# Integral Action

1. Nov 26, 2005

### PowerWill

Left my integral table at home, could someone tell me the value of these integrals?
$$\int_0^{\infty} \frac{dx}{(x^2+b^2)^2}$$

$$\int_0^{\infty} \frac{x^2}{(x^2+b^2)^2} dx$$
and the same as the latter but with the denominator raised to the power 4. Thanks!

Last edited: Nov 26, 2005
2. Nov 26, 2005

### johnw188

Just solve them yourself, they're both pretty standard trig substitutions. You should always solve the integral forms from the table yourself, at least once, before you use them. They're a nice tool for saving yourself time and effort, not a crutch to avoid learning your integration techniques properly.

3. Nov 27, 2005

### benorin

www.integrals.com

4. Nov 27, 2005

### PowerWill

What do I do with the arctan at infinity? Do I just use $$\frac{\pi}{2}$$ or do I have to be saucy about it?

Last edited: Nov 27, 2005
5. Nov 27, 2005

### HallsofIvy

Staff Emeritus
Well, I wouldn't recommend be "saucy" about homework! Technically, you should take the limit of arctan t as t goes to $\infty$ but that is, of course, $\frac{\pi}{2}$.

6. Nov 27, 2005

### benorin

EDIT: I was going to post the below, but I just realized the it doesn't matter what $\mbox{sgn}(b)$ is, assuming $b\neq 0$, take it to be positive since b is only given by b2, we may assume it is positive.

For $$\int_0^{\infty} \frac{dx}{(x^2+b^2)^2}=\lim_{M\rightarrow\infty} \frac{1}{2b^3} \left(\frac{bM}{b^2+M^2} + \tan^{-1}\left( \frac{M}{b}\right) -0\right)$$
$$=0+\lim_{M\rightarrow\infty} \frac{1}{2b^3} \tan^{-1}\left( \frac{M}{b}\right)=\mbox{sgn}(b) \frac{\pi}{4b^3}$$

Last edited: Nov 27, 2005