How to evaluate this integral:
$$ \int_{0}^{1} \left(1+\frac{1}{x} \right) e^{-1/x}dx $$
Please help
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1Have you tried substituting $u=-1/x$ ? – MasB Jul 20 '18 at 00:29
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1Are you trying to decide if the improper integral exists? You don't need to calculate a value to do that. – Ted Shifrin Jul 20 '18 at 00:38
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Let $$u=-\frac1x$$ Then $$x^2du=dx$$
So \begin{align} I &= \int_{\infty}^{1} \left(1+\frac{1}{x} \right) e^{-1/x}dx \\ &=\int_{-\infty}^{-1} (1-u)u^{-2}e^u du \end{align}
This is easily calculated using integration by parts.
\begin{align} \int u^{-2}e^u du &= -u^{-1}e^u - \int-u^{-1}e^u du \\ \int u^{-2}e^u du - \int u^{-1}e^u du &= -u^{-1}e^u \\ \int_{-\infty}^{-1} (1-u)u^{-2}e^u du &= \left[-u^{-1}e^u\right]_{-\infty}^{-1}\\ &=e^{-1} \end{align}
Karn Watcharasupat
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Not sure I see the integration by parts argument here. Also there appear to be issues with the bounds of your integrals. – Grant B. Jul 20 '18 at 01:46
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@GrantB. my careless mistake on the limits. Thanks for pointing out. The integration by parts is a bit tricky because of the negative exponent but doable. – Karn Watcharasupat Jul 20 '18 at 01:48
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