Solving Limits Problem with Beta Function

[MisterMan]

Homework Statement

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

Homework Equations

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

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

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.

 

[jegues]

I tried to continue on with my calculation pretending that I had upper limit 1 and lower limit 0.

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!

 

[MisterMan]

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!

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.

 

[jegues]

I thought b was always supposed to be greater than a

What makes you think that?

 

[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.

 

[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".

 

[Dickfore]

Make the substitution

<br /> x = \frac{1}{t}<br />

 

[MisterMan]

Make the substitution

<br /> x = \frac{1}{t}<br />

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.

 

[MisterMan]

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".

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

 

[jegues]

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

b and a are just numbers.

 

[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?

 

[jegues]

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

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

 

[Dickfore]

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

 

[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} =&gt; 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.

 

[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

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

I got:

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

 

[MisterMan]

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