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Hello I wonder if there is any asymptotics known for such integral: $$ I(x) = \int_2^x \frac{e^t}{t} dt \qquad\text{when $ x\to+\infty $}. $$

Thank you very much.

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Take a look at the Exponential integral on wikipedia. –  Harald Hanche-Olsen Dec 4 '12 at 20:31

1 Answer 1

up vote 7 down vote accepted

Apply partial integration: $$I(x) = \int_2^x \frac{e^t}{t} dt = \frac{e^x}{x}-\frac{e^2}{2}+\int_2^x \frac{e^t}{t^2}dt=\frac{e^x}{x}+\frac{e^x}{x^2}-\frac{3e^2}{4}-2\int_2^x \frac{e^t}{t^3}dt,$$ so $$\frac{e^x}{x}-\frac{e^2}{2}<I(x) < \frac{e^x}{x}+\frac{e^x}{x^2}-\frac{3e^2}{4}$$ and $$\lim\limits_{x \rightarrow \infty} I(x) / (e^x/x)=1.$$

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