Finding Integral Convergence Points

[BraedenP]

Homework Statement

I'm tasked with integrating the following functions, and values for t where the function converges:

\int_{0}^{1}x^p\cdot ln(x) dx

Homework Equations

Integration by Parts Formula: \int udv=uv-\int vdu

The Attempt at a Solution

I found a definite integral:

\frac{x^{p+1}((p+1)ln(x)-1)}{(p+1)^2}+C

With that, I can tell that the function is undefined at p = -1, but that's about it. How would I go about nailing down certain values where it can be evaluated?

Any advice to point me in the right direction would be great!

 

[Dick]

The integral is also undefined at x=0 for ANY value of p. It's an indefinite integral. You need to figure for which values of p the limit as x->0 of your integral exists.

 

[BraedenP]

How would I go about doing that? The only way I would know how to get rid of the ln(x) to solve the limit would be to apply L'Hopital's rule. However, it's not in the proper form to apply that theorem..

 

[Dick]

How would I go about doing that? The only way I would know how to get rid of the ln(x) to solve the limit would be to apply L'Hopital's rule. However, it's not in the proper form to apply that theorem..

Why not? log(x) goes to -infinity at x=0. x^(p+1) goes to 0 at x=0 if p+1>0. That's -infinity*0. You can make it an infinity/infinity limit by moving one of the functions into the denominator. BTW I didn't check your integration by parts, I just assumed you knew what you were doing.

 

[BraedenP]

Oh, okay! I didn't notice that. Thanks a heap! (I'm pretty sure I did the IBP correctly. That wasn't the challenging part.)

 

[BraedenP]

Well I rearranged it with \frac{1}{x^{P=1}} in the denominator, then applied L'Hopital's Rule... It yielded another indeterminate limit, so I did it again, and got yet another indeterminate. I'm pretty sure this pattern is just going to continue indefinitely.

 

[Dick]

If I start with log(x)/(x^(-(p+1)) and take the derivative of numerator and denominator and simplify I just get a power of x. How can that be indeterminant?

 

[BraedenP]

What are you doing with the (p+1) and -1 in the numerator, and the (p+1)^2 in the denominator?

 

[Dick]

What are you doing with the (p+1) and -1 in the numerator, and the (p+1)^2 in the denominator?

Nothing. They are just nonzero constants unless p=(-1). They don't affect convergence or divergence. And if p=(-1) then we know it diverges, yes? Do the p=(-1) as a special case.

 

[BraedenP]

I can find if it diverges that way, but if I'm removing all of the p terms how am I going to find values of p which are defined for the integral? Remember, I'm supposed to solve this for certain (valid) values of p.

 

[Dick]

I can find if it diverges that way, but if I'm removing all of the p terms how am I going to find values of p which are defined for the integral? Remember, I'm supposed to solve this for certain (valid) values of p.

You can remove the p terms that just appear as multiplicative constants. The p's that appear in the powers of x are the important p's. Can you show what you got using l'Hopital?

 

[BraedenP]

After rearranging I had (without removing the constant terms) in my limit:

-\frac{(p+1)ln(t))-1)}{\frac{1}{t^{p+1}(p+1)^2}}

Then after l'Hopital:

-\frac{(p+1)\frac{1}{t}}{-(p+1)t^{-p-1}[(p+1)ln(t)-2]}

Which is also indeterminate as t->0, and applying l'Hopital's again doesn't help.

(I'm still thinking I'll need to keep all of the p's since I'm evaluating it for certain values of p.. Once I move the constants out of the integrals I'll still need to have them all for my final evaluation, will I not?)

 

[Dick]

What are you doing? There is no ln(t) left after you do l'Hopital. Once you take the derivative of ln(t)-1 in the numerator, all that's left is 1/t in the numerator. How did ln(t) get into the denominator?

 

[BraedenP]

Oh my... I really have no clue what I was doing... For some reason when I took the derivative I treated p as a variable (t) for some reason...

I'm so retarded lol. Thanks.

 

[BraedenP]

So now after l'Hopital I get a 1/\infty limit with (p+1) in the numerator... Does this mean that I need to set p = -1 to give me another indeterminate form, and then re-apply l'Hopital?

In that case, p = -1 would be the value the question is looking for..

 

[Dick]

So now after l'Hopital I get a 1/\infty limit with (p+1) in the numerator... Does this mean that I need to set p = -1 to give me another indeterminate form, and then re-apply l'Hopital?

In that case, p = -1 would be the value the question is looking for..

No. p=(-1) is just one case where it diverges. I'm really going to have to ask you to show your work again, since you keep not doing it. Don't you just get a power of t out of l'Hopital?? It's pretty easy to say when a power of t diverges at t=0.

 

[BraedenP]

Yeah, I just get a power of t and the (p+1) constant:

-\frac{t^{p+1}}{(p+1)^2}

But then how can I tell for which values of p this converges? All I can discern is that it converges for every value but -1.

 

[Dick]

Yeah, I just get a power of t and the (p+1) constant:

-\frac{t^{p+1}}{(p+1)^2}

But then how can I tell for which values of p this converges? All I can discern is that it converges for every value but -1.

If p+1 is positive the limit as t->0 is zero, isn't it? If p+1 is negative, it diverges, doesn't it? I'll remind you once more I didn't check your steps leading up to here. I usually try to, but on this one I'm trusting you.

 

[BraedenP]

Not necessarily... (p+1) just can't be zero, right?

If p = -2, then (p+1) = -1 --> (-1)^2 = 1 --> 0/1 = 0

I need to solve this integral for some specific value(s) of p for which it converges, but like you I can only get huge intervals for which it converges.

 

[Dick]

Not necessarily... (p+1) just can't be zero, right?

If p = -2, then (p+1) = -1 --> (-1)^2 = 1 --> 0/1 = 0

I need to solve this integral for some specific value(s) of p for which it converges, but like you I can only get huge intervals for which it converges.

If p=(-2) then -t^(p+1)/(p+1)^2 is -t^(-1)/1. That diverges, right? If p=2 then you get -t^3/9. That converges as t->0, also right? You seem to be somehow scrambling p's and t's here. p is a constant. t is the variable. Where does it split between converging and diverging?

 

[BraedenP]

Oh, right! That makes sense..

However, there doesn't seem to be a positive value for which it diverges. It just gets larger and larger.

 

[Dick]

Oh, right! That makes sense..

However, there doesn't seem to be a positive value for which it diverges. It just gets larger and larger.

I'm not sure I get that but at what p value does it split between converging and diverging? I think that's what the question is asking.

 

[BraedenP]

It splits between converging and diverging at p = -1 but the question asks me to evaluate the integral for the "values of p for which it converges".

Hmmm...

 

[Dick]

It splits between converging and diverging at p = -1 but the question asks me to evaluate the integral for the "values of p for which it converges".

Hmmm...

Well, then the answer might be p>(-1), mightn't it? Assuming everything before (that I didn't check) is correct? Those would be p values where it converges. And yes, there are a lot of them.

 

[BraedenP]

Yeah, that's what I was thinking. I'm just going to put that down and hope for the best :P

Thanks a million for all the help! I didn't mean to be so confusing -- for some reason my head just isn't working properly today.