Question

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

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

Answer 1

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

Explanation:

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)`