# Where am i screwing up? Improper Integral

1. Oct 19, 2007

### rocomath

the SM shows a different method but i figure i should arrive at the same answer either way. what step am i screwing up on? thanks!

$$\int_{ - \infty}^{\infty}\frac {x^{2}dx}{9 + x^6}$$

$$\int_{ - \infty}^{0}\frac {x^{2}dx}{9 + x^6} + \int_{0}^{\infty}\frac {x^{2}dx}{9 + x^6}$$

$$u = x^{3}$$
$$du = 3x^{2}dx$$

$$\frac {1}{3}[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{9 + u^{2}} + \lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{9 + u^{2}}]$$

$$3[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} + \lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}}]$$

$$3[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \arctan{\frac {t^{3}}{3}] + 3[\lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{\frac {t^{3}}{3} - \arctan{0}]$$

Last edited: Oct 19, 2007
2. Oct 19, 2007

### Dick

The integral isn't singular at x=0. You don't have to split it up. Just do it.

3. Oct 20, 2007

### bob1182006

you don't need to go from u back to x and then plug-in t. since the limits of u are the same limits for x.

I think your integral should be divided by 1/9...since it's:
$$\int \frac{dx}{a^2+x^2}=\frac{1}{a}arctan\frac{x}{a}$$

other than that I don't see a problem. what is the solution from SM?

4. Oct 20, 2007

### rocomath

my first u-sub is where i got the 1\3 ... then i pulled out a 1\9, but if i pull out that 1\9 it becomes 3.

since i pulled it out from the denominator, simplifying the complex fraction gives me 3 as my constant infront, or no?

Last edited: Oct 20, 2007
5. Oct 20, 2007

### neutrino

If you pull out a 9 from the denominator, wouldn't there be a 1/27 at the front?
In fact, you need not get the 9 out at all. As bob pointed out, there is a standard procedure for dealing with such integrals.

6. Oct 20, 2007

### rocomath

ah damnit. i been pulling out a 1\9 for some odd reason. i pull out a 9 which gives me the (x\a)^2, lol. whew finally.

7. Oct 20, 2007

### bob1182006

no I think it goes like this:
$$\frac{1}{9+x^2}=\frac{1}{9}\frac{9}{9+x^2}=\frac{1}{9}\frac{1}{\frac{9+x^2}{9}}=\frac{1}{9}\frac{1}{1+(\frac{x}{3})^2}$$
and doing a substitution when you integrate that you have to multiply by 3 so you'd get 1/3 * arctan x/3

what do you get when you plug in for t? I get the same answer as the book.

Last edited: Oct 20, 2007
8. Oct 20, 2007

### rocomath

i get that answer now too! w00t :-] thanks again bob and neutrino!

9. Oct 20, 2007

### rocomath

is it ok if i quit using T and just have my integral going from x to 0 etc? it won't be "improper" in terms of notation to do so?

10. Oct 20, 2007

### bob1182006

do you mean just replacing t by x and have no limit? or have x -> infinity?

$$3[\lim_{x\rightarrow - \infty}\int_{x}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \arctan{\frac {x^{3}}{3}] + 3[\lim_{x\rightarrow \infty}\int_{0}^{x}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{\frac {x^{3}}{3} - \arctan{0}]$$

that would just lead to confusion, is the x in the limit the same as the original x in the problem?

11. Oct 20, 2007

### rocomath

12. Oct 20, 2007

### bob1182006

o whops ><

you still need the limit sign with t:

$$\frac{1}{9}[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \lim_{t\rightarrow - \infty}\arctan{\frac {t^{3}}{3}] + \frac{1}{9}[\lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}} = \lim_{t\rightarrow - \infty}\arctan{\frac {t^{3}}{3} - \arctan{0}]$$

I was taught that you could do it but just to remember that by pluggin in you don't mean infinity is a number.

$$\frac{1}{9}(-\arctan\frac{-\infty^3}{3})+\frac{1}{9}\arctan\frac{\infty^3}{3}=\frac{1}{9}(-\arctan (-\infty)+\arctan\infty)=\frac{1}{9}(\frac{\pi}{2}+\frac{\pi}{2})$$