Calculating Improper Integral of x^3/(x^4-3)^1/2

[FancyNut]

I stopped at the last step while calculating this improper integral:

integral of x^3\ ( x^4 - 3)^1\2 with limits from 1 to infinity...

that's x cubed over the square root of x raised to 4 minus 3...



after replacing infinity with b and taking the limits it seems that I have to take limits of two expressions one that goes to infinity which is ( (b^4 - 3)^1\2) / 8 but the other is undefined since I have a negative 2 inside a square root. The second expression is the square root of negative two all over 8.


Thanks for any help XD

 

[NateTG]

Just to make sure, that's:
$$\int_1^{\infty} \frac{x^3}{\sqrt{x^4-3}} dx$$
Which is divergent.

Perhaps there was an algebra error leading up to this?

 

[arildno]

Not only is it divergent, but it makes no sense outside the field of complex analysis..

 

[FancyNut]

Yeah that's it.

I end up with two expressions and when taking the limit of the first it's infinity but the second has a negative inside the sqaure root. I guessed it's divergent since whatever mistake I did doesn't change (I think..) that the second expression is a constant anyway... I guess. :p

Thanks dudes. ^_^