# Integration with limit of zero giving infinity - help please

1. Mar 9, 2015

### Steven Thomas

1. The problem statement, all variables and given/known data
Integral of ∫1/x^2 (or ∫x^-2) between 1 and 0.

3. The attempt at a solution
I can integrate it no problem to give me -1/x or x^-1, but when I put it between the limits of 1 and 0 I get ∞-1 which is just ∞.

Is this right or do I need to use L'Hopital's rule. If so, how? I'm unsure of how it really works.

Thanks

Steven Thomas.

2. Mar 9, 2015

### LCKurtz

When you try to put in $x=0$ in $-\frac 1 x$ you don't get $\infty^{-1}$. What you get is attempted division by $0$ which just means the integral is divergent.

3. Mar 9, 2015

### Steven Thomas

Sorry that is infinity minus 1, not infinity to the power of minus 1, which in turn equals infinity.

Is this an acceptable complete integration, or should I apply L'Hopital's rule?

4. Mar 9, 2015

### LCKurtz

You don't have an indeterminate form like $\frac 0 0$ so there is nothing to apply L'Hospital's rule to. Just say the integral is divergent.

5. Mar 9, 2015

### Ray Vickson

This is an improper integral, so is customarily defined through a limiting process. In particular, since the integrand blows up at $x = 0$ we have
$$\int_0^1 \frac{1}{x^2} \, dx = \lim_{a \to 0+} \int_a^1 \frac{1}{x^2} \, dx \; \Leftarrow \:\text{definition!}$$
What do you get when you compute the limit, if anything?

6. Mar 10, 2015

### Steven Thomas

I got 1/a - 1

As a-> infinity I get negative infinity. Is that the answer? Or is the answer that it doesn't diverge?

Also, how do you insert the equation so nicely on this forum? I have the feeling I'm going to be posting a few like this. I do have another one actually if you'd be so kind as to give me a hand please ray.

Regards,

Steven Thomas.

7. Mar 10, 2015

### Steven Thomas

My other one was integrating the product of x and e^-x between 0 and infinity. I can do the indefinite integral which gives me:

-x.e^-x - e^-x
between 0 and ∞, I just can't evaluate the first term when I use the infinity limit because I get negative infinity multiplied by zero....

Steven Thomas

8. Mar 10, 2015

### SteamKing

Staff Emeritus
This is one of the indeterminate forms you should recognize from studying L'Hopital's Rule:

http://en.wikipedia.org/wiki/Indeterminate_form

If you're going to be doing a lot of similar integrals, you should review this material.

9. Mar 10, 2015

### Steven Thomas

So then I'm pretty sure that the -1/a term goes to minus infinity and hence my whole integral does.

As for my -x.e^-x, do I take it as f(x) is -x and let g(x) = e^x as then I'd have f(x)/g(x)? Differentiating those would give me -1/e^x so it would tend to 0?

10. Mar 10, 2015

### Ray Vickson

Why do you care what happens as $a \to \infty$? This issue involves $a \to 0$.

BTW: I used LaTeX; Somewhere in this Forum there is a tutorial on how to use LaTeX. Sorry: I don't recall what it is called or where it is, exactly, but you can set up a search to find it. (I did not need the tutorial, since I had already been using LaTeX for more than 20 years before joining the Forum.)

11. Mar 10, 2015

### fourier jr

That's what I got, but don't forget the other part. Also, the right way to write it is $$\int_{0}^{\infty}xe^{-x}dx = \lim_{M \rightarrow\infty}\int_{0}^{M}xe^{-x}dx$$ for which I got
$$\lim_{M \rightarrow\infty}[\frac{-(x+1)}{e^{x}}]_{0}^{M} = \lim_{M \rightarrow\infty}[\frac{-(M+1)}{e^{M}} - \frac{-(0+1)}{e^{0}}]$$

& then as you pointed out, you use L'Hopital's rule on the first summand to get the answer.

edit: below the reply window at the bottom of the screen there's a button labelled "Latex preview". If you click it the Latex editor should expand, and then on the right there's a link to the how-to guide. It loads really slowly (on my computer anyway) so I googled the link below instead. Maybe you'll have better luck with that link. Just remember to include  on each end of th Latex code.
http://estudijas.lu.lv/pluginfile.p...trukcijas/matematika_moodle/LaTeX_Symbols.pdf

Last edited: Mar 10, 2015
12. Mar 10, 2015

### Steven Thomas

Thanks for everyones help, I've managed to sort these and the other ones that I had a problem with because of the help y'all supplied.

I'll also try that latex thing (I did wonder why it was asking if I'd like to preview some rubber :p) next time I post. Thanks all.