Integrating Exponential Function with Infinite Upper Boundary

[dj023102]

if f(x) = pi*xe^(-x^2)
integrating this function if the lower boundary is 0 and the upper boundary is infinity is the answer pi*(2e-1). is this right?

 

[d_leet]

Can you think of an appropriate substitution that may help with this integral?

 

[dj023102]

i think i worked out.
Is the answer (2e-1)*pi

 

[d_leet]

Could you show your work? I'm pretty sure that isn't the correct answer, but it is difficult to tell if you don't show how you arrived at it.

 

[dj023102]

sure i used substitution method.
let u = -x^2
then du/dx=-2x, dx=du/-2x
this gives the integral of xe^u/-2xdu
=pi/2 integrate e^u lower boundary 0 upper boundary infinity
then separating the two integral, one integral with lower boundary 0 and upper boundary 1 and the second integral with lower boundary 1 and the upper boundary infinty.
the first integral gives e^1-e^0 and the second integral we get e^1.
then adding them together we get
(pi/2) * (e-1+e) is that right?

 

[Defennder]

sure i used substitution method.
let u = -x^2
then du/dx=-2x, dx=du/-2x
this gives the integral of xe^u/-2xdu
=pi/2 integrate e^u lower boundary 0 upper boundary infinity
then separating the two integral, one integral with lower boundary 0 and upper boundary 1 and the second integral with lower boundary 1 and the upper boundary infinty.
the first integral gives e^1-e^0 and the second integral we get e^1.
then adding them together we get
(pi/2) * (e-1+e) is that right?

It's hard to follow your work. Add some empty lines to make reading easier. Using Latex helps a whole lot.

It shouldn't be e^u, it's slightly off. And what do you mean by "separating the integral"? You mean integration by parts? It's not needed here.

 

[HallsofIvy]

sure i used substitution method.
let u = -x^2
then du/dx=-2x, dx=du/-2x

More simply, if u= -x², then du= -2x dx or -(1/2)du= dx

[itex]xe^{-x^2}dx[/itex] becomes [itex]e^{-x^2}(xdx)= -(1/2)e^u du[/itex]

this gives the integral of xe^u/-2xdu
=pi/2 integrate e^u lower boundary 0 upper boundary infinity
then separating the two integral, one integral with lower boundary 0 and upper boundary 1 and the second integral with lower boundary 1 and the upper boundary infinty.
the first integral gives e^1-e^0 and the second integral we get e^1.
then adding them together we get
(pi/2) * (e-1+e) is that right?

When x= 0, u= 0 and when x="infinity", u is -"infinity" The integral becomes
[tex]\int_0^\infty xe^{-x^2} dx= -\frac{1}{2}\int_0^{-\infty} e^u du[/tex]
[tex]= \frac{1}{2}\int_{-\infty}^0 e^u du[/itex]

I can see no reason to "separate" at x= 1. In any case, since x=1 is an upper limit for one integral and a lower limit for the other, those terms will subtract, not add, and will cancel.

And there certainly is no reason to have a "pi" in there!

 

[dj023102]

So the answer is -1/2?

 

[HallsofIvy]

No, I just looked back and realized you had a "pi" in the original integral that you then dropped.

Including that pi,
[itex]\frac{1}{2}\pi \int_{-\infty}^0 e^u du= \frac{\pi}{2}e^u\right|_{-infty}^0= \frac{\pi}{2}[/tex]