# Homework Help: Finding integral convergence

1. Aug 10, 2011

### c.dube

I'm trying to refresh my knowledge of Calc II, and I'm going through improper integrals right now. The problem I am trying to solve is:

For which numbers $p\geq0$ does $\int_0^\infty \frac{e^{-x}}{x^{p}}$ converge? Justify your answer.

So far, I've split up the integral into two halves (0 to 1 and 1 to infinity) but I'm stuck because I don't know the next step to take. Any help? Just a hint would be great.

2. Aug 10, 2011

### jbunniii

Good start breaking it into pieces. Let's look at the $x \geq 1$ piece first.

Does

$$\int_{1}^{\infty}e^{-x}dx$$

converge?

Then, can you show that

$$\frac{1}{x^p} \leq 1$$

for all $x \geq 1$ and $p \geq 0$?

3. Aug 10, 2011

### Harrisonized

For x ≥ 1:
0 < e-x < 1/xp

Therefore, the integral ∫ e-x/xp dx converges in the interval [1,∞] if ∫ 1/xp dx converges.

Now look at the interval [0,1]. Again we have:

0 < e-x < 1/xp

(This is because e-x=1 at x=0 and e-x=1/e at x=1, but 1/xp=+∞ at x=0 and 1/xp=1 at x=1.)

Therefore, again the integral ∫ e-x/xp dx converges if ∫ 1/xp dx converges.

Look at the p-test to decide the value of p.

Last edited: Aug 10, 2011
4. Aug 10, 2011

### c.dube

OK, so $\int_1^\infty e^{-x}$ converges because the limit as x goes to infinity is finite. Now, because $\frac{1}{x^{p}}$ is decreasing and the largest case would be one, it is always less than or equal to one (is there a more rigorous way of saying this?).
Anyways, moving on to case between 0 and 1, we can prove the top converges by seeing that the limit as x goes to infinity is finite. Can I use the same logic as the first half, and, if so, what is a more rigorous way of saying it? And finally, does this mean the answer is any p greater than or equal to 0? Thanks in advance.
EDIT: Wait, so if the bottom must converge, does that mean, by the p-test, that p must be greater than 1? I'm a little confused.

5. Aug 10, 2011

### Harrisonized

Never mind.

Last edited: Aug 10, 2011
6. Aug 10, 2011

### jbunniii

If $x \geq 1$, then $x^p \geq 1$ for any $p \geq 0$. This is equivalent to

$$\frac{1}{x^p} \leq 1$$

Now multiply both sides by $e^{-x}$ (which is positive for all x) to get

$$\frac{e^{-x}}{x^p} \leq e^{-x}$$

Thus since the right hand side is integrable from 1 to $\infty$, the same is true of the left hand side.

7. Aug 10, 2011

### c.dube

OK that makes sense to me, but where does it leave me? How do I bridge the gap to finding what, if any, p works? Thanks

8. Aug 10, 2011

### jbunniii

My previous post shows that the integral from 1 to $\infty$ converges for all $p \geq 0$. So the result depends entirely on the behavior from 0 to 1.

9. Aug 10, 2011

### c.dube

From 0 to 1, then, don't we have the exact opposite logic, e.g. $\frac{1}{x^p}$ will always be greater than or equal to one? And therefore it diverges for all p, leaving me with an answer p = null set? Haha thanks for bearing with me I don't know why my brain isn't getting this one.

10. Aug 10, 2011

### Harrisonized

Have you tried p=0?

Last edited: Aug 10, 2011
11. Aug 10, 2011

### jbunniii

Yes, that's correct. In fact, if $p > 0$, then $\frac{1}{x^p} \rightarrow \infty$ as $x$ decreases toward 0. However, that doesn't automatically mean you can't integrate it!

No. For some values of p, the integral will be finite even though the function shoots off to infinity near the origin. Loosely speaking, it depends on how "thick" the tall part of the function is, and that is controlled by p.

Try computing the indefinite integral of $1/x^p$, and see what values of p will cause the result to blow up at x = 0, and which ones won't.

12. Aug 10, 2011

### c.dube

Oh ok so if p=0 then the $x^p$ just becomes one, leaving only $e^{-x}$, which converges, so the answer would be p=0?

13. Aug 10, 2011

### c.dube

So the indefinite integral is:
$\frac{x^{-p+1}}{-p+1} + c$
When x=0, doesn't the integral become 0 for every p except p=1, where you get the indeterminate form $0^0$? Gah I really don't know why I'm not getting this.

14. Aug 10, 2011

### jbunniii

No, consider separately these two cases:

$$-p + 1 > 0$$
$$-p + 1 < 0$$

And also observe (important point) that your formula for the integral is wrong if $-p + 1 = 0$.

15. Aug 10, 2011

### c.dube

Ohhhhh ok so we have:
$1-p<0$
Now, that will go to infinity. The other case:
$1-p>0$
Will always be zero. So, the function doesn't blow up when p<1. Would that indicate that the answer is $0\leq p<1$? And what about the case where p=1? Thanks again for your patience it is very much appreciated.

16. Aug 10, 2011

### Harrisonized

For p=1, we have:

∫ e-xx-1 dx
= e-xlog(x) + ∫ log(x)e-x dx
= e-xelog(log(x)) + ∫ log(x)e-x dx
= e-x+log(log(x)) + ∫ log(x)e-x dx

I'm pretty sure this diverges, since log(0) isn't defined. If this diverges, then every p for p>1 also diverges.

17. Aug 10, 2011

### jbunniii

Yes, that's right. That shows that you can integrate $1/x^p$ if $0 \leq p < 1$, but it blows up if $p > 1$.

So what can you say about integrating $e^{-x}/x^p$ in these two cases? (Hint: $e^{-x}$ is bounded between 1 and $e^{-1} \approx 0.368$ in the interval [0,1]. It never goes anywhere near 0, nor does it shoot off to infinity.)

Finally, p = 1 is a special case. In this case, $1/x^p = 1/x$. What's the indefinite integral of $1/x$?

18. Aug 10, 2011

### c.dube

So, on $0\leq p<1$, is $\frac{e^{-x}}{x^p}$ integrable and therefore converges?
As for p=1, we have the indefinite integral of lnx. But that's undefined at x=0, so am I missing your hint?

19. Aug 10, 2011

### jbunniii

Correct.

OK, but look at the behavior as x approaches zero. $\ln x$ approaches $-\infty$. So the integral of $1/x$ diverges. What does this imply about the integral of $e^{-x}/x$?

20. Aug 10, 2011

### c.dube

That it diverges, yes? And so the final answer would be $0\leq p<1$?