How do you evaluate the integral $int e^(-sqrtx)/sqrtx$ from 0 to $oo$ ?

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Answer 1 · 2016-09-22T07:58:54

$2$

start with the observation that $d/dx (e^(f(x)) ) = f'(x) e^(f(x))$

and so $d/dx (e^( -sqrt x)) = - 1/(2 sqrt x) e^(- sqrt x)$

So

$int_0^oo e^(-sqrtx)/sqrtx dx$

$= -2 int_0^oo - e^(-sqrtx)/(2 sqrtx) dx$

$= -2 int_0^oo d/dx (e^( -sqrt x)) dx$

$= -2 [ e^( -sqrt x) ]_0^oo$

$= -2 (0 -1) = 2$

How do you evaluate the integral int e^(-sqrtx)/sqrtxint e^(-sqrtx)/sqrtx from 0 to oooo?