Integration help

I decided to answer the question by using mth moments with m = 3. But I got stuck with the following integral (there is a fraction that exists in the normal distribution but I decided to put it to the left of the integral):

\int x^3 exp(-(x^2/2o^2)) dx

I started doing it with integration by parts.

u = x^3[br]du = 3x^2[br]dv =exp(-(x^2/2o^2))[br]v = ??

So, not sure how to do the integral of dv. Any help would be appreciated :smile:

(edited 9 years ago)

Reply 1

TeeEm

Original post by SecretDuck

\int x^3 exp(-(x^2/2o^2)) dx

I started doing it with integration by parts.

u = x^3[br]du = 3x^2[br]dv =exp(-(x^2/2o^2))[br]v = ??

So, not sure how to do the integral of dv. Any help would be appreciated :smile:

this requires a substitution first (let t be equal to the exponent of the exponential)
followed by integration by parts

It can be done by parts directly but it is a bit harder to do that way

(edited 9 years ago)

Reply 2

SecretDuck

Original post by TeeEm

this requires a substitution first (let t be equal to the exponent of the exponential)
followed by integration by parts

It can be done by parts directly but it is a bit harder to do that way

OK, I've tried what you said and while I did get closer to showing that E(X^3) = 0, I think I've gone wrong somewhere.

I got this:

I=[2o^4 (u-e^u )]

And evaluating it with both infinities gives me negative infinity.

Reply 3

TeeEm

Original post by SecretDuck

OK, I've tried what you said and while I did get closer to showing that E(X^3) = 0, I think I've gone wrong somewhere.

I got this:

I=[2o^4 (u-e^u )]

And evaluating it with both infinities gives me negative infinity.

Assuming o is a constant

you use the substitution t = -x²/(2a²)

Then you get very simple integration by parts (of the form kte^t

If your limits are from -inf to +inf, the integral is definitely zero since the integrand is odd

Reply 4

TeeEm

give a minute I will do it.

Reply 5

TeeEm

Original post by SecretDuck

OK, I've tried what you said and while I did get closer to showing that E(X^3) = 0, I think I've gone wrong somewhere.

I got this:

I=[2o^4 (u-e^u )]

And evaluating it with both infinities gives me negative infinity.

after the first substitution I got

INT(2o⁴ t e^t dt)

then parts

2a⁴ (t-1)e^t + C

then work back to x

Reply 6

SecretDuck

Original post by TeeEm

after the first substitution I got

INT(2o⁴ t e^t dt)

then parts

2a⁴ (t-1)e^t + C

then work back to x

Thank you - it works :smile:

I managed to get E(X^3) = 0.

Reply 7

TeeEm

Original post by SecretDuck

Thank you - it works :smile:

I managed to get E(X^3) = 0.

all good

Reply 8

ghostwalker

Original post by SecretDuck

...

Since you're limits are symmetrical about zero and this is an "odd" function, there's no requirement to integrate at all; the integral is zero.

An odd function is one where f(-x) = -f(x)

\displaystyle\int_{-a}^{a}f(x) \;dx = 0

Edit: By a similar argument E(X^{2n-1})=0 for all

n\in\mathbb{N}

(edited 9 years ago)

Reply 9

TeeEm

Original post by ghostwalker

Since you're limits are symmetrical about zero and this is an "odd" function, there's no requirement to integrate at all; the integral is zero.

An odd function is one where f(-x) = -f(x)

\displaystyle\int_{-a}^{a}f(x) \;dx = 0

I mentioned this in post 4, but since I have no idea what this statistical integral represented I left it to the poster to continue the way they felt appropriate. :smile:

Reply 10

SecretDuck

Original post by ghostwalker

Since you're limits are symmetrical about zero and this is an "odd" function, there's no requirement to integrate at all; the integral is zero.

An odd function is one where f(-x) = -f(x)

\displaystyle\int_{-a}^{a}f(x) \;dx = 0

Edit: By a similar argument E(X^{2n-1})=0 for all

n\in\mathbb{N}

I do understand that the normal distribution is symmetrical but the lecturer has said that he wants the full integration to be done.

Reply 11

ghostwalker

Original post by TeeEm

I mentioned this in post 4, but since I have no idea what this statistical integral represented I left it to the poster to continue the way they felt appropriate. :smile:

Sorry, I missed that.

Original post by SecretDuck

I do understand that the normal distribution is symmetrical but the lecturer has said that he wants the full integration to be done.