# Improper integral =\

1. Oct 15, 2004

### 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

2. Oct 15, 2004

### 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?

3. Oct 15, 2004

### arildno

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

4. Oct 15, 2004

### 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 doesnt change (I think..) that the second expression is a constant anyway... I guess. :p

Thanks dudes. ^_^