Improper Integrals

Relevant wiki: L'Hopital's Rule - Basic

The improper integral can be solved by integration by parts using $f' = e^x$ and $g = x$ .

$\begin{aligned} I & = \int_{-\infty}^0 x e^x \ dx \\ & = xe^x \ \big|_{-\infty}^0 - \int_{-\infty}^0 e^x \ dx \\ & = 0 - {\color{#3D99F6}0} - e^x \ \big|_{-\infty}^0 & \small {\color{#3D99F6}\text{See Note.}} \\ & = \boxed{- 1} \end{aligned}$

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Note:

$\begin{aligned} \lim_{x \to -\infty} xe^x & = \lim_{x \to -\infty} \frac x{e^{-x}} & \small {\color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to -\infty} \frac 1{-e^{-x}} & \small {\color{#3D99F6} \text{Differentiate up and down w.r.t. }x} \\ & = 0 \end{aligned}$