Improper integral $\int_{0}^{1}(1+\frac{1}{x})e^\frac{-1}{x}dx$

Question

How to evaluate this integral:

$$ \int_{0}^{1} \left(1+\frac{1}{x} \right) e^{-1/x}dx $$

Please help

Answer 1

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.

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\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}