Convergence of Improper Integrals: Exploring the Power of p

[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.

 

[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 $p \geq 0$?

 

[Harrisonized]

For x ≥ 1:
0 < e^-x < 1/x^p

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

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

0 < e^-x < 1/x^p

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

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

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

 

[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.

 

[Harrisonized]

Never mind.

 

[jbunniii]

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?).

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.

 

[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

 

[jbunniii]

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

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.

 

[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.

 

[Harrisonized]

Have you tried p=0?

 

[jbunniii]

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?

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!

And therefore it diverges for all p, leaving me with an answer p = null set?

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.

 

[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?

 

[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.

 

[jbunniii]

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.

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 $-p + 1 = 0$.

 

[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.

 

[Harrisonized]

For p=1, we have:

∫ e^-xx^-1 dx
= e^-xlog(x) + ∫ log(x)e^-x dx
= e^-xe^log(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.

 

[jbunniii]

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.

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 $1/x$?

 

[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?

 

[jbunniii]

So, on $0\leq p<1$, is $\frac{e^{-x}}{x^p}$ integrable and therefore converges?

Correct.

As for p=1, we have the indefinite integral of lnx. But that's undefined at x=0, so am I missing your hint?

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 $e^{-x}/x$?

 

[c.dube]

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

 

[jbunniii]

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

Yes, that's correct.

 

[c.dube]

Ah sweet! Thanks so much for your help it was much appreciated.

 

[jbunniii]

Ah sweet! Thanks so much for your help it was much appreciated.

You're welcome, good luck.