Integral of Sqrt(x)*e^-x: Step-by-Step Solution

[ooohffff]

Homework Statement

Evaluate the following integral:
∫₀^∞ √(x)* e^-x dx

Homework Equations

∫₀^∞ e^-x2 dx = (√π)/2

The Attempt at a Solution

So far this is what I've done:
u = x^1/2
du = 1/2 x^-1/2

2 ∫ e^-u2 u² du

Now, I'm not really sure what to do? Or if what I've done so far is leading me down the right path

 

[andrewkirk]

Try writing it as ##2\int_0^\infty (e^{-u^2}u)\cdot u\,du## and doing integration by parts. If you can get to where the only remaining integral to do is like ##\int_0^\infty e^{-u^2}\,du## you can use the fact that that integrand is a constant multiplied by the pdf of a standard normal distribution, for which the integral from 0 to ##\infty## is known.

 

[ooohffff]

Okay, so I did (I'm don't remember how to do this very well so bear with me):

w = u dw = du
dv = e^-u2 u du v = -1/2 e^-u2

(-u/2) e^-u2 - ∫ (-1/2) e^-u2 du
-(u/2) e^-u2 + (1/2)∫ e^-u2 du

For the left side of the equation, would I plug u back in? If so then the limit as x approaches infinity would make that part 0. For the right side, I'm unsure as to whether I just replace the integral with √π /2. If this is all true, then 2( 0 + (1/2) (√π/2)) and would the final answer be √π/2 ?

 

[andrewkirk]

Why have your integration limits disappeared?

The integration by parts rule for definite integrals is:

$$\int_a^b f'(t)g(t)\,dt=\left[f(t)g(t)\right]_a^b-\int_a^b f(t)g'(t)\,dt$$

 

[ooohffff]

Why have your integration limits disappeared?

The integration by parts rule for definite integrals is:

$$\int_a^b f'(t)g(t)\,dt=\left[f(t)g(t)\right]_a^b-\int_a^b f(t)g'(t)\,dt$$

Okay, so:

2[[(-u/2) e^-u2]₀^∞ - ∫₀^∞ (-1/2) e^-u2 du]
2[[(-u/2) e^-u2]₀^∞ + (1/2)∫₀^∞ e^-u2 du]
2[(0) + (1/2) (√π/2)] = √π/2

 

[PeroK]

Okay, so:

2[[(-u/2) e^-u2]₀^∞ - ∫₀^∞ (-1/2) e^-u2 du]
2[[(-u/2) e^-u2]₀^∞ + (1/2)∫₀^∞ e^-u2 du]
2[(0) + (1/2) (√π/2)] = √π/2

Is that ##\frac{\sqrt{\pi}}{2}## or ##\sqrt{\frac{\pi}{2}}##?

 

[ooohffff]

Is that ##\frac{\sqrt{\pi}}{2}## or ##\sqrt{\frac{\pi}{2}}##?

The first one, (√π)/2

 

[PeroK]

The first one, (√π)/2

Looks right.