# Improper Fractions: Integral ln(x)/sqrt(x) convergency question

1. Sep 15, 2009

### NastyAccident

1. The problem statement, all variables and given/known data
$$\int^{1}_{0} \frac{ln(x)}{sqrt(x)}$$

2. Relevant equations
Integration by Parts
Improper Fractions: Type II Integral (discontinuous at a)

3. The attempt at a solution
Now, this might seem odd, but I'm having a devil of a time understanding how this is converging when the limit goes to 0+.

Also, I'm trying to get the hang of LaTeX so, please excuse any improper usage.

$$\int^{1}_{0} \frac{ln(x)}{sqrt(x)}$$
u = ln(x) dv = 1/sqrt(x)
du = 1/x v = 2sqrt(x)

$$2sqrt(x)ln(x)\right|^{1}_{0} - \int^{1}_{0} \frac {(2sqrt(x))}{x}$$
$$2sqrt(x)ln(x)\right|^{1}_{0} - 2*\int^{1}_{0} \frac {1}{sqrt(x)}$$
$$2sqrt(x)ln(x)\right|^{1}_{0} - 4sqrt(x)\right|^{1}_{0}$$

lim $$[2sqrt(x)*ln(x) - 4sqrt(x)]^{1}_{t}$$
t->0+

Now, here is where my question lies with why this is converging....

lim 2sqrt(1)*ln(1) - 4sqrt(1) - (2sqrt(t)*ln(t) - 4sqrt(t))
t->0+

When you plug t in, you receive ln(t) which is -infinity... Thus, the integral would be diverging and not converging.

Does plugging the t into a square root prior to a natural log cause the term to then be zero instead of -infinity and thus converging?

Thanks!

Sincerely,

NastyAccident

2. Sep 15, 2009

### Office_Shredder

Staff Emeritus
When you plug in t, you get sqrt(t) which is 0. So how do you find the limit of something of the form 0*infinity?

3. Sep 15, 2009

### NastyAccident

Limit multiplication property indicates that the product of 0*infinity = 0. Hehe, thanks for the course correction... Sometimes, I just 'get lost' with all the math around me.

4. Sep 15, 2009

### Staff: Mentor

No. Take a look at the "fine print" in the limit multiplication theorem and you'll see that it says that both limits have to exist, which excludes limits that are infinite.

The 0 * infinity form is one of several indeterminate forms. These are called indeterminate because you can't determine beforehand what their values will be. For example, the following limits are all of this form, but have completely different values:
$$\lim_{x \rightarrow \infty} x\frac{1}{x}$$
$$\lim_{x \rightarrow \infty} x^2\frac{1}{x}$$
$$\lim_{x \rightarrow \infty} x\frac{1}{x^2}$$

The values are, respectively, 1, infinity, and zero.

5. Sep 15, 2009

### Gregg

Important to note that:

$$\displaystyle \lim_{x\to 0} x^kln(x) = 0$$

Consider

$$\displaystyle \lim_{x\to \infty} x^ke^{-x}=\frac{x^k}{e^x}$$

For k>0, let n be any integer larger than k.

$$\displaystyle \lim_{x\to \infty}\frac{x^k}{e^x}=\displaystyle \lim_{x\to \infty}\frac{x^k}{1+x+\frac{x^2}{2}+\cdots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\cdots}$$

From this we can show that:

$$\displaystyle \lim_{x\to \infty}\frac{x^k}{e^x}.\frac{x^{-n}}{x^{-n}}=\displaystyle \lim_{x\to \infty}\frac{x^{k-n}}{x^{-n}+x^{1-n}+\frac{x^{2-n}}{2}+\cdots+\frac{1}{n!}+\frac{x}{(n+1)!}+\cdots}$$

From this it's clear that in the limit the numerator tends to 0 since k-n is negative. All terms to the left of 1/n! tend to 0 but all terms to the right of it tend to infinity, hence:

$$\displaystyle\lim_{x\to\infty}x^ke^{-x} = 0$$ (1)

Next, let:

$$x=e^{-\frac{y}{k}}$$

From this it is clear that

$$x\to 0 \iff y\to \infty$$

$$\displaystyle\lim_{x\to 0}x^kln(x) = -\frac{1}{k} \lim_{y\to \infty} ye^{-y}$$

$$\displaystyle\lim_{x\to 0}x^kln(x) = 0$$ From (1)

6. Sep 15, 2009

### NastyAccident

First of all, thank you all very much for responding to this! I truly appreciate it.

Okay, I looked up the indeterminate forms of limits and I noticed the use of the L'Hospital rule so I'm sort of seeing how this is true.

lim sqrt(x)ln(x)
t->0+

f(x) = sqrt(x)
g(x) = ln(x)

I want to use the transformation to infinity/infinity so:

g(x)/1/f(x)

Take the derivative of the numerator and denominator:

g'(x) / 1/f'(x)

(1/x) / (1/-2x^(3/2))

which cleans up to be -2sqrt(x).

If you plug 0+ into -2sqrt(x) you end up with 0.

Just for practice and my understanding plus the fact that I have math sources right here (You folks rock!)...

I'm going to try the transformation to 0/0:

f(x)/1/g(x)

Take the derivative of the numerator and denominator:

[1/(2*sqrt(x))]/(-1/(x*log^2(x))
-1/2 sqrt(x)*log^2(x)

Which still gives me negative infinity (and brings me a step backwards).

So, by going through this process, I'm guessing that you probably want to take whatever term isn't giving you infinity and place it in the denominator with one over it. Then, take the derivatives of both the top and bottom of the function.

I follow the proof up till the point where you set x = e^(-y/k). So, if you wouldn't mind, could you explain that part in a little bit more detail.

Thank you!

Sincerely,

NastyAccident

7. Sep 15, 2009

### Gregg

We have established that

$$\lim_{x\to \infty} x^ke^{-x} = 0$$

Now we could for example set out the following (the choice of variable is arbitrary)

$$u=e^{-\frac{v}{k}}$$

$$\Rightarrow u\to 0 \iff v\to \infty$$

As u gets small (tends to 0) it is implied that v is very large (tends to infinity).

$$\Rightarrow ln(u)=-\frac{v}{k}$$

Now we have this relationship we can show that our limit tends to 0. Notice this equality:

$$u^k = e^{-v} \Rightarrow u^k (ln(u)) = e^{-v}.(-\frac{v}{k})=-\frac{1}{k}(ve^{-v})$$

Notice that the $$ve^{-v}$$ is similar to $$x^ke^{-x}$$ in the case k=1 (and of course the variable is v not x). We know three things at this point.

$$\displaystyle\lim_{v\to \infty}ve^{-v}=0$$

$$v\to \infty \iff u\to 0$$

and,

$$u^k ln(u) =\frac{1}{k}(ve^{-v})$$

From this it can be seen that,

$$-\frac{1}{k}\displaystyle\lim_{v\to \infty}ve^{-v}=\displaystyle\lim_{u\to 0}u^kln(u)=0$$

8. Sep 16, 2009

### Staff: Mentor

That's what I get, also.
I think you have made a mistake here, which is why you're getting a different value for the limit. In this case you are dealing with
$$\frac{x^{1/2}}{\frac{1}{ln x}}$$

You didn't use the chain rule correctly when you differentiated the denominator fraction.