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

Question

918 views around the world

You can reuse this answer
Creative Commons License

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#