Is there a way to gotget the closed form approximate result of $\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$

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edited Dec 4, 2016 at 3:09

Alexey Ustinov

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It is known that $$\int_ 0^{\infty}\frac {e^{-x - \frac {1} {x}}} {x}\, dx=2 K_0(2)$$, ,but$$\int_ 0^{\infty}\frac {e^{-x - \frac {1} {x}}} {x} dx=2 K_0(2),$$ but now I want to gotget the closed form approximate result of $$\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$$,$$\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x} dx.$$ I have searched the classic Table of Integrals, Series, and Products, but there is no pattern match this situation. Is there some approaches to the problem?

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asked Dec 4, 2016 at 2:48

Eric Lou

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Is there a way to got the closed form approximate result of $\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$

It is known that $$\int_ 0^{\infty}\frac {e^{-x - \frac {1} {x}}} {x}\, dx=2 K_0(2)$$, ,but now I want to got the closed form approximate result of $$\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$$, I have searched the classic Table of Integrals, Series, and Products, but there is no pattern match this situation. Is there some approaches to the problem?

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Is there a way to gotget the closed form approximate result of $\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$

Is there a way to got the closed form approximate result of $\int_ 0^a\frac {e^{-x - \frac {1} {x}}} {x}\, dx$