Improper Integral – An exponential divided by a polynomial on an infinite interval – Exercise 1541

Exercise

Evaluate the integral

0e1xx2dx\int_{-\infty}^0 \frac{e^{\frac{1}{x}}}{x^2} dx

Final Answer


0e1xx2dx=1\int_{-\infty}^0 \frac{e^{\frac{1}{x}}}{x^2} dx=1

Solution

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