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.

Scarlett Allison · Accepted Answer

# 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) #

Charles Anctil · Answer

undefined To evaluate the integral ∫₀^∞ e^(-x)/√x dx, we can use the technique of integration by parts. Let's denote:

u = √x   and   dv = e^(-x) dx

Then, find du and v:

du = (1/2)x^(-1/2) dx   and   v = -e^(-x)

Now, apply integration by parts formula:

∫ u dv = uv - ∫ v du

Substitute the values:

∫₀^∞ e^(-x)/√x dx = [-e^(-x)√x]₀^∞ - ∫₀^∞ (-e^(-x))(1/2)x^(-1/2) dx

Evaluate the limits:

lim_(a→∞) -e^(-a)√a + e^0√0 - lim_(a→0) -e^(-a)√a + e^0√0

Simplify:

= 0 + 0 + lim_(a→0) e^(-a)√a - e^0√0

= lim_(a→0) e^(-a)√a

Now, evaluate this limit:

= lim_(a→0) (e^(-a)/a^(1/2))

Since e^(-a) approaches 1 as a approaches 0, and √a also approaches 0 as a approaches 0, we can apply L'Hôpital's rule:

= lim_(a→0) (-(1/2)e^(-a)/a^(-1/2))

= lim_(a→0) (-e^(-a)/(1/2)a^(-1/2))

= -2

Therefore, the value of the integral is -2.