MHB Evaluating the Improper Integral (II)

[shamieh]

Evaluate the Integral.

$$ \int^1_{-1} \frac{1}{\sqrt{|x|}} \, dx$$

I know that there is a discontinuity at 0

When they change the limits how are they getting $$\int^0_{-1} \frac{1}{\sqrt{-x}} \, dx + \int ^1_0 \frac{1}{\sqrt{x}}
$$
I understand the limit changing part, but I don't understand why one x is -x and the other is positive when the problem clearly states |x|

 

[Evgeny.Makarov]

I don't understand why one x is -x and the other is positive when the problem clearly states |x|

Because $|x|=-x$ for $x<0$.

 

[MarkFL]

I would employ the even symmetry of the integrand and write the integral as:

$$I=2\int_0^1 x^{-\frac{1}{2}}\,dx$$

Now take the appropriate limit. :D

 

[shamieh]

I would employ the even symmetry of the integrand and write the integral as:

$$I=2\int_0^1 x^{-\frac{1}{2}}\,dx$$

Now take the appropriate limit. :D

Wow that's brilliant to notice that. I didn't even think of it being symmetrical. Oh I see so then:

Taking the A.D. I obtain: $$2[2\sqrt{x}] |^1_0 = 4$$

$$\therefore$$ both converge because both are $$\ge 1$$ ?

 

[Evgeny.Makarov]

$$\therefore$$ both converge because both are $$\ge 1$$ ?

What exactly is $$\ge 1$$?

 

[shamieh]

What exactly is $$\ge 1$$?

I'm not sure. How would I write that? Each integrand essentially. (Or at least I think, I'm just guessing).

 

[shamieh]

Like would it be something like $$lim_{x\to\infty} = 4$$

 

[Evgeny.Makarov]

I agree that $\displaystyle \int^1_{-1} \frac{1}{\sqrt{|x|}} \, dx=4$, but I don't understand what you were saying next.

What exactly is $\ge 1$?

I'm not sure.

Well, you wanted to say something. What was it?

How would I write that?

Write what?

Each integrand essentially.

The integrand is $|x|^{-1/2}$, and if you split the integral into two, the integrands are $(-x)^{-1/2}$ and $x^{-1/2}$. You yourself wrote it in the first post.

I suspect you were saying that $\int_0^1x^{-1/2}\,dx$ converges because the exponent $-1/2>-1$, not $\ge1$. Why not say so?

 

[shamieh]

So if

$$lim_{n\to\infty}$$ of some integral if $$n > 1$$ then as $$ lim_{n\to\infty}$$ it diverges?

similarly if

$$lim_{n\to\infty}$$ of some integral if $$n < 1$$ then as $$ lim_{n\to\infty}$$ it converges?

but in this case aren't i saying
$$lim_{n\to 1}$$ ? I know it's an improper integral but I'm only going to some finite number as the upper bound?

 

[MarkFL]

Wow that's brilliant to notice that. I didn't even think of it being symmetrical. Oh I see so then:

Taking the A.D. I obtain: $$2[2\sqrt{x}] |^1_0 = 4$$

$$\therefore$$ both converge because both are $$\ge 1$$ ?

What I wrote was:

$$I=2\lim_{t\to0^{+}}\left(\int_t^1 x^{-\frac{1}{2}}\,dx \right)$$

 

[shamieh]

Oh I see. I was looking at it the wrong way and forgetting to use $$lim_{t\to0}$$ as well in my problem..

So am I correct in saying this?

so $$lim_{t\to 0} \, 4\sqrt{x} |^1_t = [4] - [0]$$ and for t we plugged in $$ 0 $$ which is $$< 1 \therefore$$ converges ?

 

[MarkFL]

I would use a one-sided limit as I showed you in my previous post. It converges because the limit exists and is finite, not because of the value of $t$.

 

[shamieh]

So you're saying it has nothing to do with t if we use a one sided limit because the limit exists and is going to some finite number, being $$\to$$ 1. so as $$ lim_{t\to 0} = 4$$ we know that it converges?

but if we had something approaching infinity or a 2 sided limit(or something not symmetrical) we would consider what the value of t was?