How do I evaluate $int_0^oo e^-x/sqrtxdx$ ?

Question

I've gotten this far, but I obviously messed up somewhere because I can't evaluate the second integral.

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Answer 1 · 2018-03-28T17:49:12

$int_0^oo \ e^(-x)/sqrt(x) \ dx = sqrt(pi)$

We seek:

$I = int_0^oo \ e^(-x)/sqrt(x) \ dx$

We can perform a substitution. Let:

$u = sqrt(x) <=> x = u^2 => dx/(du) = 2u$

So that if we substitute into the integral, and change the limits accordingly, we get:

$I = int_0^oo \ e^(-u^2)/u \ 2u \ du$
$\ \ = 2 \ int_0^oo \ e^(-u^2) \ du$

This is related to well studied Gaussian integral, with well known result:

$int_(-oo)^oo \ e^(-u^2) \ du = sqrt(pi) => int_(0)^oo \ e^(-u^2) \ du = sqrt(pi)/2$

Thus, we have:

$I = 2 * sqrt(pi)/2$

Hence:

$int_0^oo \ e^(-x)/sqrt(x) \ dx = sqrt(pi)$