Integrating e^-ax^2


by sanitykey
Tags: eax2, integrating
sanitykey
sanitykey is offline
#1
Nov9-06, 10:42 AM
P: 93
Hi, i have a problem which is confusing me

Question:

Given that

[tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

What is

(i) [tex]\int_{0}^{\infty}e^{-ax^2} x^2 dx[/tex]
(ii) [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx[/tex]
(iii) [tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx[/tex]

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

I tried (ii):

[tex]u=x^3[/tex]

[tex]\frac{du}{dx} = 3x^2[/tex]

[tex]\frac{dv}{dx} = e^{-ax^2}[/tex]

[tex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]


[tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \int_{0}^{\infty}\frac{3x^2 \sqrt{\pi}}{2\sqrt{a}}[/tex]

[tex]= \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} = 0[/tex]

I know limits should be in that last bit and the bit before but i thought it wouldn't matter as both parts are the same, but then i thought i can't say [itex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/itex] anyway because the given expression wasn't general it was between limits, or does it not matter because i'm using the same limits?

I don't know what the question means when it tells me to "use differentiation for (i) and (iii) with respect to the a paramter" how will that help me get to a solution?

Thanks in advance
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
StatusX
StatusX is offline
#2
Nov9-06, 12:17 PM
HW Helper
P: 2,566
What happened to the exponential after you integrated by parts? And have you tried their suggestion for i and iii? You can move the d/da inside the integral (justifying this is a little tricky, but I'm sure you don't need to worry about that).
Office_Shredder
Office_Shredder is offline
#3
Nov9-06, 12:50 PM
Mentor
P: 4,499
You can't say [itex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/itex] for the part inside the integral

HallsofIvy
HallsofIvy is offline
#4
Nov9-06, 06:04 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

Integrating e^-ax^2


Quote Quote by sanitykey
Hi, i have a problem which is confusing me

Question:

Given that

[tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

What is

(i) [tex]\int_{0}^{\infty}e^{-ax^2} x^2 dx[/tex]
(ii) [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx[/tex]
(iii) [tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx[/tex]

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

I tried (ii):

[tex]u=x^3[/tex]

[tex]\frac{du}{dx} = 3x^2[/tex]

[tex]\frac{dv}{dx} = e^{-ax^2}[/tex]

[tex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]
This is the definite integral from [itex]-\infty[/itex] to [itex]\infty[/itex], not the anti-derivative! [itex]e^{-ax^2}[/itex] has no simple anti-derivative. Try [itex] u= x^2[/itex], [itex]dv= xe^{-ax^2}dx[/itex] instead.


[tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \int_{0}^{\infty}\frac{3x^2 \sqrt{\pi}}{2\sqrt{a}}[/tex]

[tex]= \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} = 0[/tex]

I know limits should be in that last bit and the bit before but i thought it wouldn't matter as both parts are the same, but then i thought i can't say [itex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/itex] anyway because the given expression wasn't general it was between limits, or does it not matter because i'm using the same limits?

I don't know what the question means when it tells me to "use differentiation for (i) and (iii) with respect to the a paramter" how will that help me get to a solution?

Thanks in advance
dextercioby
dextercioby is offline
#5
Nov10-06, 01:37 AM
Sci Advisor
HW Helper
P: 11,863
Quote Quote by sanitykey
Hi, i have a problem which is confusing me

Question:

Given that

[tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

What is

(i) [tex]\int_{0}^{\infty}e^{-ax^2} x^2 dx[/tex]
(ii) [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx[/tex]
(iii) [tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx[/tex]

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)
For (ii) compute first

[tex] \int_{0}^{\infty} e^{-ax^2} \ x \ dx[/tex]

And when you get the result, you can differentiate wrt "a" to get the needed integral.

HINT:Make the sub x^{2}=t and of course assume a>0.

Daniel.
(
sanitykey
sanitykey is offline
#6
Nov10-06, 04:16 AM
P: 93
Thanks for all these replies! :D

I would of responded sooner but i thought i'd best come back with something to show i've been quite busy.

I'll just show you what my answers are not sure if they're right but if they are credit to you all for helping me.

[tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

[tex]\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2}\right) dx = \frac{d}{da}\left(\frac{\sqrt{\pi}}{2\sqrt{a}}\right)[/tex]

[tex]\int_{0}^{\infty}x^2 e^{-ax^2} dx = \frac{\sqrt{\pi}}{4(\sqrt{a})^3}[/tex]

[tex]\int_{0}^{\infty}e^{-ax^2} x dx = \frac{1}{2a}[/tex]

[tex]\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2} x \right) dx = \frac{d}{da}\left(\frac{1}{2a}\right)[/tex]

[tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{1}{2a^2}[/tex]

[tex]\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2} x^2 \right) dx = \frac{d}{da}\left(\frac{\sqrt{\pi}}{4(\sqrt{a})^3}\right)[/tex]

[tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx = \frac{3\sqrt{\pi}}{8(\sqrt{a})^5}[/tex]
Tomsk
Tomsk is offline
#7
Nov10-06, 04:33 AM
P: 227
You're missing a half in the last line, otherwise that looks fine.
sanitykey
sanitykey is offline
#8
Nov10-06, 05:03 AM
P: 93
Ok edited to correct that thank you :)


Register to reply

Related Discussions
integrating x^x Calculus 11
Need help, soon please!! Integrating Calculus 12
integrating (e^4x)/x Calculus 12
Integrating (4x-3)^2 Calculus & Beyond Homework 10
Integrating csc(x) General Math 9