so i need to integrate y^3*(e^-(y^2/2v))/(2piv)^0.5 between + and - infinity and show its 0. How to compute the integral?
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Originally Posted by Duke so i need to integrate y^3*(e^-(y^2/2v))/(2piv)^0.5 between + and - infinity and show its 0. How to compute the integral? You don't, you look at the symmetry/antisymmetry about y=0. Call the integrand I(y) so: I(y)=y^3*(e^-(y^2/2v))/(2piv)^0.5 Now what is I(-y) in relation to I(y)? CB
I(-y)= -I(y) due to the y^3.
Originally Posted by Duke I(-y)= -I(y) due to the y^3. Then the integral we may write $\displaystyle \int_{-\infty}^{\infty} I(y)dy =\int_{-\infty}^0 I(y) dy + \int_{0}^{\infty} I(y) dy$= .. or you may just know what that means for the integral. CB
=0
Originally Posted by Duke =0 Yes. CB
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