# Beta Function, Limits Problem

1. Jul 25, 2010

### MisterMan

1. The problem statement, all variables and given/known data

$$\int_1^{\infty}\frac{dx}{x^2(x-1)^{1/2}}$$

2. Relevant equations

$$\int_0^1t^{x-1}(1-t)^{y-1}\,dt$$

$$\int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt,$$

3. The attempt at a solution

Hi all, I have another beta function problem. This time I'm unsure how to deal with the limits, as the book states I have to make the substitution : x = 1/y. Putting in the values of x, gives:

$$x = \infty => y = \frac{1}{\infty}$$

$$x = 1 => y = 1$$

I'm unsure how to deal with the one over infinity. It doesn't conform to the beta forms I'm aware of ( see relevant equations ). I tried to continue on with my calculation pretending that I had upper limit 1 and lower limit 0. But I got a negative answer as opposed to the positive one I should get. Any help one this question will be appreciated.

Last edited: Jul 25, 2010
2. Jul 25, 2010

### jegues

Shouldn't your upper bound be 0, and your lower bound be 1? After all,

$$\frac{1}{\infty} = 0$$

If you simply had the bounds mixed up, that will be the culprit of your unwanted negative sign!

3. Jul 25, 2010

### MisterMan

That's what I thought, but can I have an upper bound of 0 and lower bound of 1? I thought b was always supposed to be greater than a:

$$\int_a^bf(x)\,dx$$

Also, I get the negative sign from differentiating x = 1/y, not from my mixing the bounds around.

4. Jul 25, 2010

### jegues

What makes you think that?

5. Jul 25, 2010

### MisterMan

I've never seen an example in which it was less than a.

EDIT: Also when getting bounds from a graph you always choose b as the largest positive x value and a as the smallest x value.

6. Jul 25, 2010

### jegues

I don't think there's any rule to dictate that the upper bound of the integral has to larger than the lower bound.

$$\int_a^bf(x)\,dx = F(b) - F(a)$$

So,

$$\int_b^af(x)\,dx = F(a) - F(b)$$

What's wrong with that? I think of it this way, instead of integrating "left to right" we are now integration "right to left".

7. Jul 25, 2010

### Dickfore

Make the substitution

$$x = \frac{1}{t}$$

8. Jul 25, 2010

### MisterMan

That's already been done. I made the substitution x = 1/y as I stated in my initial post.

What I'm confused about is the limits of the integration and the fact that I'm getting a negative answer.

9. Jul 25, 2010

### MisterMan

I'm not sure that's correct. The upper limit is b, so really that should be the upper limit, right?

10. Jul 25, 2010

### jegues

b and a are just numbers.

11. Jul 25, 2010

### MisterMan

Do you happen to have an example to show me that that has a smaller number on the upper limit than the lower limit?

12. Jul 25, 2010

### jegues

Here's an example to show you that when you the bounds of integration are swapped, the answer is simply the negative of each other.

$$\int_0^{\pi}sin(x)\,dx = 2$$

Swapping the bounds,

$$\int_\pi^0sin(x)\,dx = -2$$

13. Jul 25, 2010

### Dickfore

Show your work. We will tell you where you made a mistake.

14. Jul 25, 2010

### MisterMan

Thanks. I'm not familiar with this sort of work ( swapping the bounds and applying a negative ) I have never done it before, and nothing was stated in the answer section or accompanying question text. So, this is what I get applying your helpful advice :

$$x = \frac{1}{y} => dx = \frac{-1}{y^2}\,dy$$

$$-\int_1^0y^{-\frac{1}{2}}(1-y)^{\frac{1}{2}}\,dy$$

And swapping the limits around gives :

$$\int_0^1y^{-\frac{1}{2}}(1-y)^{\frac{1}{2}}\,dy$$

$$B\left(\frac{1}{2},\frac{3}{2}\right) = \frac{\sqrt{\pi}\frac{1}{2}\sqrt{\pi}}{1}$$

$$B\left(\frac{1}{2},\frac{3}{2}\right) = \frac{1}{2}\pi$$

I really wish the text had stated this "bound switch" was needed to achieve the answer. Thanks again, for your help jegues.

15. Jul 25, 2010

### Dickfore

Although the end result is correct, since the Beta function is symmetric with respect to its arguments, you made a mistake in the second step

I got:

$$-\int_1^0y^{\frac{1}{2}}(1-y)^{-\frac{1}{2}}\,dy$$

16. Jul 25, 2010

### MisterMan

My mistake, I forgot the part I used to get my answer was under 1 as in, a fraction. Thanks.