\begingroup$ This is quite nice too. Thanks $\endgroup$
    – Cihan
    Commented Dec 7, 2012 at 8:58
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Note that $$\int_0^Re^{x^2-R^2}dx=\int_0^{R-\epsilon}e^{x^2-R^2}dx+\int_{R-\epsilon}^Re^{x^2-R^2}dx\le \int_0^{R-\epsilon}e^{x^2-R^2}dx+\epsilon.$$

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