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Question Number 212169 by universe last updated on 04/Oct/24

$$\underset{\lambda \to 0^{+}}{\lim }{\int }_{\lambda }^{2\lambda }\frac{e^{{(x-1)}^2}}{x}dx=?$$

Answered by MrGaster last updated on 03/Nov/24

$$=\underset{\lambda \to 0^{+}}{\lim }{\int }_1^2\frac{e^{{(\lambda u-1)}^2}}{\lambda u}\lambda du$$$$=\underset{\lambda \to 0^{+}}{\lim }\frac{e^{{\lambda }^2{u}^2-2\lambda u+1}}{u}du$$$$={\int }_1^2\frac{e^1}{u}du$$$$=e{\left[\ln u\right]}_1^2$$$$=e(\ln 2-\ln 1)$$$$=e\ln 2$$

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