# Integration with limit of zero giving infinity - help please

## Homework Statement

Integral of ∫1/x^2 (or ∫x^-2) between 1 and 0.

## 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.

LCKurtz
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## Homework Statement

Integral of ∫1/x^2 (or ∫x^-2) between 1 and 0.

## 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.
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.

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?

LCKurtz
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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.

Ray Vickson
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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.

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?

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.

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

SteamKing
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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

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.

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?

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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.

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.

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.)

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?

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:
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.